Swapping values using pointers without a temporary variable is an interesting technique, but it's important to note that it's not always the most readable or maintainable approach. However, it can be useful in certain situations where performance is critical. Here's how you can do it:
#include <iostream>
void swap(int* a, int* b) {
// Check if the pointers are not the same
if (a != b) {
// XOR the values
*a = *a ^ *b;
// XOR again to get the original value of a
*b = *a ^ *b;
// XOR one more time to get the
// original value of b
*a = *a ^ *b;
}
}
int main() {
int x{5};
int y{10};
std::cout << "Before swap: x = "
<< x << ", y = " << y << "\n";
swap(&x, &y);
std::cout << "After swap: x = "
<< x << ", y = " << y << "\n";
}
Before swap: x = 5, y = 10
After swap: x = 10, y = 5
This technique uses the XOR operation to swap the values. Here's how it works:
a = *a ^ *b
: This XORs the values of a
and b
and stores the result in a
.b = *a ^ *b
: This XORs the new value of a
with b
, which gives us the original value of a
, stored in b
.a = *a ^ *b
: This XORs the value in a
with the new value in b
, giving us the original value of b
, now stored in a
.While this method is clever, it's worth noting that modern compilers are very good at optimizing standard swap operations. In most cases, using a temporary variable or std::swap()
will be just as efficient and much more readable:
#include <iostream>
#include <utility>
int main() {
int x{5};
int y{10};
std::cout << "Before swap: x = "
<< x << ", y = " << y << "\n";
std::swap(x, y);
std::cout << "After swap: x = "
<< x << ", y = " << y << "\n";
}
Remember, code readability is often more important than micro-optimizations. Use the pointer-based swap only when you have a specific reason to do so and when you're sure it provides a significant performance benefit in your specific use case.
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