@Dave: You said *a ^= *b ^= *a ^= *b violates the sequence point rule
in swap(). But it's the same as in main function. Why doesn't it
violates the rule in main() ?
On Jul 13, 1:30 am, Don <dondod...@gmail.com> wrote:
> To make it into valid code, break it into three operations:
>
> void swap(int *a, int *b)
> {
> *a ^= *b;
> *b ^= *a;
> *a ^= *b;
> }
>
> On Jul 12, 3:25 pm, Dave <dave_and_da...@juno.com> wrote:
>
>
>
>
>
>
>
> > @Tendua: The statement *a ^= *b ^= *a ^= *b violates the sequence
> > point rule, which says that results are undefined if a variable is
> > assigned more than one value between sequence points.
>
> > Dave
>
> > On Jul 12, 3:15 pm, tendua <6fae1ce6347...@gmail.com> wrote:
>
> > > # include <stdio.h>
>
> > > void swap(int *a, int *b)
> > > {
> > > *a ^= *b ^= *a ^= *b;
> > > }
>
> > > int main()
> > > {
> > > int a=45, b= 56;
> > > // a ^= b ^= a ^= b;
> > > swap(&a,&b);
> > > printf("%d %d",a,b);
> > > }
>
> > > This code gives output 0, 45
> > > While if we uncomment the line in main function and don't use swap
> > > function, we get correct value. Explain why the same line when used in
> > > swap function gives such output.
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