UnrealScript XOR XAND

['Zac]

Hello, in some vanilla UT code, I see return statements that look like this:

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....
return Foo & Bar

or

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....
return Foo | Bar

I understand the usage in an if statement, meaning that

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Foo | Bar

is true if only one is false and one is true (Is this still correct UScript? I never use these one liners )

So my Question is, does....

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....
return Foo & Bar
// returns true if both are true? false if not?

or

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....
return Foo | Bar
// return true if one is false and one is true?
// or return the true value?

It's just always stuck with me since seeing this syntax, and I never see it anywhere else. Discuss below

[Barbie]

A single "&" respective "|" means a bitwise AND respective OR, the double ones mean logical AND and OR.
Example:

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var byte a, b;

a = 3; // in dual notation 0000.0011
b = 5; // in dual notation 0000.0101

return a & b; // will return 0000.0001

See also

• https://en.wikipedia.org/wiki/Bitwise_operation
• https://wiki.beyondunreal.com/Operators

If used in double variant (logical), just imagine some braces around the expression:

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var bool a, b;

a = true;
b = false;
return a && b; //same as return (a && b)  same as return FALSE;

I understand the usage in an if statement, meaning that

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Foo | Bar

is true if only one is false and one is true

Hmm, if you have meant

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Foo || Bar

then the expression is TRUE if one of the members is TRUE.

['Zac]

Thanks for the clearing up! and the links!

[Higor]

In UnrealScript | and & are used as bitwise operators for integer types, and both RHS and LHS (right hand side, left hand side) are tested.

When it comes to || and &&, these are boolean operators with a 'skip' qualifier, LHS is checked first and depending on the result RHS may or may not be checked.

In other low level languages | and & may be used for many other types, and & may not yield the same results as &&, just as using & may crash your program.


==
Btw, in the case of booleans Exclusive-OR can be evaluated using !=

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if ( bool1 != bool2 )