Potential problem in "Exchange of values ​​of two variables without using a third variable"

Question

I recently came up with this method to replace the values ​​of two variables without using a third variable.

a^=b^=a^=b

But when I tried the above code on different compilers, I got different results, some gave the correct results, and some did not.

Is something terribly wrong with the code?

Answer 1

Is something terribly wrong with the code?

Yes!

a^=b^=a^=b actually causes Undefined Behavior in C and C ++, because you are trying to change the value of a more than once between two points in a sequence.

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Try writing (although not reliable)

 a ^= b; b ^= a; a ^= b; 

instead of a^=b^=a^=b .

PS : never try to change the values ​​of two variables without using a third. Always use the third variable.

EDIT :

As @caf b^=a^=b noted, even if the order of evaluation of the arguments of the ^= operator ^= not specified, since all calls to b in the expression are used to calculate the final value that is stored in b , the behavior is well defined.

Answer 2

If you use C ++, why not use the swap algorithm in STL? It is ideal for this purpose, and it is very clear what it does:

 #include <algorithm> using namespace std; // ... int x=5, y=10; // x:5 y:10 swap(x,y); // x:10 y:5 

Answer 3

Based on R. and sellibitze materials:

Use the comma operator:

  (a^=b,b^=a,a^=b); 

From the text and Wikipedia:

"The comma operator can be used to link related expressions together. The list of expressions associated with a comma is evaluated from left to right, and the value of the rightmost expression is the value of the combined expression. It acts like a sequence point."

“The sequence point ensures that all side effects of previous evaluations are fulfilled, and no side effects from subsequent evaluations have yet been performed. This eliminates undefined behavior resulting from a fuzzy execution order of the original expression.”

Answer 4

I suggest you use std :: swap () for C ++.

For c use this macro. Note that first you need to compare a and b, otherwise, when they point to the same memory location, you will destroy the value and it will become 0.

 #define swap(a, b) ((a) == (b) || (a) ^= (b), (b) ^= (a), (a) ^= (b)) 

Answer 5

Do it like this:

 a ^= b; b ^= a; a ^= b; 

Answer 6

How about this?

 a = a + b; b = a - b; a = a - b; 

Answer 7

I wondered why no one suggested bracketing the expression. This does not seem to be UB anymore.

 a^=(b^=(a^=b)); 

Answer 8

You can also try the following, but if for numbers sufficiently large the value will overflow

 a=a*b; b=a/b; a=a/b;