Bruce Eckel's Thinking in C++, 2nd Ed Contents | Prev | Next

Taking the address of a generated function template

There are a number of situations where you need to take the address of a function. For example, you may have a function that takes an argument of a pointer to another function. Of course it’s possible that this other function might be generated from a template function so you need some way to take that kind of address [57]:

//: C19:TemplateFunctionAddress.cpp
// Taking the address of a function generated
// from a template.

template <typename T> void f(T*) {}

void h(void (*pf)(int*)) {}

template <class T> 
  void g(void (*pf)(T*)) {}

int main() {
  // Full type exposition:
  h(&f<int>);
  // Type induction:
  h(&f);
  // Full type exposition:
  g<int>(&f<int>);
  // Type inductions:
  g(&f<int>);
  g<int>(&f);
} ///:~

This example demonstrates a number of different issues. First, even though you’re using templates, the signatures must match – the function h( ) takes a pointer to a function that takes an int* and returns void, and that’s what the template f produces. Second, the function that wants the function pointer as an argument can itself be a template, as in the case of the template g.

In main( ) you can see that type induction works here, too. The first call to h( ) explicitly gives the template argument for f, but since h( ) says that it will only take the address of a function that takes an int*, that part can be induced by the compiler. With g( ) the situation is even more interesting because there are two templates involved. The compiler cannot induce the type with nothing to go on, but if either f or g is given int, then the rest can be induced.

[57] I am indebted to Nathan Myers for this example.

Contents | Prev | Next

Contact: webmaster@codeguru.com
CodeGuru - the website for developers.