Problem evaluating conditional operator

[Richard Tierney]

I'd like to do something like this:

cout << (A? x : B? y:z) << endl;

where x, y, and z are signed and unsigned versions of the same type.
However, this doesn't work. When the expression in brackets is
evaluated, it does something more complicated than simply returning
one of x, y, or z. Can someone tell me what the line above actually
does? Complete (short) test program below.

Thanks -

Richard

----------------------------------------------------------
#include <iostream>
#include <iomanip>

using std::dec;
using std::cout;
using std::endl;

main() {
bool A = false;
bool B = true;
unsigned x = 10;
int y = -20;
int z = 30;
cout << (A? x : B? y:z) << endl; // this prints '4294967276'..
if(A)
cout << dec << x << endl;
else if(B)
cout << dec << y << endl; // but this prints '-20'!
else
cout << dec << z << endl;
}

 

[Rolf Magnus]

I'd like to do something like this:

cout << (A? x : B? y:z) << endl;

where x, y, and z are signed and unsigned versions of the same type.

C++ is a statically typed language. The return type of operator ?: is
fixed at compile time and cannot depend on the value of A.

However, this doesn't work. When the expression in brackets is
evaluated, it does something more complicated than simply returning
one of x, y, or z.

It also converts them all to the same type.

Can someone tell me what the line above actually does? Complete
(short) test program below.

Thanks -

Richard

----------------------------------------------------------
#include <iostream>
#include <iomanip>

using std::dec;
using std::cout;
using std::endl;

main() {
bool A = false;
bool B = true;
unsigned x = 10;
int y = -20;
int z = 30;
cout << (A? x : B? y:z) << endl; // this prints '4294967276'..

x is an unsigned int, therefore the return type of the outer operator ?:
becomes unsigned int. From the second one, y is chosen, which is then
converted to unsigned int.

if(A)
cout << dec << x << endl;
else if(B)
cout << dec << y << endl; // but this prints '-20'!

Of course. That's what the value of y is.

 

[marbac]

I'd like to do something like this:

cout << (A? x : B? y:z) << endl;

where x, y, and z are signed and unsigned versions of the same type.
However, this doesn't work. When the expression in brackets is
evaluated, it does something more complicated than simply returning
one of x, y, or z. Can someone tell me what the line above actually
does? Complete (short) test program below.

this should work correctly:

#include <iostream>
#include <iomanip>

using std::dec;
using std::cout;
using std::endl;

main() {
bool A = false;
bool B = true;
unsigned x = 10;
int y = -20;
int z = 30;

cout << (int)(A? x : B? y:z) << endl; // this prints '4294967276'..
if(A)
cout << dec << x << endl;
else if(B)
cout << dec << y << endl; // but this prints '-20'!
else
cout << dec << z << endl;
}

 

[Richard Tierney]

<snipped>

That's what I call service - an answer in 10 minutes!



Richard