C operator -- gives two different results

Question

#import <stdio.h>
int main()
{
    int a,d,b,c;
    a = 10;
    c = 10;
    d = --c + --c+1; 
    b = --a +1+ --a ;
    printf("b= %d, d = %d" , b,d);
    return 0;
}

b= 18, d = 17

the code and it's run

Why d&b are not equal? can you please explain why d=17?

[Undefined behaviour](http://en.wikipedia.org/wiki/Undefined_behavior) — bitcell, Nov 13 '14 at 10:53
Interesting. In codepad b and d are both equal to 17. http://codepad.org/xfCWITcq — twasbrillig, Nov 13 '14 at 10:57
possible duplicate of [Why are these constructs (using ++) undefined behavior?](http://stackoverflow.com/questions/949433/why-are-these-constructs-using-undefined-behavior) — Klas Lindbäck, Nov 13 '14 at 11:04

score 0 · Accepted Answer · answered Nov 13 '14 at 11:05

0

Oh right, it's because it's undefined when the value of a and c will decrement.

d = --c + --c+1;

If c is decremented twice and then d is calculated, then it gives

d = 8 + 8 + 1 = 17

But if c is decremented as we proceed from left to right, then it gives

d = 9 + 8 + 1 = 18

answered Nov 13 '14 at 11:05

twasbrillig

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Following the equation the way the OP wrote it, `d = --c` (which is 9) `+ --c` (which is 8) `+ 1`; that's why I went in that order. – twasbrillig Nov 13 '14 at 11:13

C operator -- gives two different results

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