For example, there're 3 files, sum.h, sum.cpp, and main.cpp.
sum.h --
...
int sum (int, int, int);
...
sum.cpp
...
int sum (int a, int b, int c=10) {
return a + b + c;
}
main.cpp
...
cout << sum (1, 2) << endl;
...
The compiler throws an error saying too few arguments to function....
It works fine if I code as cout << sum (1,2,3) << endl;
But how to just pass only 2 argument?
Default function arguments must go in the function declaration that the call site sees.
int sum (int, int, int = 10);
They are required at the expression where the function is called. The implementation shouldn't care if it is passed the default or not.
Furthermore, you can re-declare the function in a smaller scope, and give completely different default arguments. This code snippet is taken from the c++17 standard draft and demonstrates what I mean:
void f(int, int);
void f(int, int = 7);
void h() {
f(3); // OK, calls f(3, 7)
void f(int = 1, int); // error: does not use default
// from surrounding scope
}
void m() {
void f(int, int); // has no defaults
f(4); // error: wrong number of arguments
void f(int, int = 5); // OK
f(4); // OK, calls f(4, 5);
void f(int, int = 5); // error: cannot redefine, even to
// same value
}
void n() {
f(6); // OK, calls f(6, 7)
}
Theoretically (don't do this in practice), you can even have different headers that declare the same function with a different default parameter value. And it will work as expected, so long as they aren't both included in the same translation unit.
You must specify the default value in the prototype (the function definition in the .h file):
int sum (int, int, int=10);
No need to specify it in the function implementation.
