Function as a parameter

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Here, what is the meaning of this?

Can we pass the function as a parameter to the other function ?

like this---

int func2(parameter3,parameter4)
{
// code
}

void func1(parameter1,parameter2,func2) // why it is not int func2( )?
{
// code
}

Here Are we calling fun2( ), then why there are no brackets and actual parameters to func2( )?

What I have tried:

How could we pass like that ?
.........

Posted 26-Jun-21 19:23pm

_-_-_-me

Updated 26-Jun-21 22:15pm

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KarstenK · Answer 1 · 2021-06-26T22:15:00

Solution 2

it is a common pattern in the Apple world to work with such constructs. Such functions are called completion handler. The main reason is improve usage of threads and use the function result of lengthly operations like downloads. Read this article for details.

Posted 26-Jun-21 22:15pm

KarstenK

OriginalGriff · Answer 2 · 2021-06-26T19:32:00

C

int foo ( int (*f)(int) )
   {
   return f(666);
   }

This means that foo is a function returning an integer that takes a function f as a parameter. f will be a pointer to a function which returns an integer and which takes a single int parameter.

When calling a function with a function parameter, the value passed must be a pointer to a function. Just use the function's name (without parentheses) for this:

C

int myFunct(int x)
   {
   return x * 2;
   }
...
printf("%u\n", foo(myFunct));