error assignment to expression with array type c

How to solve “Error: Assignment to expression with array type” in C?

In C programming, you might have encountered the “ Error: Assignment to expression with array type “. This error occurs when trying to assign a value to an already initialized  array , which can lead to unexpected behavior and program crashes. In this article, we will explore the root causes of this error and provide examples of how to avoid it in the future. We will also learn the basic concepts involving the array to have a deeper understanding of it in C. Let’s dive in!

What is “Error: Assignment to expression with array type”? What is the cause of it?

An array is a collection of elements of the same data type that are stored in contiguous memory locations. Each element in an array is accessed using an index number. However, when trying to assign a value to an entire array or attempting to assign an array to another array, you may encounter the “Error: Assignment to expression with array type”. This error occurs because arrays in C are not assignable types, meaning you cannot assign a value to an entire array using the assignment operator.

First example: String

Here is an example that may trigger the error:

In this example, we have declared a char array “name” of size 10 and initialized it with the string “John”. Then, we are trying to assign a new string “Mary” to the entire array. However, this is not allowed in C because arrays are not assignable types. As a result, the compiler will throw the “Error: Assignment to expression with array type”.

Initialization

When you declare a char array in C, you can initialize it with a string literal or by specifying each character separately. In our example, we initialized the char array “name” with the string literal “John” as follows:

This creates a char array of size 10 and initializes the first four elements with the characters ‘J’, ‘o’, ‘h’, and ‘n’, followed by a null terminator ‘\0’. It’s important to note that initializing the array in this way does not cause the “Error: Assignment to expression with array type”.

On the other hand, if you declare a char array without initializing it, you will need to assign values to each element of the array separately before you can use it. Failure to do so may lead to undefined behavior. Considering the following code snippet:

We declared a char array “name” of size 10 without initializing it. Then, we attempted to assign a new string “Mary” to the entire array, which will result in the error we are currently facing.

When you declare a char array in C, you need to specify its size. The size determines the maximum number of characters the array can hold. In our example, we declared the char array “name” with a fixed size of 10, which can hold up to 9 characters plus a null terminator ‘\0’.

If you declare a char array without specifying its size, the compiler will automatically determine the size based on the number of characters in the string literal you use to initialize it. For instance:

This code creates a char array “name” with a size of 5, which is the number of characters in the string literal “John” plus a null terminator. It’s important to note that if you assign a string that is longer than the size of the array, you may encounter a buffer overflow.

Second example: Struct

We have known, from the first example, what is the cause of the error with string, after that, we dived into the definition of string, its properties, and the method on how to initialize it properly. Now, we can look at a more complex structure:

This struct “struct_type” contains an integer variable “id” and a character array variable “string” with a fixed size of 10. Now let’s create an instance of this struct and try to assign a value to the “string” variable as follows:

As expected, this will result in the same “Error: Assignment to expression with array type” that we encountered in the previous example. If we compare them together:

• The similarity between the two examples is that both involve assigning a value to an initialized array, which is not allowed in C.
• The main difference between the two examples is the scope and type of the variable being assigned. In the first example, we were dealing with a standalone char array, while in the second example, we are dealing with a char array that is a member of a struct. This means that we need to access the “string” variable through the struct “s1”.

So basically, different context, but the same problem. But before dealing with the big problem, we should learn, for this context, how to initialize a struct first. About methods, we can either declare and initialize a new struct variable in one line or initialize the members of an existing struct variable separately.

Take the example from before, to declare and initialize a new struct variable in one line, use the following syntax:

To initialize the members of an existing struct variable separately, you can do like this:

Both of these methods will initialize the “id” member to 1 and the “struct_name” member to “structname”. The first one is using a brace-enclosed initializer list to provide the initial values of the object, following the law of initialization. The second one is specifically using strcpy() function, which will be mentioned in the next section.

How to solve “Error: Assignment to expression with array type”?

Initialize the array type member during the declaration.

As we saw in the first examples, one way to avoid this error is to initialize the array type member during declaration. For example:

This approach works well if we know the value of the array type member at the time of declaration. This is also the basic method.

Use the strcpy() function

We have seen the use of this in the second example. Another way is to use the strcpy() function to copy a string to the array type member. For example:

Remember to add the string.h library to use the strcpy() function. I recommend going for this approach if we don’t know the value of the array type member at the time of declaration or if we need to copy a string to the member dynamically during runtime. Consider using strncpy() instead if you are not sure whether the destination string is large enough to hold the entire source string plus the null character.

