variable function parameters functionname(int a, ...)

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Views: 2474 | Replies: 1

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Join Date: Aug 2004

Posts: 138

Reputation: chound is an unknown quantity at this point

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Solved Threads: 1

chound

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Junior Poster

variable function parameters functionname(int a, ...)

Aug 28th, 2004

How do I create a function with variable parameters. Will "functionname(float a, ...);" work? If it does how do I use the variables.
Eg.

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void f1(int i, ...);
void main()
{
int a, b,c;
cout<<"Enter 3 values";
cin>>a>>b>>c;
f1(a, b, c);
}
void fi()
{
//How do I access the variables that are passed as parameters
}

:cry: :lol: :rolleyes:

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Join Date: Jun 2004

Location: Marin, CA, USA

Posts: 434

Reputation: Chainsaw is an unknown quantity at this point

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Chainsaw

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Unprevaricator

Re: variable function parameters functionname(int a, ...)

Aug 28th, 2004

You need at least one real parameter, as you have in your example. Then you use some special macros to access the parameters:

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void AddSomeNumbers( int howMany, .... ) // any number of 'real' params, folowed by ...
{
    va_list  argptr;
    int  ret = 0;

    va_start( argptr, howMany ); // use the last 'real' param here

    for (int i = 0; i < howMany; i++)
        ret += va_arg( argptr, int );  // the next arg is an int, we hope

    va_end(argptr);

    return ret;
}

The caviats are:
1) You do not know how many parameters were supplied. Thats why in my sample I had that be the one 'real' parameter.
2) You do not know the type of parameters, so you have to 'guess' and hope the caller knows what they are doing!

printf() and its variants use this, and the format string specifies the number and type of parameters:

printf( "%s %d %x\n", "string", number, number );

but if you screw up you get fun results!

printf( "%s %d %d\n", number, "string", 1.2 ); // ick!