Mysterious default argument.

Hello,

I'm a beginner in C++, so even default argument gives me troubles! Heh!

Basically I have this little test program:

#include <iostream.h>

class defarg
{
    public:              
        char d;
        int i;
        float f;
   
        defarg(int iArg, char dArg = 'p', float fArg = 5.0)
             {
                 i = iArg;
                 d = dArg;
                 f = fArg;
             }          

};

int main()
{
    defarg test(5, 9.8);
    cout << test.i << " " << test.d << " " << test.f;
}

Now, before we begin, we all know that a default argument gets used only when we don't supply our own argument right? In other words, if we supply our own argument, then the default argument is disregarded, and if we don't supply, then the default argument gets used. Am I right here or what?

Now, my problem with the above program is, I want to supply my own i, use the default d (which is 'p'), and supply my own f.

In the main method I gave my own i (5), supplied nothing for d, and gave my own f (9.8), and so I expect the program to output 5 p 9.8.
However, that's not the case. It compiles fine, but its output is either 5 [blank] 9.8 or sometimes 5 [a weird looking character] 9.8.

Can anyone help me get the output I want? Or is it impossible?

Thanks.
   
zpivatAsked:
Who is Participating?
 
snehanshuConnect With a Mentor Commented:
zpivat,
  You have 5 open questions. It is a good practice to accept answers.
...Shu
0
 
AlexFMCommented:
defarg test(5, 9.8);

Result of this call is:

iArg = 5
dArg = (char) 9.8;   // 9.8 casted to char = 9
fArg = 5.0;              // default

Weird looking character in output is dArg (9).

It's impossible in C++ to skip parameter in the middle of the list.
0
 
snehanshuCommented:
zpivat,
  You can only skip the last "n" arguments: You cannot skip an argument in between.
  So, if you have a function

int myfunc(int a, char b='1', char c='2', int d = 3);

then you can only call it in one of the following ways only:
myfunc(1)
myfunc(1, 'a')// -> b = 'a', others default
myfunc(1, 'a', 'b') // -> b = 'a', c = 'b', others default
myfunc(1, 'a', 'b', 0)// -> b = 'a', c = 'b', no default

Does that clarify?

And to get your output, you can change the function to
 defarg(int iArg, float fArg = 5.0, char dArg = 'p')

...Snehanshu
0
 
zpivatAuthor Commented:
Hi AlexFM and Snehanshu,

I'm clear now. So I was right; it's impossible afterall to skip the middle argument(s)!

Thank you!
0
 
an_menCommented:
When using default arguments, only the trailing arguments may be defaulted.
That is, you cannot have a default argument followed by a nondefault argument.
Also once you start using default arguments, all the remaining arguments must be defaulted.
0
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