Use pointers

We can also use pointers to avoid this error. Instead of assigning a value to the array type member, we can assign a pointer to the member and use malloc() to dynamically allocate memory for the member. Like the example below:

Before using malloc(), the stdlib.h library needs to be added. This approach is also working well for the struct type. In the next approach, we will talk about an ‘upgrade-version’ of this solution.

Use structures with flexible array members (FAMs)

If we are working with variable-length arrays, we can use structures with FAMs to avoid this error. FAMs allow us to declare an array type member without specifying its size, and then allocate memory for the member dynamically during runtime. For example:

The code is really easy to follow. It is a combination of the struct defined in the second example, and the use of pointers as the third solution. The only thing you need to pay attention to is the size of memory allocation to “s”. Because we didn’t specify the size of the “string” array, so at the allocating memory process, the specific size of the array(10) will need to be added.

This a small insight for anyone interested in this example. As many people know, the “sizeof” operator in C returns the size of the operand in bytes. So when calculating the size of the structure that it points to, we usually use sizeof(*), or in this case: sizeof(*s).

But what happens when the structure contains a flexible array member like in our example? Assume that sizeof(int) is 4 and sizeof(char) is 1, the output will be 4. You might think that sizeof(*s) would give the total size of the structure, including the flexible array member, but not exactly. While sizeof is expected to compute the size of its operand at runtime, it is actually a compile-time operator. So, when the compiler sees sizeof(*s), it only considers the fixed-size members of the structure and ignores the flexible array member. That’s why in our example, sizeof(*s) returns 4 and not 14.

How to avoid “Error: Assignment to expression with array type”?

Summarizing all the things we have discussed before, there are a few things you can do to avoid this error:

• Make sure you understand the basics of arrays, strings, and structures in C.
• Always initialize arrays and structures properly.
• Be careful when assigning values to arrays and structures.
• Consider using pointers instead of arrays in certain cases.

The “ Error: Assignment to expression with array type ” in C occurs when trying to assign a value to an already initialized  array . This error can also occur when attempting to assign a value to an array within a  struct . To avoid this error, we need to initialize arrays with a specific size, use the strcpy() function when copying strings, and properly initialize arrays within structures. Make sure to review the article many times to guarantee that you can understand the underlying concepts. That way you will be able to write more efficient and effective code in C. Have fun coding!

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How To Solve Error “Assignment To Expression With Array Type” In C

“ Error: assignment to expression with array type ” is a common error related to the unsafe assignment to an array in C. The below explanations can help you know more about the cause of error and solutions.

Table of Contents

How does the “error: assignment to expression with array type” happen?

This error happens due to the following reasons: First, the error happens when attempting to assign an array variable. Technically, the expression (called left-hand expression) before the assignment token (‘=’) has the type of array, and C language does not allow any assignment whose left-hand expression is of array type like that. For example:

Or another example:

Output: 

The example above shows that the left hand of the assignment is the arr variable, which is of the type array declared above. However, the left type does not have to be an array variable. Instead, it can be a call expression that returns an array, such as a  string :

In C, the text that is put between two double quotes will be compiled as an array of characters, not a string type, just like other programming languages.

How to solve the error?

Solution 1: using array initializer.

To solve “error: assignment to expression with array type” (which is because of the array assignment in C) you will have to use the following command to create an array:

type arrayname[N] = {arrayelement1, arrayelement2,..arrayelementN};

For example:

Another example:

The first example guides you to create an array with 4 given elements with the value 1,2,3,4 respectively. The second example helps you create an array of char type, each char is a letter which is same as in the string in the double quotes. As we have explained, a string in double quotes is actually of array type. Hence, using it has the same effects as using the brackets.

Solution 2: Using a for loop

Sometimes you cannot just initialize an array because it has been already initialized before, with different elements. In this case, you might want to change the elements value in the array, so you should consider using a for loop to change it. 

Or, if this is an integer array:

When you want to change the array elements, you have to declare a new array representing exactly the result you expect to change. You can create that new array using the array initialiser we guided above. After that, you loop through the elements in that new array to assign them to the elements in the original array you want to change.

We have learned how to deal with the error “ error: assignment to expression with array type ” in C. By avoiding array assignment, as long as finding out the reasons causing this problem in our  tutorial , you can quickly solve it.

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I’m Edward Anderson. My current job is as a programmer. I’m majoring in information technology and 5 years of programming expertise. Python, C, C++, Javascript, Java, HTML, CSS, and R are my strong suits. Let me know if you have any questions about these programming languages.

