No, you cannot overload functions that differ only in their default arguments. If you try to do this, you will get a compilation error due to ambiguity. For example:
#include <iostream>
void Print(int x) {
std::cout << "Print(int)";
}
void Print(int x, int y = 0) {
std::cout << "Print(int, int)";
}
int main() {
Print(10);// error: ambiguous call
}error: 'Print': ambiguous call to overloaded functionIn this case, both Print(int) and Print(int, int) are equally valid for the call Print(10), because the second parameter of Print(int, int) has a default value. This creates an ambiguity, and the compiler will emit an error.
To avoid this, ensure that your overloaded functions differ in their parameter types or number of parameters, not just in their default arguments.
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This Question is from the Lesson:
Understanding Overload Resolution
Learn how the compiler decides which function to call based on the arguments we provide.
