内存分配和char数组

Question

I still do not quite understand, what exactly will happen in the situation:

int i = 0;
for(i; i <100; i ++)
{
    char some_array[24];
     //...
    strcpy(some_array,"abcdefg");

}

Will the some_array act as:

some_array = malloc(24);

At the beginning of the cycle and free(some_array) at the end of the cycle?

Or those variables are gonna be allocated in the stack, and after the function ends destroyed?

Answer 1

some_array is local to the block , so it's created at the beginning of each iteration of the loop, and destroyed again at the end of each iteration of the loop.

In the case of a simple array, "create" and "destroy" don't mean much. If (in C++) you replace it with (for example) an object that prints something out when it's created and destroyed, you'll see those side effects happen though.

Answer 2

There are two ways of storing character strings. 1. Creating some space using malloc and then storing the char string In this case memory is allocated from heap and will not be freed until you free it explicitly. Here memory is not deallocated even after the scope 2. Creating an array and storing in it Here memory is allocated from stack and is freed implicitly after the scope

Answer 3

There is no implicit malloc() or free calls introduced by the example shown.

You can demonstrate that by adding a "free(some_array)" statement within the loop body. The result will be a compilation error.

The array is - as far as the program is concerned - created at the start of the block and destroyed at the end. Which means the C programmer must assume it is created and destroyed for every iteration of the loop.

It is up to the compiler as to whether the array is created on the stack - or how it optimises the repeated creation and destruction of the array. In practice, it often will be created on the stack.