Lazy Evaluation

[OBXKH]

i would think this is simple...

let a = 0; // declare the variable a and assign the value of 0<< 0

false && (a = 1); // (a = 1) is truthy, but it won't be evaluated, since the first operand is false<< false

a // the value of a is still 0<< 0
false || (a = 1); // this will evaluate both operands, so a will be assigned the value of 1, which is returned<< 1

if

a = 0

, declared first, how can

a = 1

be truthy?

 

[BigDady]

In this statement: false || (a = 1); you set a equal to 1. The single '=' assigns a value to the variable 'a'.