Would it be safe to do swap by "b=(a+b)-(a=b);"?

Question

In Cprogramming.com I found this piece of code:

int a,b;
scanf("%d %d",&a,&b);
b=(a+b)-(a=b);
printf("%d %d",a,b);

It is claimed to be a tip/trick to "swap without using temporary". My tests on Linux gcc prove it. However, wouldn't the order how different compilers or platforms computing this expression matters here? Is it safe to use such code?

Yeah, 'cos using a temp variable is such a terrible thing to do!!! — David Heffernan, Feb 25 '12 at 18:21
Recent discussion on this ["trick"](http://stackoverflow.com/questions/8265641/switch-two-values-without-temp-storage), expanding on what @DavidHeffernan mentions. — Brett Hale, Feb 25 '12 at 21:00

score 12 · Accepted Answer · answered Feb 25 '12 at 16:41

12

No. In the expression (a+b)-(a=b) there is no sequence point between a being written to and a being read in the (a+b) sub-expression to determine the value to be stored to b so the behaviour is undefined.

answered Feb 25 '12 at 16:41

CB Bailey

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even if there were a sequence point, it still wouldn't be portable because of potential integer overflow; see also http://en.wikipedia.org/wiki/XOR_swap_algorithm#Variations – Christoph Feb 25 '12 at 16:49

score 8 · Answer 2 · answered Feb 25 '12 at 16:41

8

However, wouldn't the order how different compilers or platforms computing this expression matters here?

Yes.

Is it safe to use such code?

No, it's undefined behavior.

answered Feb 25 '12 at 16:41

sepp2k

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Would it be safe to do swap by "b=(a+b)-(a=b);"?

2 Answers2