Scope and lifetime of local variables in C

Question

I would like to understand the difference between the following two C programs.

First program:

void main()
{
    int *a;
    {
        int b = 10;
        a=&b;
    }
    printf("%d\n", *a);
}

Second program:

void main()
{
    int *a;
    a = foo();
    printf("%d\n", *a);
}

int* foo()
{
    int b = 10;
    return &b;
}

In both cases, the address of a local variable (b) is returned to and assigned to a. I know that the memory a is pointing should not be accessed when b goes out of scope. However, when compiling the above two programs, I receive the following warning for the second program only:

warning C4172: returning address of local variable or temporary

Why do I not get a similar warning for the first program?

Answer 1

As you already know that b goes out of scope in each instance, and accessing that memory is illegal, I am only dumping my thoughts on why only one case throws the warning and other doesn't.

In the second case, you're returning the address of a variable stored on Stack memory. Thus, the compiler detects the issue and warns you about it.

The first case, however skips the compiler checking because the compiler sees that a valid initialized address is assigned to a. The compilers depends in many cases on the intellect of the coder.

Similar examples for depicting your first case could be,

char temp[3] ;
strcpy( temp, "abc" ) ;

The compiler sees that the temp have a memory space but it depends on the coder intellect on how many chars, they are going to copy in that memory region.