I am trying to create an array of function pointers, so that I can print out output from the functions by using the pointers. (For an exercise from the Effective C book.)
#include <stdio.h>
#include <stdlib.h>
int a(void) { return 1; }
int b(void) { return 2; }
int c(void) { return 3; }
int main(void)
{
int(*func_arr)[3] = {&a, &b, &c};
printf("%d\n", (*func_arr)[0]);
}
However, when I compile it, I get the warnings
starting.c:10:26: warning: incompatible pointer types initializing 'int (*)[3]' with an expression of type
'int (*)(void)' [-Wincompatible-pointer-types]
int(*func_arr)[3] = {&a, &b, &c};
^~
starting.c:10:30: warning: excess elements in scalar initializer
int(*func_arr)[3] = {&a, &b, &c};
^~
And, when I run the program, I get an output of -443987883, while it should just be 1.
Does anyone know the solution for this? Thank you!
This declaration
int(*func_arr)[3] = {&a, &b, &c};
declares a pointer to the array type int[3].
So there is declared a scalar object of a pointer type that you are trying to initialize using a brace enclosed list of initializers with more than one initializer of an incompatible pointer type.
Instead you need to write
int ( *func_arr[3] )( void ) = { a, b, c };
The function designators used as initializers are implicitly converted to pointers to the functions.
So there is no need to write as for example &a though such an expression is also valid to be used as an initializer.
To simplify the array declaration you could introduce a typeded name for the function pointer type.
For example
typedef int ( *FnPtr )( void );
Now the array declaration can look the following way
FnPtr func_arr[3] = { a, b, c };
And to output results of function calls you need to write calls of printf like
printf("%d\n", func_arr[0]() );