C 已初始化和未初始化的可变大小数组

Question

I have the next two code examples:

const char *val = strchr(ch, ' ');
const int diff = (int)(val - ch);

char arr[diff];

and

const char *val = strchr(ch, ' ');
const int diff = (int)(val - ch);

char arr[diff] = {0};

The second one generates the error like

error: variable-sized object may not be initialized

It is correct error and I understand why it happens.

I wonder why the first code snippet doesn't generate the error?

Update: Also regarding sizeof(arr) at first snippet gives the size of array, but I thought that sizeof is a compile time operator (?)

Answer 1

In the second case you are trying to initialize a variable length array (because the size of the array is not specified with an integer constant expression; the presence of the qualifier const in the declaration of the variable diff does not make it an integer constant expression in the array declaration)

char arr[diff] = {0};

that is not allowed for variable length arrays.

From the C Standard (6.7.9 Initialization)

3 The type of the entity to be initialized shall be an array of unknown size or a complete object type that is not a variable length array type

You could set all elements of the array to zero the following way

#include <string.h>

//...

char arr[diff];
memset( arr, 0, diff );

As for the operator sizeof then for variable length arrays it is calculated at run-time.

From the C Standard (6.5.3.4 The sizeof and alignof operators)

2 The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant

Answer 2

This definition:

char arr[diff];

Creates a variable length array . Such an array has its size determined at runtime. Because of this, it may not be initialized.