If the compiler defines the macro constant __STDC_NO_VLA__ to integer constant 1, then VLA objects with automatic storage duration are not supported.
The support for VM types and VLAs with allocated storage durations is mandated.
If expression in an array declarator is omitted, it declares an array of unknown size. Except in function parameter lists (where such arrays are transformed to pointers) and when an initializer is available, such type is an incomplete type (note that VLA of unspecified size, declared with * as the size, is a complete type) (since C99) :
Within a definition, an array of unknown size may appear as the last member (as long as there is at least one other named member), in which case it is a special case known as . See for details: s { int n; double d[]; }; // s.d is a flexible array member struct s *s1 = (sizeof (struct s) + (sizeof (double) * 8)); // as if d was double d[8]
| (since C99) |
If an array type is declared with a , , or (since C99) qualifier (which is possible through the use of ), the array type is not qualified, but its element type is: | (until C23) |
An array type and its element type are always considered to be identically qualified, except that an array type is never considered to be -qualified. | (since C23) |
is not allowed to be applied to an array type, although an array of atomic type is allowed. int A[2]; // _Atomic A a0 = {0}; // Error // _Atomic(A) a1 = {0}; // Error _Atomic int a2[2] = {0}; // OK _Atomic(int) a3[2] = {0}; // OK | (since C11) |
Objects of array type are not modifiable lvalues , and although their address may be taken, they cannot appear on the left hand side of an assignment operator. However, structs with array members are modifiable lvalues and can be assigned:
Any lvalue expression of array type, when used in any context other than
(since C11) |
undergoes an implicit conversion to the pointer to its first element. The result is not an lvalue.
If the array was declared register , the behavior of the program that attempts such conversion is undefined.
When an array type is used in a function parameter list, it is transformed to the corresponding pointer type: int f ( int a [ 2 ] ) and int f ( int * a ) declare the same function. Since the function's actual parameter type is pointer type, a function call with an array argument performs array-to-pointer conversion; the size of the argument array is not available to the called function and must be passed explicitly:
When the element type of an array is another array, it is said that the array is multidimensional:
Note that when array-to-pointer conversion is applied, a multidimensional array is converted to a pointer to its first element, e.g., pointer to the first row:
Multidimensional arrays may be variably modified in every dimension if VLAs are supported(since C11): n = 10; int a[n][2*n]; | (since C99) |
Zero-length array declarations are not allowed, even though some compilers offer them as extensions (typically as a pre-C99 implementation of flexible array members ).
If the size expression of a VLA has side effects, they are guaranteed to be produced except when it is a part of a sizeof expression whose result doesn't depend on it:
for Array declaration |
Hello I am a beginner in C and I was working with structures when i have this problem:
With this one I have this error ( error: assignment to expression with array type ). error: assignment to expression with array type )。--> So I decided to do it using pointers and it actually working this is the code I used:
So the question is what is the difference between both of them since strings are pointers. thank you..
First part, you try to copy two array of character (string is not a pointer, it is array of character that is terminated by null character \0 ). \0 终止的字符数组)。-->
If you want to copy value of an array to another, you can use memcpy , but for string, you can also use strcpy . memcpy ,但对于字符串,你也可以使用 strcpy 。-->
Second part, you make the pointer point to string literal. The pointer points to first character of the string ( r in this case). r )。-->
You can see How to copy a string using a pointer 如何使用指针复制字符串 -->
And Assigning strings to pointer in C Language 字符串分配给 C 语言中的指针 -->
Strings are not pointers, they are sequences of characters terminated with a null byte. These sequences are stored in arrays, which are not pointers either, but objects that live in memory and to which a pointer can point. A pointer is a variable that contains the address of an object in memory.
In the first example, you try and store a string into an array with = . = 将字符串存储到数组中。--> C does not support this type of assignment and complains with an explicit error message. You can copy the string into the array with strcpy , assuming the destination array is large enough to store the bytes of the string, including the null terminator. strcpy 将字符串复制到数组中,假设目标数组足够大以存储字符串的字节,包括 null 终止符。--> "reda" uses 5 bytes, so it is OK to copy it into E[0].nom : "reda" 使用 5 个字节,因此可以将其复制到 E[0].nom :-->
The second example uses a different approach: nom is now defined as a pointer to char . nom 现在定义为指向 char 的指针。--> Defining the array E in main leaves it uninitialized, so you can only read from the pointers after you set their value to point to actual arrays of char somewhere in memory. E[0].nom = "reda"; main 中定义数组 E 使其未初始化,因此您只能在将指针的值设置为指向 memory 中某处的 char 的实际 arrays 后才能读取指针 E[0].nom = "reda"; --> does just that: set the address of the string literal "reda" into the member nom of E[0] . "reda" 的地址设置为 E[0] 的成员 nom 。--> The string literal will be placed by the compiler into an array of its own, including a null terminator, and that must not be changed by the program.
It is a lower-level concept actually.
If we say char nom[20] than its memory location will be decided at compile time and it will use stack memory (see the difference between stack memory and heap memory to gain more grasp on lover level programming). char nom[20] ,那么它的 memory 位置将在编译时确定,它将使用堆栈 memory(请参阅堆栈 memory 和堆 memory 之间的区别,以更好地掌握情人级编程)。--> so we need to use it with indexes.
