Validity of forcing at least one iteration of for loop with goto statement

Question

Disclaimer: I know it is obscure, and I would not program like that. I am aware of the preferred do-while statement, rather than that, the question is more about validity of a specific language construct.

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Is goto always supposed to omit conditional expression of the for loop? From what I observed it skips both the first (i.e. initializing) and second expressions. Will this always happen this way or is this behavior purely compiler dependent?

#include <stdio.h>

int main(void)
{
    int m = 5;

    goto BODY;
    for (m = 0; m < 5; m++)
        BODY: puts("Message"); // prints "Message" once

    printf("m = %d\n", m); // prints m = 6

    return 0;
}

Answer 1

Yes, you're jumping over both m = 0 and m < 5, and this is as it should be.

for (A; B; C)
    D;

is equivalent to

{
    A;
  loop:
    if (B)
    {
        D;
        C;
        goto loop;
    }
}

There is no way to transfer control to a point between A and B.

The semantics of your loop are exactly like this "pure goto" version:

int m = 5;
goto BODY;
m = 0;
loop:
if (m < 5) 
{
  BODY: puts("Message"); // prints "Message" once
   m++;
   goto loop;
}
printf("m = %d\n", m); // prints m = 6

Answer 2

Does goto is always supposed to omit conditional expression of the for loop?

Yes. A goto jump is always an unconditional jump and the execution will carry from the label from there.

From C11 draft, § 6.8.6.1 The goto statement:

A goto statement causes an unconditional jump to the statement prefixed by the named label in the enclosing function.

With the only exception being:

The identifier in a goto statement shall name a label located somewhere in the enclosing function. A goto statement shall not jump from outside the scope of an identifier having a variably modified type to inside the scope of that identifier.