why i^=j^=i^=j isn't equal to *i^=*j^=*i^=*j

Posted by klvoek on Stack Overflow See other posts from Stack Overflow or by klvoek
Published on 2012-09-08T14:16:15Z Indexed on 2012/09/08 15:38 UTC
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In c , when there is variables (assume both as int) i less than j , we can use the equation
i^=j^=i^=j
to exchange the value of the two variables. For example, let int i = 3, j = 5; after computed i^=j^=i^=j, I got i = 5, j = 3 . What is so amazing to me. But, if i use two int pointers to re-do this , with *i^=*j^=*i^=*j , use the example above what i got will be i = 0 and j = 3. Then, describe it simply:


In C

1

    int i=3, j=5;
    i^=j^=i^=j; // after this i = 5, j=3

2

    int i = 3, j= 5;
    int *pi = &i, *pj = &j;
    *pi^=*pj^=*pi^=*pj; // after this, $pi = 0, *pj = 5

In JavaScript

    var i=3, j=5;
    i^=j^=i^=j; // after this, i = 0, j= 3

the result in JavaScript makes this more interesting to me

my sample code , on ubuntu server 11.0 & gcc

    #include <stdio.h>
    int main(){
        int i=7, j=9;
        int *pi=&i, *pj=&j;
        i^=j^=i^=j;
        printf("i=%d j=%d\n", i, j);
        i=7, j==9;
        *pi^=*pj^=*pi^=*pj
        printf("i=%d j=%d\n", *pi, *pj);
    }


however, i had spent hours to test and find out why, but nothing means. So, please help me. Or, just only i made some mistake???

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