Do I have the right answer to my question in C?

Wrong reasoning. i++ increases i by one, but yields the old value. Each of the three loops will be executed exactly ten times (unless there is something in the loop body).

But consider the value of i within the body of the loop: What would be the output of

Code:

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

?

 

C is the correct answer, but not for the reason you stated. The short of it is, A and B increment 'i' after printing it, C increments it before printing.

Set up a small test program, compile it and test it so you can see what happens. That is the best way to learn.

Code:

#include <stdio.h>

int main () {

    int i;
    for (i =0; i < 10; i++) {
        printf("%d ",i);
    }
    printf("\n");
    for (i =0; i < 10; ++i) {
        printf("%d ",i);
    }
    printf("\n");
    for (i =0; i++ < 10; ) {
        printf("%d ",i);
    }
    printf("\n");
    return 0;
}

There are 100 different ways of doing different types of counting in C (and related languages). Obviously answer C in this case is not very clear, so you would want to pick a method that is easy to read and understand while accomplishing what you want.

 

Write this out on paper:

a) for (i =0; i < 10; i++)
First step: i is assigned the value 0
Second step: i is compared to 10, i is 0, pass.
Third step: Execute loop body, i is 0
Fourth step: i++, i is now 1
Fifth step: i is compared to 10, i is 1, pass
Sixth step: execute loop body, i is 1
Seventh step: i++, i is set to 2

b) for (i =0; i < 10; ++i)
First step: i is assigned the value 0
Second step: i is compared to 10, i is 0, pass.
Third step: Execute loop body, i is 0
Fourth step: ++i, i is now 1
Fifth step: i is compared to 10, i is 1, pass
Sixth step: execute loop body, i is 1
Seventh step: ++i, i is set to 2

c) for (i =0; i++ < 10; )
First step: i is assigned the value 0
Second step: i++ is compared to 10, i++ evaluates to 0, so 0 < 10 , pass. i set to 1
Third step: Execute loop body, i is 1
Fourth step: empty statement, i is now 1
Fifth step: i++ is compared to 10, i++ evaluates to 1, so 1 < 10 , pass. i set to 2


You can tell by the time you execute the loop body for the first time that C is different, as i is 1 for the first iteration, not 0 like the other two. I stopped, but it should take a few minutes for you to go through all 100 or so steps. At the end you can see what i's value is, and what the value of i was while executing the loop body each time.

-Lee

 

As has been mentioned in another thread, for is just convenient syntax for a while loop. Like this:

Code:

for ( [I][COLOR="Green"]start[/COLOR][/I] ; [I][COLOR="DarkOrange"]test[/COLOR][/I] ; [I][COLOR="Purple"]cycle[/COLOR][/I] ) { [I][COLOR="blue"]loop body[/COLOR][/I]; )

// expands to:

[I][COLOR="green"]start[/COLOR][/I]; // initialize the variable
while ( [I][COLOR="DarkOrange"]test[/COLOR][/I] ) {
    [I][COLOR="Blue"]loop body[/COLOR][/I]; // all the statements in the loop
    [I][COLOR="purple"]cycle[/COLOR][/I]; // increment the variable, or whatever
}

This probably does not help you, but then again, it might.