Does (4 > y > 1) Always Evaluate to False in C ?

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Evaluating the Validity of (4 > y > 1) in C

In C , the expression (4 > y > 1) may raise questions about its validity and evaluation.

To understand its behavior, it's essential to know that chained comparison operators, such as (4 > y > 1), are parsed as nested comparisons from left to right. Therefore, the statement:

(4 > y > 1)

is parsed as:

((4 > y) > 1)

The comparison operators (> and <) evaluate in the order of the expression, which means that the result of (4 > y) is either 0 (false) or 1 (true) depending on the comparison.

Then, the result of (4 > y) is compared to 1 using the greater-than operator (>). However, since the result of (4 > y) is always either 0 or 1, it will never be greater than 1. Therefore, the entire statement always returns false.

Exception for Operator Overloading:

However, there is an exception to this behavior. If y is an object of a class with an overloaded greater-than operator (>), the evaluation may change. In such cases, the overloaded operator's implementation dictates the behavior of the expression.

Example:

Consider the following code:

class mytype {
};

mytype operator>(int x, const mytype &y) {
    return mytype();
}

int main() {
    mytype y;

    cout << (4 > y > 1) << endl;

    return 0;
}

In this example, the custom operator operator> for the mytype class is used, which may result in different behavior according to the class implementation.

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