Is there any way to force java to test two condition regardless the first condition? [duplicate]

Question

let's say I have a statement

if(stack.pop() == 1 && stack.pop() == 1)

if top of stack is 0, then the second condition won't be implemented, which means it just pops one value at the top. What I want is to pop both, the top and the value after top. Is there any way to do that without using another if-else statement in it?

Answer 1

int first = stack.pop();
int second = stack.pop();
if (first == 1 && second == 1)

Answer 2

Use the bitwise AND operator**:

if( stack.pop() == 1 & stack.pop() == 1 )

This will force the evaluation of both sides.

** I know it by "non-short-circuiting" of logical AND, but it is indeed a bitwise operator that acts on boolean operands (documentation here).

Update: As JBNizet said in his comment below, this code is "fragile", since some developer may "correct" the operator to the short-circuit version. If you choose to use & instead of storing the values of the method calls (forcing them to run) likewise JBNizet answer, you should write a comment before your if statement.

Answer 3

Yet another one-line and slightly obfuscated way to do this:

if( stack.pop() == 1 ? stack.pop() == 1 : stack.pop() == 1 && false )

Personally I'd go for JB Nizet's way. Makes it as clear as can be exactly what you're trying to do.