Function overloading in C++

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List of C++ name resolution (and overloading) rules

What are the rules in C++ for how the compiler decides which function to choose ? (that's is given two functions with the same name - how does the compiler pick/prioritize one function over the other, mainly I want to know what types of casting the compiler is more willing to do when he chooses)

The rules are many... and complicated. All of this is covered in section **3.4** of the C++ standard — parapura rajkumar, Nov 17 '11 at 20:03
@Belgi - I see, I took what you meant by prioritizing in a different context. — MGZero, Nov 17 '11 at 20:05
@phs - had to be, vote to close (Belgi, learn to search before you ask) — KevinDTimm, Nov 17 '11 at 20:06

score 4 · Accepted Answer · answered Nov 17 '11 at 20:08

As already stated, the rules are fully described in the standard. As a basic rule of thumb, the compiler will select the overload that requires the least automatic conversions, with the caveat that it will never apply 2 user-defined conversions.

Integer types get automatically cast around a lot. So if you have a function overloaded on an int and a double, the compile will pick the int function if called with a constant that is an integer. If you didn't have the int version, the compiler would select the double one. And among various integer types, the compiler prefers int for integer constants, because that is their type. If you overloaded on short and unsigned short, but called with a constant of 5, the compiler would complain that it couldn't figure out which overload to use.

Scott Meyers' book does indeed have the best explanation I have ever read.

"Scott Meyers' book". Could you be more specific by naming the title of the book? — Robᵩ, Nov 17 '11 at 23:10
For the life of me, I cannot find it now. +1 to anyone with the reference? — drdwilcox, Nov 17 '11 at 23:25
Scott Meyers' books: http://www.aristeia.com/books.html ; the reference is probably to either Effective C++ or possibly More Effective C++. — pnkfelix, Feb 20 '14 at 17:28

score 0 · Answer 2 · answered Nov 17 '11 at 20:06

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The entire function name is comprised of what you called the function as well as the parameter list. So, logically, 2 functions called the same thing but with different parameter lists both have different "full names." My terminology is probably a bit off, so if someone wants to correct that, feel free.

answered Nov 17 '11 at 20:06

MGZero

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say you have doSomething(int x) and doSomething(short x), which does the compiler choose when you call doSomething(5) ? – Tom Nov 17 '11 at 20:08
Ahh, that makes it more interesting! Good question, and also beyond the scope of my knowledge. I feel like it would depend on what the compiler considers 5 to be. I'm inclined to say int, but I'm not certain at all. – MGZero Nov 17 '11 at 20:09
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It will choose the `int` version. – drdwilcox Nov 17 '11 at 20:09

score 0 · Answer 3 · answered Nov 17 '11 at 20:07

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It is based on the type of the argument(s). No casting involved if the type does not match it simply will not compile.

answered Nov 17 '11 at 20:07

Miguel Garcia

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See my comment on MGZero's post – Tom Nov 17 '11 at 20:08
For what I know if you have void f(double x) and you call f(5) it comiles and 5 is casted to 5.0 – Belgi Nov 17 '11 at 20:08

Function overloading in C++

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