How does this program get 5 finally?

Question

Can anyone explain this program? How does it print '5' at the end? Thanks in advance.

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int a = 1;
    while (a++ <= 1)
    while (a++ <= 2);
    printf("%d",a);  //5
    return 0;
}

You just have to play "c-interpreter" for this program to see, what it does, step by step. Remind the semantics of the ++-postfix operator (first evaluate a, then increment). Alternatively, step through it with a debugger line by line and have the variable "a" watched — Ctx, Nov 25 '19 at 09:47
@StoryTeller Looks like the indention is missing, not the semicolon — Ctx, Nov 25 '19 at 09:48
Time to fire up your debugger and single step while watching the variable. — Lundin, Nov 25 '19 at 09:49
@Ctx - I guess we're both assuming different things about the OPs code. Goes towards this question needing some clarifications. — StoryTeller - Unslander Monica, Nov 25 '19 at 09:50
And yes this is why we never write semicolon at the same line as a loop condition. Either write the semicolon on a line of its own or use empty `{}`. Otherwise there's no telling if the semicolon is accidental or intentional. — Lundin, Nov 25 '19 at 09:51
@Ctx - Exactly. That's what the OP expects. Which to me means they did not notice the missing semi-colon. — StoryTeller - Unslander Monica, Nov 25 '19 at 09:52
@StoryTeller You interpret too much into this, he just wants to know, how this program produces the 5. There is no hint in any way, that he expected another result, so why assume it? — Ctx, Nov 25 '19 at 09:53
@Ctx - I guess I'm gonna have to lead by example here. Cheers. — StoryTeller - Unslander Monica, Nov 25 '19 at 09:58
@StoryTeller It always positively affects credibility if one sticks to his own rules. However, I do not have anything against a discussion with good arguments. What else are you here for? — Ctx, Nov 25 '19 at 09:59

Rishikesh Raje · Answer 1 · 2019-11-25T09:57:07.753

Note the absence of the ; after the first while statement. This means that there are nested while loops.

while (a++ <= 1)
    while (a++ <= 2);

Let us check the lines one by one.

a = 1; // initialize 
while (a++ <= 1)  // condition is TRUE, after this statement a === 2
   while (a++ <= 2); // condition is TRUE, after this a == 3
   while (a++ <= 2); // condition is FALSE, 
                     // but `a++` is still evaluated. So after this statement a == 4. 
                     // Inner loop exits
while (a++ <= 1)  // condition is FALSE, 
                  // `a++` is evaluated and after this a==5
                  // Outer Loop exits
printf("%d",a);   // print value of a i.e. print 5.

score 1 · Answer 2 · answered Nov 25 '19 at 10:10

when it is in the first while the value that is checked is 1 then it gets increased by 1 and goes into the next while in there it's 2 so 2<=2 is true and it gets increased by 1 to 3 while checking 3<=2 it gets increased by 1 anyway cause a++ does not care if the condition is true so now we have a=4 it jumps back to the first while for checking as you know 4<=1 is sill false but it gets increased by one anyway and 5 comes out.

score 0 · Answer 3 · answered Nov 25 '19 at 10:00

a = 1;                      First iteration         Second iteration
while (a++ <= 1) {          (1 <= 1) True           (4 <= 1) Fail                
                            (a = a + 1) == 2        (a = a + 1) == 5    
    while (a++ <= 2) { };   (a == 2) <= 2 True        
                            (a == 3) first time
                            (a == 4) When while evalutes to fail
}

score 0 · Answer 4 · answered Nov 25 '19 at 10:02

If you for reasons unknown can't use a debugger and single step while watching the variable, then you could add some printf debugging code along the way. And fix the indention while you are at it:

#include <stdio.h>

int main()
{
    int a = 1;
    while (printf("Outer while %d\n", a), a++ <= 1)
      while (printf("Inner while %d\n", a), a++ <= 2)
        ;
    printf("Result %d\n", a);
    return 0;
}

Output:

Outer while 1
Inner while 2
Inner while 3
Outer while 4
Result 5

This prints the value each time it is checked, before it is increased. Basically both loops have to check the value a first time when it is in range, then a second time when it is out of range.

Please note that mixing ++ with other operators in the same expression is bad practice. The above is some fishy, artificial code, just used to illustrate the execution path.

(And no, there is no UB here.) – Lundin Nov 25 '19 at 10:04 — Lundin, Nov 25 '19 at 10:04

How does this program get 5 finally?

4 Answers4