function parameter that is pointer to another function

Question

This is a more specific question based on a question I asked earlier...

If I have a function that takes two parameters (one required, one optional):

an STL container such as a vector
an "optional" comparison function that serves as a relational overload and returns the maximum value, whatever that is, in the vector...

Code:

template <typename Type>
Type FindMax (std::vector<Type> &myVec, int (*cmp)(Type one, Type two) = CallBack)
/.../

WHAT exactly, does the "int (*cmp)(Type one...)" say to the compiler? I want it to say, here's a function to use when comparing two of type Type...ie when using the relational operators <, >, =, etc. If no function is supplied by the user then use the default, otherwise, use what the user provides...

What exactly does the (*cmp)(Type one, Type two) say? Here's a pointer to a function that takes two parameters Type one and Type two? Is there any significance as to what comes after the *, ie could I write (*titsmagee)(Type one, Type two)? I'm assuming the naming convention is to help future readers?

For this to work with a "struct" does anything specific to the potential comparisons to be made need to be stored within said struct?

Thanks!

Luchian Grigore · Accepted Answer · 2012-03-12 14:06:48Z

5

int (*cmp)(Type one, Type two)

The parameter named cmp is a pointer to a function returning int that takes two parameters of type Type.

struct Foo
{
   int x;
};
int compare(Foo x, Foo y)
{
   return (x.x == y.x) ? 0 : (x.x > y.x ? 1 : -1);
}

std::vector<Foo> vec;

FindMax<Foo>(vec, &compare);

You need cmp so you can call the comparison function inside FindMax:

template <typename Type>
Type FindMax (std::vector<Type> &myVec, int (*cmp)(Type one, Type two) = CallBack)
{
    //whatever loop
    max = cmp(myVec[i],myVec[j]) >= 0 ? myVec[i] : myVec[j];
}

EDIT Breaking down the return:

return (x.x == y.x) ? 0 : (x.x > y.x ? 1 : -1);

?: is the ternary conditional operator.

condition ? expression1 : expression2

returns (loosely speaking) expression1 if condition is true, expression2 otherwise.

So what that means is:

if (x.x == y.x) 
   return 0; 
else 
   if (x.x > y.x) 
       return 1; 
   else 
       return -1;

It's what you expect the comparison function to do. Return 0 for equality, 1 if the first element is bigger than the second, and -1 for the inverse.

EDIT 2

struct Foo
{
   int x;
};
//Foo has a member x.
Foo f;
//Create a Foo object called f.
f.x;
//Access the member x of the object

edited Mar 12, 2012 at 14:06

answered Mar 12, 2012 at 12:43

Luchian Grigore

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3 Comments

MCP Over a year ago

that makes sense and I can see partly where I've gone wrong. I'm not familiar with the way you wrote a few things. Could you clarify or at least tell me the name of the syntax so that I can read about it? Particularly: "return(x.x == y.x)" How does the x.x, y.x work? What does the "? 0 :" mean?... Basically, if you could break down that "return" line that would be HUGE. Thanks!

MCP Over a year ago

Thanks for the edit. That helps a ton. Can someone explain what the "x.x and y.x" is?

MCP Over a year ago

damnit, can't believe I didn't see that. You're right, basic. It was that both the instance and the variable were the same name...it threw me. I was thinking, "why is the same variable getting called twice?!" Thanks!

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