memory allocation for string and array of char

Question

I can't understand how memory is allocated in the following code:

#include<stdio.h>
#include<string.h>

int main()
{
    char a[]={"text"};
    char b[]={'t','e','x','t'};
    printf(":%s: sizeof(a)=%d, strlen(a)=%d\n",a, sizeof(a), strlen(a));
    printf(":%s: sizeof(b)=%d, strlen(b)=%d\n",b, sizeof(b), strlen(b));
    return 0;
}

The output is

:text: sizeof(a)=5, strlen(a)=4
:texttext: sizeof(b)=4, strlen(b)=8

By looking into memory addresses and the output code it seems that variable b is placed before variable a, and that's why strlen(b), by looking for \0, returns 8. Why does this happen? I expected variable a to be declared first.

Answer 1

The language makes no guarantees about what is placed where. So, your experiment make very little sense. It might work, it might not. The behavior is undefined. Your b is not a string and it is UB to use strlen with something that is not a string.

From the purely practical point of view though, local variables are usually allocated on the stack, and the stack on may moderns platforms (like x86) grows backwards, i.e. from higher addresses to lower addresses. So, if you are using one of these platforms, it is possible that your compiler decided to allocate variables in the order of their declaration (a first and b second), but because stack grows backwards b ended up at lower addresses in the memory than a. I.e. b ended up before a in memory.

One can note though that a typical implementation does not normally allocate stack space for local variables one-by-one. Instead, the entire block of memory for all local variables (stack frame) is allocated at once, meaning that the logic I described above does not necessarily apply. Yet, it is still possible that the compiler still follows the "reverse" approach to local variable layout anyway, i.e. variables declared earlier are placed later in the local memory frame, "as if" they were allocated one-by-one in the order of their declaration.