[SOLVED] XOR like function in C/C++

[WitchCraft]

Question:

if i want to xor 2 values in C/C++, i can do this like this using:
( (p || q) && !(p && q) )
which is equivalent to p xor q.

How do i implement an xor function, that looks like p xor q?
Example:

int main()
{
     bool a,b;
     a=true; b=true;
     if (a xor b) 
          printf("true\n");
     else
          printf("false\n");
     return EXIT_SUCCESS ;
}

I can of course do a function like:

bool xor(bool p, bool q)
{
     return ( (p || q) && !(p && q) ) 
}

But then the function call looks like:
if( xor(a,b) )

and i want: if ( a xor b )

Can I do that at all in C/C++?

[shaggy999]

I don't think you can do that in C++ itself. C++ does have operator overloading, but that's only for particular operators. You can't create any operator of your choosing. My suggestion would be to create some preprocessor syntax that gets parsed before getting handed off to the compiler. That way you can have your simple function, but it will convert to C++ code at compile time.

[slavik]

^ is the xor operator in C/C++

[shaggy999]

Oh damn, it hurts. Hurts to the CORE. I was wrong. Damn.

[WitchCraft]

^ is the xor operator in C/C++

Oh, i thought ^ is only bitwise. Obviously that also works non bitwise; funny

But it would still be interesting to know how to do a function that looks like:
argument1 OPERAND argument2

[slavik]

Oh, i thought ^ is only bitwise. Obviously that also works non bitwise; funny

But it would still be interesting to know how to do a function that looks like:
argument1 OPERAND argument2

while in C you won't be able to do that (unless you change the language/compiler) in C++ you can overload the ^ operator to work on other things.

(un)fortunately you cannot define your own operators in C/C++ (whether this is good or not can be debated till infinity).

[Zwack]

Ummm, Depends on the types you are using...

in C a bitwise xor is just ^

thus a ^ b is a xor b

unfortunately you can't overload ^ with

bool operator^ (bool p, bool q)

As at least one of the data types needs to be user defined...

Z.

[slavik]

Ummm, Depends on the types you are using...

in C a bitwise xor is just ^

thus a ^ b is a xor b

unfortunately you can't overload ^ with

bool operator^ (bool p, bool q)

As at least one of the data types needs to be user defined...

Z.

I am pretty sure operator^(bool,bool) is defined already

[Zwack]

I'd be surprised if bool ^ bool isn't defined... But my point is more that you can't really redefine it for built in types. So if you want a function that sits in the middle like a binary operator then you have to be redefining a current operator for a user defined type.

If you decided you wanted a function to operate on two built-in types and you wanted it to look like a binary operator you can't... But then how would the compiler know what precedence to give it anyway?

a ^ b & c - d has a set precedence, it will always act in the same way...
but supposing I defined a new operator "

what order would a ^ b " c - d be done in? What should the expected result be? While quote(a ^ b, c -d) is something that does have a recognised precedence.

Equally being able to overload operators for built-in types would be dangerous... After all if you've redefined ^ to being something else then later you use it because you need to xor something how are you ever going to track down the cause of your odd results?

Z.

[WitchCraft]

a ^ b & c - d has a set precedence, it will always act in the same way...
but supposing I defined a new operator "

That would be simple if you could set the precedence for the operator.
But i see, that's something for (C++)++.