Name of the university: HCMUT Major : CS Programming Languages : Python, C, C++, Javascript, Java, HTML, CSS, R

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[Fixed] "error: assignment to expression with array type error" when I assign a struct field (C) – C

Quick Fix: In the provided code, the issue lies when assigning a value to a struct field, resulting in an "error: assignment to expression with array type error". To resolve this, use strcpy() to copy the value into the array instead of direct assignment. Additionally, you can initialize the struct using a brace-enclosed initializer list, where each member is assigned a value in order.

The Problem:

While using C structs, assigning values to struct fields directly using the dot operator (.) generates an error, such as assignment to expression with array type error , specifically when trying to assign string literals. This error occurs when attempting to assign a string literal, like "Paolo" , to a character array within a struct field, like s1.name . The compiler doesn’t allow such direct assignment because character arrays are treated as pointers to their first element, making the assignment invalid.

The Solutions:

Solution 1: modifying lvalue and brace-enclosed initializer list.

In C, an array is considered an lvalue, but unlike other data types, arrays are not directly modifiable. Therefore, you cannot directly assign a value to an array element using the assignment operator (=). Instead, you must use a function like strcpy() to copy the value into the array.

In your code, you attempted to directly assign the string "Paolo" to s1.name , which resulted in the compilation error. The correct way to assign a string to s1.name is to use strcpy() , like this:

You can also use a brace-enclosed initializer list to set the values of a struct when it is declared. This is a more concise way to initialize a struct, and it avoids the need to use strcpy() . For example, you could rewrite your main function as follows:

Using a brace-enclosed initializer list is generally the preferred way to initialize a struct, as it is more concise and easier to read. However, there are times when you may need to use strcpy() to assign a value to an array element, such as when you are dynamically allocating memory for the array.

Solution 2: Further Interpretation of Data Types and Assignments in C Structs

C structs are user-defined types that allow you to group related data together. When you declare a struct variable, a block of memory is allocated on the stack, and the compiler knows the size of this memory block based on the type of data you’re storing.

In your specific case, the issue was with this line of code:

You can’t assign a string directly to a char array inside a struct because it tries to assign a pointer to a string to a string. The char array s1.name is a fixed-size array with 30 characters, while " Paolo " is a pointer to a string literal.

To copy the string "Paolo" into the name field of the data struct, you need to use strcpy or memcpy to manually copy the characters into the array. Here’s an example using strcpy :

Alternatively, you can define your struct to have pointers to character arrays, like this:

In this case, you would assign the strings directly to the pointers:

This works because you’re now assigning pointers to strings, which is valid in C.

In summary, when working with structs in C, it’s important to understand the data types of the struct members and how assignments work. Always ensure you’re assigning the correct type of data to the struct members to avoid compilation errors.

Solution 3: Initialize the struct fields using strcpy

In C, strings are arrays of characters, and arrays are not assignable. This means that you can’t assign a string literal directly to a struct field that is an array of characters. Instead, you need to use the strcpy() function to copy the string literal into the struct field.

The corrected code below uses the strcpy() function to initialize the name and surname fields of the s1 struct:

Now, the code will compile and run without errors, and it will print the following output:

Solution 4: Using strcpy() function

The assignment error occurs because you cannot directly assign a string literal to a character array in C. Instead, you need to use the strcpy() function to copy the string literal into the character array. The strcpy() function takes two arguments: the destination character array and the source string literal. It copies the characters from the source string literal into the destination character array, overwriting any existing characters in the destination array.

Here is the modified code using the strcpy() function:

In this code, we first include the string.h header file, which contains the declaration of the strcpy() function. Then, we use the strcpy() function to copy the string literals “Paolo” and “Rossi” into the name and surname fields of the s1 structure, respectively.

With this modification, the code should compile and run without errors.

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C语言中出现[Error] assignment to expression with array type

数组不能直接给数组赋值 指针不能直接给数组赋值

需要使用strcpy或者strncpy函数拷贝

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How to Fix Error: Assignment To Expression With Array Type

This error shows the error as not assignable due to the value associated with the string; or the value is not feasible in C programming language. 

In order to solve this, you have to use a normal stcpy() function to modify the value.

However, there are other  possible ways to fix this issue.

Causes of the error

Here are few reasons for error: assignment to expression with array type: 

#1. Using unassignable array type

You are using an array type that is not assignable, or one that is trying to change the constant value in the program

If you are trying to use an unsignable array in the program, here’s what you will see:

This is because s1.name is an array type that is unassignable. Therefore, the output contains an error.

#2. Error occurs because block of memory is below 60 chars

The error also appears if the block of memory is below 60 chars. The data should always  be a block of memory that fits 60 chars.

In the example above, “data s1” is responsible for allocating the memory on the stack. Assignments then copy the number, and fails because the s1.name is the start of  a struct 64 bytes long which can be detected by a compiler, also Ogechukwu is 6 bytes long char due to the trailing \0 in the C strings. So , assigning a pointer to a string into a string is impossible, and this expression error occurs.

