CODEWITHSUNDEEP
cfunction
abc
abc

Reputation: 11929

function call with different semantics

Considering this code with 3 differents function call semantics:

void f(void){
   puts("OK");
}

int main(void){
   f();
  (*f)();
  (&f)();

  return 0;
}

The first is the standard way to call f,

the second is the semantic for dereferencing function pointers,

but in the third I'm applying the & operator to the function name and it seems to work fine.

What does in the second and third case happen?

Thanks.

Upvotes: 5

Views: 630

Answers (2)

Heisenbug
Heisenbug

Reputation: 39164

In the second case you are using a function pointer. Function pointers are used to memorize a reference to a function, and be able to call it elsewhere in your call. They are typically used to implement callbacks. So if you have store a pointer to a function you should use the first notation.

I thinkg the first and the third are equivalent. In fact if you declare a function pointer you can initialize it both the following ways:

void AFunction();
void (*funcPtr)() = NULL;

funcPtr = AFunction; 
funcPtr = &AFunction;

Upvotes: 1

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272517

Function calls are always performed via function pointers. From C99 section 6.5.2.2:

The expression that denotes the called function shall have type pointer to function.

However, in almost all cases a function type decays to a function-pointer type. From C99 section 6.3.2.1:

Except when it is the operand of the sizeof operator or the unary & operator, a function designator with type "function returning type" is converted to an expression that has type "pointer to function returning type".

So your three calls are evaluated thus:

(&f)();
(&(*(&f)))();
(&f)();

All are valid. But obviously, the first one (f()) is the cleanest and easiest to read.

Upvotes: 8

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