On the other hand if we create a string using the double quotes method (or the second way in which you used pointers). The double quoted strings are identified as const char* by the compiler and the string is placed at the heap memory rather than the stack memory. moreover their memory location is not decided at compile time. const char* 并且字符串被放置在堆 memory 而不是堆栈 memory。此外,它们的 memory 位置在编译时没有确定。-->
a very efficient way of testing the difference between these two types of strings is that when you use sizeof() with char[20] nom it will return 20 * sizeof(char) which evaluates to 20 without any doubt. sizeof() 与 char[20] nom 一起使用时,它将毫无疑问地返回 20 * sizeof(char) ,其计算结果为 20 。--> it clearly states that whether you use only 5 characters or 2 characters it will always occupy space of 20 charaters (in stack memory)
but in case of const char* if you use sizeof operator on this variable than it will return the size of the pointer variable which depends on the hardware (32bit will have 4bytes pointer) and (64bit will have 8bytes pointer). const char* 的情况下,如果您在此变量上使用 sizeof 运算符,它将返回指针变量的大小,这取决于硬件(32 位将有 4 字节指针)和(64 位将有 8 字节指针)。--> So, the sizeof operator do not shows how many characters are stored in the pointer. it just shows the size of pointer. but the benefit of using const char* is that only take memory that is required if 4 characters are stored than it will consume 5 byte memory (last byte is additionally added in case of strings) and in case of 8 characters it will consume 9 bytes and so on but note that the characters will be stored in heap(memory) const char* 的好处是,如果存储 4 个字符,则只需要 memory,而不是消耗 5 个字节 memory(最后一个字节在字符串的情况下额外添加),如果是 8 个字符,它将消耗 9 个字节等等,但请注意字符将存储在堆(内存)中-->
What you have been doing in the first problem is that you were assigning a string that is only stored on heap to a data types that is only stored on stack. it is not the matter of string. it the matter of memory model
I hope you might have understood. I not i can elaborate more in comments.
an additional information. Stack Memory is faster compared to the Heap Memory.
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First part, you try to copy two array of character (string is not a pointer, it is array of character that is terminated by null character \0). If you want to copy value of an array to another, you can use memcpy, but for string, you can also use strcpy. E[0].nom = "reda"; change to: strcpy(E[0].nom,"reda"); Second part, you make the pointer ...
Then, correcting the data type, considering the char array is used, In the first case, arr = "Hello"; is an assignment, which is not allowed with an array type as LHS of assignment. OTOH, char arr[10] = "Hello"; is an initialization statement, which is perfectly valid statement. edited Oct 28, 2022 at 14:48. knittl.
Now let's create an instance of this struct and try to assign a value to the "string" variable as follows: struct struct_type s1; s1.struct_name = "structname"; // Error: Assignment to expression with array type. As expected, this will result in the same "Error: Assignment to expression with array type" that we encountered in the ...
Learn why you cannot assign a value to an array as a whole in C, and how to fix the error with examples. See how to copy array elements using loops, strcpy, or memcpy ...
Learn what causes and how to fix the error assignment to expression with array type in C and C++. The error occurs when you try to assign a value to an array or use ...
To fix the "assignment to expression with array type" error, you should avoid direct assignment to arrays by using functions like `strcpy` for strings or loops through elements for other types. When returning arrays from functions, return a pointer to the array or use dynamic memory allocation.
You cannot add 1 to an array of 5 ints, nor can you assign an array of 5 ints. A pointer to an array is not a widely-used type in C. To refer to the first element through a pointer to an array, you must use (*ptr)[0] = (*ptr)[0] + 1. This syntax is a bit cumbersome and gives you no real advantages over a using a simple int *ptr = data, which is ...
Learn how to fix the error "Error: Assignment to Expression with Array Type" in C++. Find tips and solutions for assigning values to arrays using loops, memcpy, std ...
c言語の勉強をしています subscripted value is not an array, pointer or vecterというエラーがでたのですが分かる方教えてください C言語関連 学校の宿題で、C言語で 1.以下の プロトタイプ宣言で、 result には a の 階乗 の 計算 結果が返される関数 を作りなさい。
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How to fix the error:assignment to expression with array type#syntax #c #howto #clanguage #error #codeblocks
The example code that I have given you (a summary of yours), throws the following error: error: incompatible types in assignment of 'int*' to 'int [20]'. Effectively, you are trying to assign a pointer to int (points to the first int reserved with malloc) to a pointer that points to an array of 20 positions. And this is your second problem, you ...
El código de ejemplo que te he puesto (resumen del tuyo), arroja el siguiente error: error: incompatible types in assignment of 'int*' to 'int [20]'. Efectivamente, estas intentando asignar un puntero a int (apunta al primer int reservado con malloc) a un puntero que apunta a un array de 20 posiciones.
Assignment to expression with array type error, char array value unable to be set to a variable in structure 1 C Programming: error: assignment to expression with array type
Learn how to declare arrays of constant or variable size, multidimensional arrays, and arrays of pointers in C language. See syntax, examples, and references for each array type.
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prog.c: In function 'main': prog.c:12:15: error: assignment to expression with array type 12 | s1.myString = "Some text"; // Assign a value to the string
Using the unused variable br to store that return value, and the counter variable zbroj1 as an index into the array you can build up rec using the following code: rec[zbroj1]=br; zbroj1++; Strings in C need to have a \0 character at the end to terminate them so you'll also need to have the line to finish it.
This makes sense, since the argument expects an address, and really does not know anything else about the address. 2. Main difference. One can assign to a pointer variable, but not an array variable. That is, given: int b[3]; int *a, *c; a = c; // has meaning and can compile, b = c; // has no meaning, and cannot compile.
1. You cannot assign to a whole array, only to its elements. You therefore cannot dynamically allocate memory for a variable of bona fide array type such as you are using, so it's a good thing that you do not need to do. The size of an array is fully determined by its elements type and length, and the language provides for both appropriate ...
It is a lower-level concept actually. If we say char nom[20] than its memory location will be decided at compile time and it will use stack memory (see the difference between stack memory and heap memory to gain more grasp on lover level programming). so we need to use it with indexes.. such as, nom[0] = 'a'; nom[1] = 'b'; // and so on. On the other hand if we create a string using the double ...