Why the Semicolon is not emptying the loop body?

Question

The semicolon at the end of the for loop is suppose to empty the body and create a null loop. But why this is printing 6?

void main()
{
    int i;
    for(i=1;i<=5;i++);
    {
        printf("%d\n",i);
    }
}

It makes null loop(do nothing), but it does increment `i` to 6, and so after loop ends next print statement is executed — Priyesh Kumar, May 29 '17 at 18:27
You probably wanted to ask why there was only one number printed instead of 5, or you really don't get loops. — zubergu, May 29 '17 at 18:28
It would be more accurate to say that the empty statement terminated by the semicolon *is* the loop body. This is not some kind of special case or magic significance of the semicolon. It simply creates a `for` loop whose *body* does nothing, but that works in all respects like any other `for` loop. — John Bollinger, May 29 '17 at 18:29
@zubergu no i know loops,i am asking why it is printing 6 now i got it. — Maths Maniac, May 29 '17 at 18:35
@MathsManiac what else do you expect to be printed? You have a loop that does nothing except incrementing `i`. After the loop you print `i`. What do you expect to happen **after** the loop? — Gerhardh, May 30 '17 at 07:41

score 11 · Answer 1 · answered May 29 '17 at 18:26

The loop body is empty, otherwise it would print 1, 2, 3, 4, 5. But the loop head runs nethertheless and in each iteration it increases i. When it reaches 6 which is not <=5 the loop ends. Printing i after the loop prints i as 6. Incrementing i is a side effect of the loop.

score 1 · Answer 2 · answered May 29 '17 at 18:26

1

It is.at the end of the loop i will be 6 and the printf does this.

answered May 29 '17 at 18:26

Ed Heal

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score 1 · Answer 3 · answered May 29 '17 at 18:27

1

The for loop for(i=1;i<=5;i++); will run exactly 5 times, incrementing i from 1 to 6 (even though the for loop body is a no-op). Thus, in here:

{
    printf("%d\n",i);
}

the program will print the current value of i, that is 6.

answered May 29 '17 at 18:27

syntagma

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pmg · Answer 4 · 2017-05-29T18:58:01.803

1

Try this for fun

#include <stdio.h>

int main(void)
{
    int i;
    for (i = 1; i <= 5; i++) /* void */;

    /* floating block one */
    {
        int i = 42; /* new i, hides old i */
        printf("%d\n",i);
    }

    /* floating block two */
    {
        printf("%d\n",i);
    }
}

edited May 29 '17 at 18:58

answered May 29 '17 at 18:27

pmg

106,608
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Explain me this @pmg – Maths Maniac May 29 '17 at 18:43
@Magnus see this void main() { int i; for(i=1;i<=5;i++); { int i = 42; printf("%d\n",i); } printf("%d\n",i); } – Maths Maniac May 29 '17 at 18:49
It should print 42 42,but printing 42 6 – Maths Maniac May 29 '17 at 18:50
1

The second definition of `i` (the `42`), inside the "floating" block, creates a different variable efectively hiding the first one (there are two `i`s, but you can only access the innermost). When that block finishes, the other "floating" block starts with the previous `i` in scope. – pmg May 29 '17 at 18:52
How can it get back the previous valur that is "2"?It is supposed to be replaced with 42. – Maths Maniac May 29 '17 at 18:55
how 42 is being replaced by 6. @pmg – Maths Maniac May 29 '17 at 19:07
The `6` is not replaced by anything. It "lives" at the same time as the `42`, both have the same name but different scopes (and lifetimes). The `i` with value 42 exists and is available only inside its floating block. The `i` with value 6 exists inside all of the `main()` function, but is hidden inside the floating block of the other `i`. – pmg May 29 '17 at 19:13

pastaleg · Answer 5 · 2017-05-29T18:38:26.247

1

Because you declare the int outside the null loop, the value is saved outside the increment loop.

Why the Semicolon is not emptying the loop body?

6 Answers6