Notes

Same function name, different parameter lists. The compiler picks which one you meant based on the arguments you pass.

void print(int x);
void print(float x);
void print(const std::string& x);

print(5);       // calls print(int)
print(5.0f);    // calls print(float)
print("hi");    // calls print(const string&)

What doesn’t count as “different”

int getValue();
float getValue();   // error - return type alone isn't enough to overload on

Overload resolution looks at parameters, not return type. There’s no way to know which one you wanted from auto x = getValue(); alone.

Gotcha: ambiguous calls

void f(int a);
void f(float a);
f(5.0);   // error: ambiguous - 5.0 is a double, an equally bad implicit
          // conversion to either int or float

Related

Default arguments
Function templates, overload resolution and specialization