function whith default parameters

[Rooting]

hi mans!!..

I have a problem, and "googling" ddidn't find nothing..
myu problem is that i have a function such

codeguru(int i, int y, int z)
{
     cout << i, y, z;
}

now the function is right!!..
but i don't wont to pass the y variable,
so i can do :

codeguru(int i, int y=3 , int z)
{
     cout << i, y, z;
}

but when i call it, whith

codeguru(2, ,4);

isn't write..
why is wrong??..
how can i make it better?

[laitinen]

The default parameters need to be the right-most parameters in the function.

void codeguru(int i, int z, int y=3)
{...}

Laitinen

[GNiewerth]

You can only provide default parameters from right to left, you cannot omit the center ones. When calling a function with default parameters you simply omit the parameters you don´t want to specify, you cannot use placeholders. In your example you have to specify both y and z, because z is the rightmost parameter and it has no default value, so you must specify z. Once you specify z you must specify y, too, because y is left of z.
You can reorder your parameters and move the ones which will most likely use default values to the right.

[Rooting]

i need to assign a temporaney variable, becouse when i use this function
i'm free to ommit the variable...

i use how you told me..

#include <cstdlib>
#include <iostream>

using namespace std;
void function(int x, int y=3) 
{
        cout << x << endl << y;  
}
int main(int argc, char *argv[])
{
    function(2,);
    system("PAUSE");
    return EXIT_SUCCESS;
}

but there are errors in this line:

 function(2,);

this is an error becouse i don't pass the value of variable,
but in function i asssign it!!...
if i don't wont to assign when i call, how i can do??..

[laserlight]

The function call should be: function(2);

[Graham]

cout << i, y, z;

It's entirely possible that that line does not do what you think it does.

[Rooting]

It's entirely possible that that line does not do what you think it does.

i try not this source but this one:

#include <cstdlib>
#include <iostream>

using namespace std;
void function(int x, int y=3) 
{
        cout << x << endl << y;  
}
int main(int argc, char *argv[])
{
    function(2,);
    system("PAUSE");
    return EXIT_SUCCESS;
}

in this source is right!!..
but is wrong when i call the function..

[Lindley]

The function call should be: function(2);

Repeated.

[Rooting]

mmm...ok..
when i have 2 variable now is ok..
but, in my case i have 3 variable as:

#include <cstdlib>
#include <iostream>

using namespace std;
void function(int x=1, int y=3, int z = 4) 
{
        cout << x << endl << y << endl << z;  
}
int main(int argc, char *argv[])
{
    function(,5,);
    system("PAUSE");
    return EXIT_SUCCESS;
}

in this case i have 3 variable, and i wold like to specify the y one..
the y is in the midle..
the expression

function(,5,) is wrong!.
how can i resolve??..

[Lindley]

C++ doesn't support that. It can only supply default values to those parameters on the right of the ones you specify.

And you need to stop trying to use commas that don't separate anything. As far as I know that's always a syntax error.

[laserlight]

And you need to stop trying to use commas that don't separate anything. As far as I know that's always a syntax error.

There is at least one exception where it is allowed, namely, when it comes after an initialiser list, e.g., {1, 2, 3,}. That does not apply here, though.

[Graham]

Why not just overload the functions?

void func(int x, int y, int z)
{
	cout << x << y << z;
}

void func(int y)
{
	func(1, y, 2);
}
int main()
{
	func(3);
}

[Rooting]

mmm...
i'm making a library, so I don't know if the user can use the x or y or z variable...
understand me??..
so if the user don't specify the x variable, but put the y and z , the library must go!!..
it's the function of class..
overloading as graham sad is very simple, but not elegant, becouse i have 15 function in
this class..and i don't know if he specify the x or y or z one..
any suggestion??

[code_carnage]

Perahps variadic function might help.. Do google to find out the implementation of variadic function...

[laserlight]

I think that neither function overloading nor variadic functions will cut it here. What is needed is named arguments, which is not a feature of C++. The reason is that there is nothing to differentiate foo(x, y) from a call to foo(x, z) or foo(y, z).

I am not sure if this really is a good idea, but I would propose a workaround with std::map and some function templates:

#include <iostream>
#include <map>

template<char c1>
std::map<char, int> make_values(int v1)
{
    std::map<char, int> ret;
    ret[c1] = v1;
    return ret;
}

template<char c1, char c2>
std::map<char, int> make_values(int v1, int v2)
{
    std::map<char, int> ret;
    ret[c1] = v1;
    ret[c2] = v2;
    return ret;
}

template<char c1, char c2, char c3>
std::map<char, int> make_values(int v1, int v2, int v3)
{
    std::map<char, int> ret;
    ret[c1] = v1;
    ret[c2] = v2;
    ret[c3] = v3;
    return ret;
}

void foo(const std::map<char, int>& values)
{
    std::map<char, int>::const_iterator iter = values.find('x');
    int x = (iter == values.end()) ? 0 : iter->second;
    iter = values.find('y');
    int y = (iter == values.end()) ? 0 : iter->second;
    iter = values.find('z');
    int z = (iter == values.end()) ? 0 : iter->second;

    std::cout << x << ' ' << y << ' ' << z << std::endl;
}

int main()
{
    foo(make_values<'x'>(1));
    foo(make_values<'z', 'y'>(10, 20));
    foo(make_values<'y', 'x', 'z'>(2, 3, 4));
}