You can provide default values for function parameters. For example:
#include <iostream>
using namespace std;
int a = 1;
int f(int a) { return a; }
int g(int x = f(a)) { return x; }
int h() {
a = 2;
{
int a = 3;
return g();
}
}
int main() {
cout << h() << endl;
}
This example prints 2 to standard output, because the a referred to in the declaration of g() is the one at file scope, which has the value 2 when g() is called.
The default argument must be implicitly convertible to the parameter type.
A pointer to a function must have the same type as the function. Attempts to take the address of a function by reference without specifying the type of the function will produce an error. The type of a function is not affected by arguments with default values.
The following example shows that default arguments are not considered part of a function's type. The default argument allows you to call a function without specifying all of the arguments, it does not allow you to create a pointer to the function that does not specify the types of all the arguments. Function f can be called without an explicit argument, but the pointer badpointer cannot be defined without specifying the type of the argument:
int f(int = 0);
void g()
{
int a = f(1); // ok
int b = f(); // ok, default argument used
}
int (*pointer)(int) = &f; // ok, type of f() specified (int)
int (*badpointer)() = &f; // error, badpointer and f have
// different types. badpointer must
// be initialized with a pointer to
// a function taking no arguments.
In this example, function f3 has a return type int, and takes an int argument with a default value that is the value returned from function f2:
const int j = 5;
int f3( int x = f2(j) );
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