I’ve read this from cppreference.com:

Function-to-pointer conversion:
An lvalue of function type T can be implicitly converted to a prvalue pointer to that function. This does not apply to non-static member functions because lvalues that refer to non-static member functions do not exist.

Source: https://en.cppreference.com/w/cpp/language/implicit_conversion

What does ‘function type T’ here mean? does that mean a function with a return type T?

>Solution :

In programming languages like C++, a function type describes the signature of a function. Specifically, it tells you what type of arguments the function takes and what type of value it returns, i.e. it is signature of that function.

The T serves to just give a name to that function type.

Think of

There always exist real number X that is larger than any given real number.

X is that existing real number.

For your specific example:

In C++, an lvalue of function type can indeed be implicitly converted to a prvalue (pure rvalue) pointer to that function. Here’s a simple example to demonstrate this concept.

#include <iostream>

void foo(int x) {
  std::cout << "Value of x: " << x << std::endl;
}

int main() {
  // Both of these are valid:
  void (*funcPtr1)(int) = &foo;  // Explicitly using the address-of operator
  void (*funcPtr2)(int) = foo;   // Implicit conversion to pointer - What yours example is about

  // Calling the function using the function pointers
  funcPtr1(42);  // Output: "Value of x: 42"
  funcPtr2(43);  // Output: "Value of x: 43"

  return 0;
}