Do &&
and ||
stop evaluating as soon as a result is known?
In other words will (true == true) || (true == false)
not evaluate the right side because the whole expression is known to be true
after only evaluating the left side.
Rust has a great story around short circuiting. And it's not just with return , break , and continue .
Do logical operators in the C language are evaluated with the short circuit? Explanation: None.
In Java logical operators, if the evaluation of a logical expression exits in between before complete evaluation, then it is known as Short-circuit. A short circuit happens because the result is clear even before the complete evaluation of the expression, and the result is returned.
The | and & logical operators, known as non-short circuit operators, should not be used. Using a non-short circuit operator reduces the efficiency of the program, is potentially confusing and can even lead to the program crashing if the first operand acts as a safety check for the second.
Yes.
From the Rust reference:
fn main() {
let x = false || true; // true
let y = false && panic!(); // false, doesn't evaluate `panic!()`
}
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