Why, the output is 6 and not 7?

Question

The following C code executes correctly but not as expected. Post increment operator here in z=z++ is creating confusion here. I may not be able to figure out silly mistake/concept, Can I have a brief explanation or some helpful link please.

 #include<stdio.h>
    int main()
    {
        int x=5,y=6,z=7;
        if(x-y)

            z=z++;
            z=--z;

        printf("%d",z);
    }

alinsoar · Answer 1 · 2019-03-06T19:20:09.577

0

You are not allowed to do z=z++; because between 2 sequence points you are not allowed to assign a variable 2 times.

This one is a full expression in which you assign z 2 times. So it can be interpreted ambigously and the result of the C abstract machine is undefined behavior.

The same for z=--z.

edited Mar 06 '19 at 19:20

answered Mar 06 '19 at 18:28

alinsoar

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Thank you, i got your point. You also mean that the answer will depend on compiler. – skyconfusion Mar 06 '19 at 18:33
Just one doubt, if I have to pick one value from 6, 7 and 8. Which one will be more general and expected answer? – skyconfusion Mar 06 '19 at 18:35
1

@skyconfusion This is [undefined behavior](https://en.wikipedia.org/wiki/Undefined_behavior). You can't rely on any value. – interjay Mar 06 '19 at 18:39

Why, the output is 6 and not 7?

1 Answers1