I was asked by a friend about validity of following function prototypes,
void func1(int = 0, int*);
void func2(int = 1, int& = 2);
void func3(int*, int& = 3);
void func4(int& = 4, int* = 0);
void func5(int& = 0, int = 1);
void func6(int = 5, int& = 6, int* = 0);
I think the only prototype that is invalid is func1 because it does not have default parameter on the far right.
I am not sure if I am right, though.
Default parameters are places one the right far..
Next time post your code in [code]/*your code*.[/code]
Welcome to the forum
Code - functions and small libraries I use
It’s 2014 and I still use printf() for debugging.
"Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson
thank you for the feedback, I'll definitely post code using that next time.Originally Posted by std10093
Default parameters are places one the right far..
Next time post your code in [code]/*your code*.[/code]
Welcome to the forum
So I am right about func1 being invalid, but how about the rest?
You could have just pasted them into a compiler, and found the answer a few seconds later.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
The compiler shows following error,
error: default argument for ‘int& <anonymous>’ has type ‘int’|
Could you explain to me what does int& <anonymous> means?
I think that <anonymous> means the identifier for the parameter is missing.
This is a guess from a C programmer.
With identifier for the parametersCode:void func2(int = 1, int& = 2);
Tim S.Code:void func2(int param1 = 1, int& param2 = 2);
Last edited by stahta01; 03-11-2013 at 04:53 PM.
"...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson
@ stahta01
thanks I understand what <anonymous> means.
error: default argument for ‘int& a’ has type ‘int’|, this is the error I get when I change the parameter name to "a".
but I still am confused about why int & is not the same as int, for which I cannot assign an interal literal to a type int& variable.
References must be constructed from existing objects.
You could do something like:
Code:int globalparam = 3; void foo (int& param = globalparam) { std::cout << param << "\n"; } int main() { int x = 4; foo(x); foo(); }
Globalparam is guaranteed to exist before the function is invoked, so this is OK... on the other hand, globalparam is a global variable. It's scope has to be considered when you think about how reliable this function is.
Only if the reference were const would you be allowed to do:
void foo(const int& param = 3);
Since integer literals are immutable.
Last edited by whiteflags; 03-11-2013 at 08:36 PM.
Came C++11, the follow is also valid:
void foo (int&& param = 1337)
Since an rvalue reference allows you to bind temporaries (such as number literals) to references.
&& is called an rvalue reference if you want to look it up.