Example: 

The output will be thus :

#3. Not filling structure with pointers

If a programmer does not define a structure which points towards char arrays of any length, an expression error occurs, however  If the programmer uses a  defined structure field, they also have to set pointers in it too. Example of when a program does not have a defined struct which points towards a char.

#4. There is a syntax error

If a Programmers uses a function in an incorrect order, or syntax.also if the programer typed in the wrong expression within the syntax.

example of a wrong syntax, where using “-” is invalid:

#5. You directly assigned a value to a string

The C programming language does not allow  users to assign the value to a string directly. So you cant directly assign value to a string, ifthey do there will receive an expression error. 

Example of such:

#6. Changing the constant value

As a programmer, if you are trying to change a constant value, you will encounter the expression error. an initialized string , such as “char sample[19]; is a constant pointer. Which  means the user can change the value of the thing that the pointer is pointing to, but cannot change the value of the pointer itself. When the user tries to change the pointer, they are actually trying to change the string’s address, causing the error.

The output gives the programmer two independent addresses. The first address is the address of s1.name while the second address is that of Ogechukwu. So a programmer cannot change s1.name address to Ogechukwu address because it is a constant.

How to fix “error: assignment to expression with array type” error?

The following will help you fix this error:

#1. Use an assignable array type

Programmers are advised to use the array type that is assignable by the arrays because an unassignable array will not be recognized by the program causing the error.  To avoid this error, make sure the correct array is being used.

Check out this example:

#2. Use a block of memory that fits 60 chars

Use the struct function to define a struct pointing to char arrays of any lenght.  This means the user does not have to worry about the length of the block of memory. Even if the length cis  below 60 chars, the program will still run smoothly.

But if you are not using the struct function, take care of the char length and that the memory block should fit 60 chars and not below to avoid this error.  

#3. Fill the structure with pointers

If the structure is not filled with pointers, you will experience this error. So ensure you fill the struct with pointers that point to char arrays.

Keep in mind that the ints and pointers can have different sizes. Example 

This example, the struct has been filled with pointers.

#4. Use the correct syntax

Double check and ensure you are using the right syntax before coding.if you are confused about the right syntax, then consult Google to know the right one.

Example 

#5. Use Strcpy() function to modify

In order to modify the pointer or overall program, it is advisable to use  the strcpy() function. This function will modify the value as well because directly assigning the value to a string is impossible in a C programming language. usin g this function  helps the user  get rid of assignable errors. Example 

Strcpy(s1.name, “New_Name”);

#6. Copy an array using Memcpy 

In order to eliminate this error,a programmer can use the memcpy function to copy an array into a string and prevent  errors.

#7. Copy into the array using  Strcpy() 

Also you can copy into the array using strcpy() because of the modifiable lvalue. A modifiable lvalue is an lvalue that  has no array type. So programmers have to  use strcpy() to copy data into an array. 

Follow these guide properly and get rid of “error: assignment to expression with array type”.

I am an experienced tech and innovation writer. It’s been 3 years since I started writing at Terecle , covering mostly Consumer electronics and Productivity. In my spare time, I enjoy reading and learning the latest happenings around the tech ecosystem.

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Error: Assignment to Expression With Array Type: A Comprehensive Guide

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To solve this, a normal strcpy() function is used for the modification of the value. In this guide, we’ll review some possible ways to get around this issue by using relevant examples.

JUMP TO TOPIC

– Using Un-assignable Array Type

– the block of memory is below 60 chars, – not filling the struct with pointers, – syntax error, – directly assigned a value to a string, – trying to change the constant, – use assignable array type, – use strcpy() to copy into the array, – block of memory that fits 60 chars, – fill the structure with pointers, – use correct syntax, – use strcpy() function for modification, – use memcpy to copy an array, why is the “error: assignment to expression with array type” error happening.

The Program error: assignment to expression with array type error usually happens when you use an array type that is not assignable or is trying to change the constant value in the program. Moreover, you will encounter the same error if the block of memory is below 60 chars.

However, these reasons are not the only ones contributing to the cause of the error. There are a few other reasons behind the occurrence of this error such as

• Not filling the struct with pointers.
• Syntax error
• Directly assigned a value to a string.
• Trying to change the constant.

It is a common array-type error the programmers make while coding. They end up accidentally using an unassignable array type that cannot be used by the program and get an expression error.

