stack variable disappears at the last action of a function?

Question

The advantage of tail recursion is that we call the function (recursion) again in the last action of the code, so the stack variables doesn't need to be saved.

So here do this will act the same? and I can't be sure that the string will stay In the stack and not will be cleaned?

void foo(char* c){
    c[0]='b';     
    printf("%s",c);

}
void foo2(){
    char str[]="aaa";
    foo(str);
}
int main(){
    foo2();
}

or I can be sure that the string in foo2 will stay until the function ends (gets to the end scope).

I am asking if str that is local variable of function foo2 will still be available, or if this local variable (str) will be cleaned because no more actions to do in this function foo2 after calling foo(str).

I want to understand if could be that in foo I will try to use a string that is not available anymore.

Output:

baa

By just this example I can see that str didn't disappear until got to the end of foo2 scope, but I want to know if this will be like this every time, and if yes how this makes sense with the idea of tail recursion that also calls a function in the last action.

As long as functions foo and foo2 are blocking, there is nothing wrong about this. — hennep
– hennep, Commented Jul 18, 2023 at 8:41
There is no recursion whatsoever in your code. Also tell us if the output you get is different from the output you expect. — Jabberwocky
– Jabberwocky, Commented Jul 18, 2023 at 8:47
It would be different when foo was in a different thread, for instance a USB send function, that returns before the data was sent. What did you try to achieve when you experienced problems with it? — hennep
– hennep, Commented Jul 18, 2023 at 8:48
The local c variable infoo2 will obviously be modified by calling foo(c). I'm not really getting what you're asking here. — Jabberwocky
– Jabberwocky, Commented Jul 18, 2023 at 8:50
All you can be sure about is the output. But you can't tell how the compiler will make a binary that generates the output. I checked gcc with -O3 and it didn't even call foo2. It just called printf directly from main. With -O2 main called foo2 but foo2 never called foo. Just printf. With -O1 and -O0 both foo2 and foo was called. — 4386427
– 4386427, Commented Jul 18, 2023 at 9:23

0___________ · Accepted Answer · 2023-07-18 09:32:31Z

1

or I can be sure that the string in foo2 will stay until the function ends (get to the end scope).

Yes as you use it before the function foo2 returns. After the call to foo foo2 will wait until foo returns.

Your example is not a recursion of any kind. Recursion is when the function calls itself. Exmaple of tail recursion:

void foo(char *x, size_t n)
{
    if(x[n])
    {
        x[n] = 'b';
        printf("%s\n", x);
        foo(x, n + 1);
    }
}

int main(void)
{
    char s[4] = "aaa";
    foo(s, 0);
}

https://godbolt.org/z/66PTrrY83

Tail recursion if the function is calling itself as the last operation in the function. This case makes compilers work much easier as they can quite easily get rid of the recursion (which is expensive operation). In the case above when you enable optimizations the compiler will remove the recursion completely:

foo:
        cmp     BYTE PTR [rdi+rsi], 0
        je      .L6
        push    rbp
        mov     rbp, rdi
        push    rbx
        mov     rbx, rsi
        sub     rsp, 8
.L3:
        mov     BYTE PTR [rbp+0+rbx], 98
        mov     rdi, rbp
        add     rbx, 1
        call    puts
        cmp     BYTE PTR [rbp+0+rbx], 0
        jne     .L3
        add     rsp, 8
        pop     rbx
        pop     rbp
        ret
.L6:
        ret
main:
        sub     rsp, 24
        xor     esi, esi
        lea     rdi, [rsp+12]
        mov     DWORD PTR [rsp+12], 6381921
        call    foo
        xor     eax, eax
        add     rsp, 24
        ret

edited Jul 18, 2023 at 9:32

answered Jul 18, 2023 at 9:02

0___________

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1 Comment

4386427 Over a year ago

Quote from your comment: "There is no recursion in your code. Question makes no sense" hmmm... nice to answer a question that makes no sense :-/

Daniel Zaken · Accepted Answer · 2023-07-20 17:04:32Z

0

My problem was that I didn't understand enough about tail call optimizations, the compiler optimizes tail recursion and this doesn't have anything about the logic that the local variable is available until leaving the scope. ty for your help.

answered Jul 20, 2023 at 17:04

Daniel Zaken

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