Returning pointer to function from function

Hi,

I am learning function pointers and I found it little bit confusing.

I have wrote a program as shown below but it's not compiling.

Please suggest me where I am wrong.

Code:

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#include<stdio.h>

int add( int a, int b) { return a+b; }

(int(*)(int,int) display();

void show( int(*)(int,int) );

int main()
{
int(*fun)(int,int) = NULL;
int(*fun)(int,int) = display();
show( int(*fun)(int,int) );
}

int(*)(int)(int) display()
{
 return &add;
}

void show( int (*ptr)(int a,int b) )
{
int result = (*ptr)(56,63);

printf("%d",result) ;
}

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Please let me know where I am wrong ?

Thanks

Quote:

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Please let me know where I am wrong ?

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You have several syntax errors. Your compiler should have pointed out where they were.

What you want to do is this:

Code:

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int (*display())(int, int)
{
    return add;
}

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But honestly, it would be much simpler with a typedef:

Code:

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typedef int (*display_function)(int, int);

display_function display()
{
    return add;
}

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Thanks for help!
I have done it with typedef , but I want to do this by normal condition.
Can you please tell me the prototype of display() function ?
Compiler is giving error in prototype declaration...

Thanks

Quote:

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Can you please tell me the prototype of display() function ?

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I have shown you a possible implementation, so the prototype can be taken directly from that:

Code:

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int (*display())(int, int);

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Thanks , Its working ....
I have one question....What was the problem with this prototype I have declared previously ?

Quote:

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int(*)(int,int) display();

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Thanks

Quote:

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What was the problem with this prototype I have declared previously ?

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It is simply syntactically incorrect. Note that you still have other errors that need to be fixed.

I wrote that prototype since this,

Quote:

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void show( int(*)(int,int) ); //THis was working fine

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Was working fine.I concluded (previously) that if we received the parameter like this then we can return it in the same fashion.

Thanks

Function pointer, array, and array pointers types wrap the declaration around the object. So declaring a local looks like this. Declaring a local int *provided for comparison:

Code:

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int *intptr;
int(*funcptr)(int,int);
int array[10];
int (*arrayptr)[10];

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Typedefs are the same:

Code:

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typedef int *intptr_t;
typedef int(*funcptr_t)(int,int);
typedef int array_t[10];
typedef int (*arrayptr_t)[10];

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Function arguments are the same, except that the name can be omitted.
Casts are look like omitted names too (except you can't cast to an array). The identifier is not part of the cast, so the cast does not wrap around it:

Code:

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(int *)intptr;
(int(*)(int,int) ) funcptr;
(int (*)[10]) arrayptr;

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Return types follow the same pattern except that the "name" is the rest of the function declaration. This looks weird, but follows the pattern (again, you can't return arrays):

Code:

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int *func1(char param);
int(*func2(char param))(int,int);
int (*func3(char param))[10];

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