CODEWITHSUNDEEP
c++function-pointerstypedef
KasraM
KasraM

Reputation: 73

Function pointer syntax

I encountered two different codes using function pointers.

Please tell me which one is better? Is there any preference?

typedef int myFunction(int a,int b);
extern myFunction *ptfunction;

and

typedef int (*ptfunction)(int a,int b);

Upvotes: 1

Views: 327

Answers (5)

Yochai Timmer
Yochai Timmer

Reputation: 49251

Neither, use std::function

std::function<int (int ,int)> ptfunction;
ptfunction = f1;

Example (it's in std in c++0x):

how to use boost bind

std::function

Upvotes: 1

Mike Seymour
Mike Seymour

Reputation: 254471

The second is invalid; it would be equivalent to the first if you fix the first line to declare a variable, not a type:

int (*ptfunction)(int a, int b);

Which is "better" depends entirely on what you want to do with these things. If you need to refer to the same type many times, then the typedef will make things clearer; if you simply want to declare a single function pointer as in your examples, then it just adds extra noise.

(Presumably, in your first example, the second parameter should be int not int*, otherwise the function call wouldn't compile).

It's often more convenient and flexible to use function objects, and std::function in particular, since these can have arbitrary state bound to them. But that's beyond the scope of this question.

Upvotes: 1

Rango
Rango

Reputation: 1105

for the 1st code, you have to define ptfunction somewhere. and i don't think 2nd code can be compiled.

try this:

typedef int *PTFUNCTION(int, int);
PTFUNCTION func = f1;
func(1, 2);

Upvotes: 0

user877329
user877329

Reputation: 6200

The first one contains another error: You cannot cast int to a pointer to int. And which one is better? They are both completely out of context.

Upvotes: 0

unwind
unwind

Reputation: 399863

Both are bad, in that they are syntactically incorrect.

You need:

typedef int (*myFunction)(int a, int *b);

Upvotes: 5

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