The array function has unassignable array types, and the programmers forget them and later use that array type; thus, the error occurs. Given below is a program used as an example of an un-assignable array type:

The issue is in the program line:

This is because s1.name is an array type which is not assignable. Thus, the output contains an error.

The error also occurs if the block of memory is below 60 chars. It is a defined fact that the data should be a block of memory that fits 60 chars.

Taking the example used above, “data s1;” allocates the memory on the stack . Assignments just copy numbers or sometimes pointers, and it fails, such as (s1.name = “Arora” ;).

This failure in the program happened because the s1.name is the start of a struct 64 bytes long can be detected by the compiler, and “Arora” is a 6 bytes long char[]. It is six bytes long due to the trailing \0 in the C strings. Therefore, assigning a pointer to a string into a string is impossible, and an expression error occurs.

Here’s an example for better understanding:

The output of this program will not be according to the desired output. It will be something like this:

Here’s an example where the programmer hasn’t defined a struct which points towards a char array.

Programmers must be careful with the functions they use and if they are using them in the correct order or syntax. This error can also occur if the user has typed in the wrong expression within the syntax.

If that’s the case, then an expression error will occur. Here is an example of the wrong syntax, where using “-” is invalid:

The C programming language does not allow its users to assign the value to a string directly. Thus, directly assigning value to a string is not possible, and if the programmer has done that, they will receive an expression error again.

Here’s an example that shows a directly assigned value to a string:

Another reason why the programmer gets the error is that they are trying to change a constant value. When a string is initialized, such as “char sample[19];” , the sample itself is a constant pointer. This means the user can change the value of the thing that the pointer is pointing to, but cannot change the value of the pointer itself.

When “char sample” is initialized, it is like the sample in which the pointer is initialized aschar* const, and when the user tries to change the pointer , they are actually trying to change the string’s address.

Moreover, when the user compiles and runs the program, the assigned string fills part of the memory independent of s1.name.

To understand better, try running the following code :

The output will give the programmer two independent addresses. The first address is the address of s1.name and the second is the address of Arora . Therefore, the programmer cannot change s1.name address to Arora address because it is a constant.

How To Fix “Error: Assignment to Expression With Array Type” Error?

However, multiple other solutions to this error are discussed in detail in this article. They are given below:

Always use the array type that is assignable by the arrays. By using an unassignable array will not be recognized by the program, and an error will occur. To avoid this, make sure the correct array is being used.

To elaborate further, the assignment operator shall have a modifiable lvalue as its left operand. Check out the following example.

Regarding the modifiable lvalue, a modifiable lvalue is an lvalue that does not have any array type. Therefore, it is said to use strcpy() to copy data into an array. With that said, (data s1 = {“Arora”, “Rayn”, 20};) worked fine because it is not a direct assignment involving assignment operator.

An error occurs when the programmer uses a brace-enclosed initializer list to provide the initial values of the object. Thus, by the law of initialization, each brace-enclosed initializer list should have an associated current object.

If no designations are present, then subobjects of the current object are initialized in order according to the type of the current object. In the above examples, the order is: array elements in increasing subscript order, the structure members in declaration order, and the first named member of a union.

However, if you is not using the struct function , it is important to take care of the char length and that the memory block should fit 60 chars and not below. If it doesn’t fit, the error will occur.

As mentioned above, the error will occur if the struct is not filled with a pointer . Therefore, make sure to fill the struct with pointers that point to char arrays.

Do keep in mind that the ints and pointers can have different sizes . Given below is a program for better understanding.

In this example, we have assigned/filled the struct with the pointers.

Make sure to use the correct syntax before coding. In case the programmer gets confused about the syntax, do a quick google search on the right syntax and go back to coding error-free.

Given below are the two examples of correct syntax while coding:

In order to make modifications in the pointer or overall program, make sure to use the strcpy() function. This function is used for modifying the value as well.

It is important to use the strcpy() function to modify value because directly assigning the value to a string is impossible in a C programming language . By using that function, the user can get rid of assignable errors. Given below is an example of how to modify value using strcpy().

Another method a programmer can use is by using the memcpy function to copy an array into a string. This will save them from errors.

Below are two examples of using the memcpy function to copy an array.

The programmer can use the memcpy() function present in the header file. Here’s how:

After reading this guide, the reader will have a clear and in-depth understanding of the “error: assignment to expression with array type” error message. This article covers almost all the reasons and solutions for this error. Here are the key points to take notes from:

• Use the correct syntax format and write down the basic, most common syntax used for the program the programmer will use.
• Carefully check the struct() function. The pointers should be filled and pointed to the char array.
• For modifying, changing, and copying an array using either memcpy or strcpy() function.

By following this guide properly , you will be able to master the art of fixing this error.

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