C Function Pointer Mishap - c

Question

Okay so I'm trying to learn function pointers. I have a basic function pointer setup like so.
Function to print out linked list:
void seq_print(Seq seq, void (* print_func)(void *)){
Node * p = seq->top;
if(p == NULL){
printf("%s %c", "There is no data to print.", '\n');
return;
}
while(p != NULL){
print_func(p->data);
p = p->next;
}
}
Testing the function:
seq_print(s, printFunc(1));
I get this error:
seq.h:113:32: error: expected declaration specifiers or ‘...’ before ‘(’ token
extern void seq_print(Seq seq, (void *) print_func(void *));
I'm really not sure what to do, any insight would be helpful.

Answer 1

You have two mistakes:
First, notice declaration in error message: in your header file seq.h, declaration of function is wrong!
extern void seq_print(Seq seq, (void *) print_func(void *));
// ^ ^ wrong = parenthesis return type
it should be:
extern void seq_print(Seq seq, void (*print_func) (void *));
// ^ correct ^ = parenthesis function name
Second, at calling place.
seq_print(s, printFunc(1));
// ^^^ you are calling function, and passes returned value
should be:
seq_print(s, printFunc);
// ^^^^^^^^^^ don't call pass function address
My following code examples will help you to understand better (read comments):
#include<stdio.h>
void my_g2(int i, (void*) f(int)); // Mistake: Notice () around void*
void f(int i){
printf("In f() i = %d\n", i);
}
int main(){
my_g2(10, f(1)); // wrong calling
return 0;
}
void my_g2(int i, void (*f)(int)){
printf("In g()\n");
f(i);
}
Check codepad for working code. You can see error is similar to what you are getting:
Line 2: error: expected declaration specifiers or '...' before '(' token
In function 'main':
Line 8: error: too many arguments to function 'my_g2'
Now correct version of this code:
#include<stdio.h>
void my_g2(int i, void (*f)(int)); // Corrected declaration
void f(int i){
printf("In f() i = %d\n", i);
}
int main(){
my_g2(10, f); // corrected calling too
return 0;
}
void my_g2(int i, void (*f) (int)){
printf("In g()\n");
f(i);
}
Now check codepade for output:
In g()
In f() i = 10
Edit: Adding on the basis of comment.
But what if it's like void (*f) (void *) how do I pass in values to that?
From calling function in main() (in my example = my_g2) you need to pass function pointer which you wants call (in my example f()) from function you calls in main (that is my_g2).
You wanted to call f() from my_g2()
We always pass parameters to function at the time of function calling. So if you wants to pass parameters to f() function you have to pass when you call this in my_g2().
A calling expression like below (read comments):
seq_print(s, printFunc(1));
^ // first printFunc(1) will be called then seq_prints
pass returned value from printFunc(1)
is wrong because if you do so seq_print will be called with second paramter value = returned value from function printFunc(1).
To pass void pointer, my following code may help you further:
#include<stdio.h>
void my_g2(void* i, void (*f)(void*));
void f(void *i){
printf("In f(), i = %d\n", *(int*)i);
*(int*)i = 20;
}
int main(){
int i = 10;
my_g2(&i, f);
printf("Im main, i = %d", i);
return 0;
}
void my_g2(void* i, void (*f)(void*)){
printf("In g()\n");
f(i);
}
Output #codepade:
In g()
In f(), i = 10
Im main, i = 20

Answer 2

There's a typo in the forward declaration quoted in the error message. It needs to match the code snippet you posted, with a parameter declaration of void (* print_func)(void *).

Answer 3

Looks like your header has a typo. The declaration should be
extern void seq_print(Seq seq, void (*print_func)(void *));
(void *)print_func(void *) is not a valid function pointer declaration. To declare a function print_func that accepts a void pointer and does not return a value, use void (*print_func)(void *)
EDIT: omitting the parens around (*print_func) does create a function pointer but for a function returning a pointer