1

The code below shows the reference to function pointer:

typedef int (*t_somefunc)(int,int);
int product(int, int);
int main(void) {
  t_somefunc afunc = &product; // "product" works fine without "&"

What do we use "&" for referencing function pointer?
See full working code:

#include <stdio.h>
typedef int (*t_somefunc)(int,int);
int product(int, int);

int main(void) {
  t_somefunc afunc = &product; // product without & works also
  int x2 = (*afunc)(123, 456); // call product() to calculate 123*456
  printf("x2 value is %d\n", x2);
  return 1;
}

int product(int u, int v) {
  return u*v;
}
chux - Reinstate Monica
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amed svensson
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  • And what is the question? – Vlad from Moscow Sep 20 '22 at 20:54
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    I've _never_ used the `&` [in 40+ years]. IMO, it's clutter. And we get consistency with `t_somefunc afunc = product; t_somefunc bfunc = afunc;` When setting `afunc` we shouldn't have to whether `product` is the a function address or another pointer. – Craig Estey Sep 20 '22 at 23:06
  • Similar question to https://stackoverflow.com/questions/73612682/referring-to-a-c-function-pointer-with-or-without/73613627#73613627 – KamilCuk Sep 20 '22 at 23:32

2 Answers2

4

Functions are automatically converted to pointers for programmer convenience.

When a function is used in an expression, it is automatically converted to a pointer to the function, unless it is the operand of sizeof or unary &. C 2018 6.3.2.1 4 says:

A function designator is an expression that has function type. 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”.

(In fact, if you attempt to convert the pointer back to a function by applying *, the automatic conversion will happen again. You can write t_somefunc afunc = *******************product; and still end up with the address of product.)

In declarations of function parameters, a function will be automatically adjusted to be a pointer to a function. C 2018 6.7.6.3 8 says:

A declaration of a parameter as “function returning type” shall be adjusted to “pointer to function returning type”, as in 6.3.2.1.

(This is called an adjustment because there is no value being converted; the declaration is changed to declare a pointer to a function instead of a function.)

Eric Postpischil
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    It may be worth to add a note that function parameters are also transformed. For example `void foo( void f(int))` is a valid declaration of function taking a function pointer argument. It may help make declarations prettier without resorting to `typedef`s – tstanisl Sep 20 '22 at 22:18
0

Nothing, just use the name of the function (without the () list of parameters) e.g.

double(*f_to_cal)(double);

f_to_cal = sin;  /* trigonometric sine function */

return f_to_cal(angle_deg * M_PI / 180.0);
Luis Colorado
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