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For loop ends one step later than expected
#include <cs50.h>
#include <stdio.h>
int main (void)
{
bool parm = false;
int i = 0;
for (i = 0; i < 7 && !parm; i++)
{
printf("loop: %d\n", i);
if (i == 4)
{
parm = true;
}
}
printf("end: %d\n", i);
}
@BaSmi
Owner

The last time line 11 prints to the screen i = 4. But When line 18 prints i = 5. I would have expected 4 again. If I break after line 14, I end with i = 4. Can somebody help me understand why this loop behaves the way it does?

@IncipientLT

It is because it finished that cycle of the for loop after i was already 4, thus it iterated i. It then found that the "!parm" condition was no longer met, then broke out of the loop.

If you wanted to break out of the loop as soon as i == 4, you should use a break statement, "break;" to exit out of the for loop. Place "break;" on the next line after "parm = true;" to see the results.

@BaSmi
Owner

Thanks! I wrongly assumed i++ would only happen after evaluating the condition.

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