Using Increment and Decrement Operators in Initialisation, Condition and Updation in For Loop-CodePudding

Today I had a test on C at my university and one of the questions asked to give the output for this block of code:

int a=9;

    for(a--;a--;a--)

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

I thought this would create an infinite loop but on checking it gave

as output. Why did it not create an infinite loop? What's going on in this program?

CodePudding user response：

If you separate the post-decrement into separate statements this is the equivalent code:

#include <stdio.h>
    
int main(void) {
    int a = 9;
    a--;
    while(a) {
        a--;
        printf("%d\n", a);
        a--;
    }
    return;
}

Program output:

This won't stop when a is even, unless the condition is changed to

while(a > 0)

CodePudding user response：

Labeling the parts of the for statement:

for(a--¹; a--²; a--³) printf("%d\n",a);

it is executed:

a is initially 9.
a--¹ is evaluated and ignored. This changes a to 8.
a--² is evaluated and tested. This changes a to 7 and evaluates as 8. Since 8 is non-zero, the for execution continues.
printf("%d\n",a); is evaluated. This prints 7.
a--³ is evaluated and ignored. This changes a to 6.
a--² is evaluated and tested. This changes a to 5 and evaluates as 6. Since 6 is non-zero, the for execution continues.
printf("%d\n",a); is evaluated. This prints 5.
a--³ is evaluated and ignored. This changes a to 4.
a--² is evaluated and tested. This changes a to 3 and evaluates as 4. Since 4 is non-zero, the for execution continues.
printf("%d\n",a); is evaluated. This prints 3.
a--³ is evaluated and ignored. This changes a to 2.
a--² is evaluated and tested. This changes a to 1 and evaluates as 2. Since 2 is non-zero, the for execution continues.
printf("%d\n",a); is evaluated. This prints 1.
a--³ is evaluated and ignored. This changes a to 0.
a--² is evaluated and tested. This changes a to −1 and evaluates as 0. Since 0 is zero, the for execution ends.