Why is a prototype required even without a class declaration?

Question

If I just do this: Ex1:

#include <iostream> int main() { //try to call doSomething function doSomething(); } void doSomething() { std::cout << "Call me now!" << std::endl; } 

I get a compilation error! Since the compiler does not know what "doSomething" is.

But if I changed the position of doSomething to come in first place, the program successfully compiles. Ex2:

 #include <iostream> void doSomething() { std::cout << "Call me now!" << std::endl; } int main() { //try to call doSomething function doSomething(); } 

I can declare a prototype as follows: Ex3:

 #include <iostream> void doSomething(void); int main() { //try to call doSomething function doSomething(); } void doSomething() { std::cout << "Call me now!" << std::endl; } 

But why does the first example not work? Why do I even need to first declare a prototype or call functions and a main function?

Thank!

Answer 1

You cannot call a function without a compiler, seeing either a definition or a declaration, at first simple. Before calling, a prototype or actual definition should appear.

Answer 2

Since the compiler did not see doSomething before using it.

Either you must prototype it, or define it first so that the compiler knows how to analyze its use.

Answer 3

This CC legacy is a one-pass language, which means that it should do everything just by reading the file once. To be able to call a function without a direct declaration / prototype, you need to read the file twice; First time find all function signatures and second time to compile.

C ++ retained this requirement for functions that were part of C, such as free functions and global variables. However, classes are new to C ++, and there was no need to keep the old way of doing something. Thus, as part of the definition of a single class, multi-pass compilation is used. This is why you can do this:

 class MyClass { void foo() {bar();} void bar() {} }; 

But you cannot do what you indicated in your question.