How to call a function from a function pointer? [duplicate] - c++

Question

This question already has answers here:
Function pointer as parameter
(3 answers)
Closed 6 months ago.
How can you call a function from only a pointer to the function? For example:
void print()
{
std::cout << "Hello World!" << std::endl;;
}
void run_func(void* func)
{
func(); // what im trying to do (doesnt actually work)
}
int main()
{
run_func(print);
}
Expected output:
Hello World!
It's a bit like how std::thread creates a thread from the pointer of a variable.

Answer 1

Function pointers in function parameter lists need to be wrapped in parentheses. Change the one line, either by wrapping with parentheses or using the typedef'ed function parameter, and your sample works.
// Simplifies arcane function parameter synax.
typedef void (*FUNC_TO_RUN)();
void print()
{
std::cout << "Hello World!" << std::endl;;
}
// Change your parameter as follows. You need to wrap function
// pointers in parantheses.Or, use a typedef.
// void run_func(void *func())
// void run_func(void (*func)()) // Works, but harder to read
void run_func(FUNC_TO_RUN func) // Works, easier to read
{
func(); // what im trying to do (and now works)
}
int main()
{
run_func(print);
}

Answer 2

You have to change your parameter type of run_func to void (*func) (), which means func is a pointer to a void function that takes no parameters.
Corrected code:
void print() {
std::cout << "Hello World!\n";
}
void run_func(void (*func)()) {
func();
// or you can call the print function by (*func)();
}
int main() {
run_func(print);
// prints "Hello World!"
}

Answer 3

You could use std::function:
void print()
{
std::cout << "Hello World!" << std::endl;;
}
void run_func(std::function<void()> f) // f returns void and takes no parameters
{
f();
}
int main()
{
run_func(print);
}