CODEWITHSUNDEEP
c++uint64
MrPontes
MrPontes

Reputation: 49

Unexpected result of logical AND operation between two uint64_t numbers

In the following code, I check if the result of the logical AND operation equals zero:

#include <iostream>
#include <cstdint>

int main() {

  uint64_t num1 = 0b0000000000000000000000000000000000000000000000000000000000000000;
  uint64_t num2 = 0b0000000000000000000000000000000000000000000000000000000000000000;
  uint64_t result = num1 & num2;

  if(result == 0) {
    std::cout << "It got here!\n";
  }

  return 0;
}

The output is: It got here!

However, if I don't use the "result" variable:

#include <iostream>
#include <cstdint>

int main() {

  uint64_t num1 = 0b0000000000000000000000000000000000000000000000000000000000000000;
  uint64_t num2 = 0b0000000000000000000000000000000000000000000000000000000000000000;

  if(num1 & num2 == 0) {
    std::cout << "It got here!\n";
  }

  return 0;
}

It doesn't print anything. Why does this happen?

Upvotes: 0

Views: 98

Answers (1)

David Brown
David Brown

Reputation: 13526

The == operator has higher precedence than the & operator. So your if statement is equivalent to

num1 & (num2 == 0)

Since num2 == 0 is true this works out to 0 & 1 which is 0.

Most compilers will warn in this situation if you have warnings enabled.

Upvotes: 1

Related Questions