CODEWITHSUNDEEP
c++
Orange Fox
Orange Fox

Reputation: 147

Weird boolean conversion (?)

Explain pleasy why second expression returns false

   cout << (4==4) << endl; //1
   cout << (4==4==4) << endl; // 0

Upvotes: 2

Views: 167

Answers (3)

user2249683
user2249683

Reputation:

You can not apply binary compare operators to more than two operants, reasonably (unless you have an overwrite making it possible, somehow).

To compare an unknown number of arguments:

#include <iostream>

template <typename A, typename B>
bool equal(const A& a, const B& b) {
    return a == b;
}

template <typename A, typename B, typename ... Other>
bool equal(const A& a, const B& b, const Other& ... other) {
    return a == b && equal(b, other ...);
}

int main() {
    std::cout << equal(1, 1, 1) << '\n';
    std::cout << equal(1, 2, 3) << '\n';
}

C++ 11

Upvotes: 1

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361252

(4==4==4) is basically ((4==4)==4) which is (true == 4) which is (1==4) 1 which is false 2 which is getting printed as 0.

Note that == has associativity left-to-right, but that doesn't matter (in this case) because even if it had associativity right-to-left, the result would have been the same.

1. Due to integral promotion.
2. Note that one might tempted to think 4 in (true==4) could be treated as true (after all 4 is non-zero, hence true). This thinking might conclude (true==4) is (true==true) which is true. But that is not how it works. It is the bool which gets promoted to int, instead of int to bool.

Upvotes: 13

mrks
mrks

Reputation: 8333

4==4 evaluates to true, which for the purpose of comparing it with 4 is converted to 1. 1 == 4 is false, which is 0.

Upvotes: 2

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