what are the unary values for unary boolean operators

lostbits (18)
I am implementing boolean operator overloads and have some strange results for unary operators. My code shows that in gcc 4.5.3:
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                         expections
   actual results        true  false
!true  = !false ==  0      0     1   (logical not of value)
~true  = ~false == -2    0xFE  0xFF  (binary  not of value)
+true  = +false ==  1                (depends if bool is an int or bool)
++true = ++false == 1                (if bool then illegal otherwise ++int)
true++ = false++ == 1                (same as above)
--true = -- false == NOT ALLOWED     (legal and treating like bool++)


Where some values depend on whether the boolean input is treated as a bool (with values true|false), or an int (with values 0|1). If it treated as a boolean then I would expect the results to be a boolean (true|false) or within the range of a boolean treated as an int (0|1).

The results are unexpected and appear wrong. Is my code wrong, gcc at fault, or my understanding faulty? Any help is appreciated.

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# include <iostream>
# include "cstdio"
# include <cstdlib>

using namespace std;
void func0(bool x) {
   string xStr = (x)? "true": "false";
   cout << "func0(" << xStr << ')' << endl
        << "!x  " << (!x)  << endl
        << "~x  " << (~x)  << endl
        << "+x  " << (+x)  << endl
        << "-x  " << (-x)  << endl
        << "++x " << (++x) << endl
        << "--x " << "error" << endl
        << "x++ " << (x++) << endl
        << "x-- " << "error" << endl
        << endl;
   
}
void func1(bool x, bool y) {
   string xStr = (x)? "true": "false";
   string yStr = (y)? "true": "false";
   cout << "func1(" << xStr << ", " << yStr << ')' << endl
        << "x + y " << (x + y) << endl
        << "x - y " << (x - y) << endl
        << "x * y " << (x * y) << endl;
   if (y == 0) {
        cout << "x / y " << x << " / 0" << endl;
   } else {
        cout << "x / y " << (x / y) << endl;
   }
}
void func2(bool x, int y) {
   string xStr = (x)? "true": "false";
   char buf[10];
   sprintf(buf, "%i", y);
   string yStr(buf);
   cout <<"func2("  << xStr << ", " << yStr << ')' << endl
        << "x + y " << (x + y) << endl
        << "x - y " << (x * y) << endl
        << "x / y " << (x / y) << endl;
   cout << yStr << ", " << xStr << endl
        << "y + x " << (y + x) << endl
        << "y - x " << (y - x) << endl
        << "y * x " << (y * x) << endl;
   if (x == 0) {
      cout << "y / x " << y << " / 0" << endl;
   } else {
      cout << "y * x " << (y / x) << endl;
   }
}
int main(int argc, char** argv) {
   bool x = false;
   func0(true);
   func0(false);
   
   func1(true, true);
   func1(true, false);
   func1(false, true);
   func1(false, false);

   func2(true, 1);
   func2(false, 1);
   return 0;
}
Cubbi (4774)
Your expectations appear to be different from how C++ defines the results of those operations, but your code is also very wrong: your func0() output statement invokes undefined behavior because it attempts to modify (via opeator++) and access (many times) the same object without any sequencing.

Note that ++ for bools is a bizarre anachronism, it was deprecated the moment it was inherited from C, and will likely be removed from the language soon.
Last edited on
aschwarz1309 (10)
Thanks for the update on the outlandish ++.

I don't understand your comments on func0() having undefined output behavior. Could you elaborate?

When I coded I thought I'd done just fine (sigh). Except of X++ and ++X, each statement presents the result of a computation without altering the value of the variable, and I assumed the variable was then immediately available for a subsequent operation. I don't see the issue.
Cubbi (4774)
I don't understand your comments on func0() having undefined output behavior. Could you elaborate?

The expression cout << x << ++x is UB (if x is of a non-class type) because it attempts to modify x and read from x on independent, unsequenced code paths.
There are some details here: http://en.cppreference.com/w/cpp/language/eval_order

g++ -Wall describes it as
test.cc:17:16: warning: operation on ‘x’ may be undefined [-Wsequence-point]


Put a semicolon before the first cout << (++x), and the output of your func0() will make more sense.
Last edited on
DeXecipher (458)
unary means that i only requires one operand
binary means that it only requires 2
tertary means 4
and so on.
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