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Feb 14th, 2006

Is this notation possible?

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Hello ladies and gents,

There's an exercise, well, more a question that goes like this in Accelerated C++:

If the following code is legal, what can we infer about the return type of f?

C++ Syntax (Toggle Plain Text)
double d = f() [n];
double d = f() [n];

What I can gather is that the return type will be a double variable, but, I don't know wether this way of writing is actually correct?

Ive tried to use it, but failed in getting anything to work with this kind of function.

Could someone tell me wether this is a legal piece of code.

Thanks.

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JoBe

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Feb 14th, 2006

Re: Is this notation possible?

Quote originally posted by JoBe ...

If the following code is legal, what can we infer about the return type of f?

That it returns a pointer to a double:

C++ Syntax (Toggle Plain Text)
double *f()
double *f()

Possible usage:

C++ Syntax (Toggle Plain Text)
#include <iostream>
 
double *f()
{
    static double value[] = {1,2,3,4,5};
    return value;
}
 
int main ( void )
{
    int n = 2;
    double d = f() [n];
    std::cout << "d = " << d << '\n';
    return 0;
}
 
/* my output
d = 3
*/
#include <iostream>

double *f()
{
    static double value[] = {1,2,3,4,5};
    return value;
}

int main ( void )
{
    int n = 2;
    double d = f() [n];
    std::cout << "d = " << d << '\n';
    return 0;
}

/* my output
d = 3
*/

Dave Sinkula

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Feb 14th, 2006

Re: Is this notation possible?

Quote originally posted by Dave Sinkula ...

That it returns a pointer to a double...

Darn, should have seen it, because of f() [n]

It couldn't have been a double.

Quote ...

C++ Syntax (Toggle Plain Text)
double *f()
double *f()
Possible usage:

Thanks for the example Dave. :!:

JoBe

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