Since my first post using the web interface seems to have been lost, I'm

From: "Markus Schöpflin" <markus.schoepflin@ginit-technology.com> trying again via email...

Is it just me or is it my compiler (MSVC6SP5)? The following code gives me

an error. Whats wrong?

#include <boost/bind.hpp> #include <functional>

using boost::bind;

void main() { bind(std::greater<int>, _1, _2)(1, 2); }

main.cpp(8) : error C2275: 'std::greater<int>' : illegal use of this type

as an expression To make it compile, you need to: * change std::greater<int> to std::greater<int>(); * change bind(... to bind<bool>(... since MSVC 6 cannot deduce the return type; * use the qualified boost::bind<bool> syntax to work around a bug in the compiler; and finally, use int x = 1; int y = 2; boost::bind<bool>(std::greater<int>(), _1, _2)(x, y); since bind cannot take rvalues (this is usually not a problem in real code, just in examples.)

On Monday 25 March 2002 10:43 am, you wrote:

Since my first post using the web interface seems to have been lost, I'm trying again via email...

Is it just me or is it my compiler (MSVC6SP5)? The following code gives me an error. Whats wrong?

#include <boost/bind.hpp> #include <functional>

using boost::bind;

void main() { bind(std::greater<int>, _1, _2)(1, 2); }

main.cpp(8) : error C2275: 'std::greater<int>' : illegal use of this type as an expression

TIA, Markus

Two problems here: - std::greater<int> is a type. You want std::greater<int>() to create an object of that type - constants can't (directly) be passed into boost::bind objects. You'll need to have: int x = 1, int y = 2; and call the boost::bind object with (x, y) instead of (1, 2). If you still get errors, then it's your compiler's fault because it can't figure out the return type. You'll have to tell bind to use an 'int' return type, so the final code would look like this: #include <boost/bind.hpp> #include <functional> using boost::bind; int main() { int x1 = 1; int x2 = 2; bind<int>(std::greater<int>(), _1, _2)(x1, x2); } Doug