Why std::bind can be assigned to argument-mismatch

问题:

I have code as follows:

#include <functional>
#include <iostream>
using namespace std;
void F(int x) {
  cout << x << endl;
}
int main() {
  std::function<void(int)> f1 = std::bind(F, std::placeholders::_1);
  f1(100);  // This works, will print 100.

  int x = 0;
  std::function<void()> f2 = std::bind(F, x);
  f2();  // This works, will print 0.

  std::function<void(int)> f3 = std::bind(F, x);
  f3(200);  // BUT WHY THIS WORKS?????? It prints 0.
  return 0;
}

My compiler info is: Apple LLVM version 6.0 (clang-600.0.56) (based on LLVM 3.5svn) Target: x86_64-apple-darwin13.4.0 Thread model: posix

回答1:

That is correct behavior.

std::bind needs this looseness to fit its own specification.

Consider std::placeholders, which is used to mark parameters that are passed through to the bound function.

using std::placeholders;
std::function<void(int)> f2 = std::bind( F, _1 );
//  Parameter 1 is passed to                ^^
//  the bound function.

f2(7); // The int 7 is passed on to F

Similarly, there is _2 for the second parameter, _3 for the third, and so on.

That brings up an interesting question. How should this function object behave?

auto f3 = std::bind( F, _3 );

As you might imagine, it follows its own promise to pass the third parameter to F. Which means it does nothing to the first two parameters.

f3(10, 20, 30); // The int 30 is passed on to F. The rest?  Ignored.

So this is expected behavior, and possibly the only "feature" that std::bind holds over lambdas, even in C++14 and C++17.

The object produced by std::bind is designed to accept and ignore any extraneous parameters.