Small brainteaser. Do you know that function below doesn't work in some cases.
Try to attack it.
#include <iostream>
using namespace std;
void swap(int &a, int &b){
a = a + b;
b = a - b;
a = a - b;
}
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attempt in edit.
Yeap, it is true. But more interesting example is swap(a[i], a[j]), and accidentally i == j.
do you mean the overflow, when a + b > 2^31 — 1
do not work if *a == *b
even it occurs overflow, it works well
it is UB in case of overflow signed integers
int a=12; swap(a,a);
best swap function without overflows :)
Does not work if value of a is equal to value of b. In this case after XOR operation a and b will be equal to zero.
Why? I think it will work.
that's why Mozilla created Rust. In this language any attempt to call
swap(a, a)
will lead to compile-time error. Prooflink: http://is.gd/m65CVw