using bind with member function [duplicate]

Question

Why are both bind versions compiling and working without problems, even though I'm using different parameter types in each call?

version 1 -> parameter foo vs.
version 2 -> parameter address of foo

I'd expected version 1 to produce a compile error...

#include <iostream>
#include <functional>

using namespace std;

class Test   
{   
public:
   bool doSomething(){ std::cout << "xxx"; return true;};   
};

int main() {

Test foo;

// Version 1 using foo
std::function<bool(void)> testFct = std::bind(&Test::doSomething, foo);

// Version 2 using &foo
std::function<bool(void)> testFct2 = std::bind(&Test::doSomething, &foo);

testFct();
testFct2();

return 0;
}

Becasue you can call a member function on both an object and pointer to an object? — jrok
– jrok, Commented Sep 11, 2013 at 11:47

Mike Seymour · Accepted Answer · 2013-09-11 11:59:11Z

1

std::function<bool(void)> testFct = std::bind(&Test::doSomething, foo);

This will bind to a copy of foo, and call the function as copy.doSomething(). Beware that this will be wrong if you're expecting the function to be called on foo itself.

std::function<bool(void)> testFct2 = std::bind(&Test::doSomething, &foo);

This will bind to a pointer to foo and call the function as pointer->doSomething(). Beware that this will be wrong if foo has been destroyed before calling the function.

I'd expected version 1 to produce a compile error...

You could disallow this behaviour by making Test uncopyable, if you wanted.

edited Sep 11, 2013 at 11:59

answered Sep 11, 2013 at 11:50

Mike Seymour

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