Why doesn't this code print the letter a?
#include <iostream>
#include <stack>
void a()
{
std::cout<<"a";
}
int main ()
{
void a();
return 0;
}
You're accidentally declaring a function inside main() instead of calling it.
int main ()
{
void a(); // <-- DECLARES a function, does not call it
return 0;
}
Here is the fix:
int main ()
{
a();
return 0;
}
Also note that you probably want a newline:
void a()
{
std::cout<<"a\n";
}
Or you can use std::endl, if you like typing.
You're declaring your function twice:
#include <iostream>
#include <stack>
void a()
{
std::cout<<"a";
}
int main ()
{
void a(); // this is a declaration
return 0;
}
Do this instead:
int main ()
{
a(); // this is a function call, which will execute your function
return 0;
}
Its ok qwertyu uytrewq, everyone have these kind of problems in start, the main thing is never hesitate to question.
The main error in your code is, that you are declaring the function but not calling it. There are three phases of functions.
Declaration i.e
void a();
Definition i.e
void a ()
{
std::cout << "a";
}
Calling a Function i.e
a();
Now the major error in your program belongs to the 3rd phase (Calling a Function) you are mentioning the function type also which is know as Deceleration, so the correct code is as following.
#include <iostream>
#include <stack>
void a()
{
std::cout<<"a";
}
int main ()
{
a();
return 0;
}
In your main function:
int main()
{
// Don't add Void
a();
return 0;
}
Related
I have a request for function pointer by C++. below is the sample what I need:
in API file:
class MyClass {
public:
void function1();
void function2();
void function3();
void function4();
};
in main file:
MyClass globalglass;
void global_function_call(???)// <---- how to do declaration of argument???
{
//Do *function
}
int main()
{
global_function_call(&globalglass.function1()); // <---- pseudocode, I need to use global class
global_function_call(&globalglass.function2());
global_function_call(&globalglass.function3());
global_function_call(&globalglass.function4());
return 1;
}
I have no idea to do declaration...
To do what you are asking for, you can use a pointer-to-member-method, eg:
MyClass globalglass;
void global_function_call(void (MyClass::*method)())
{
(globalglass.*method)();
}
int main()
{
global_function_call(&MyClass::function1);
global_function_call(&MyClass::function2);
global_function_call(&MyClass::function3);
global_function_call(&MyClass::function4);
return 1;
}
Online Demo
The following code compiles without any warning or error:
#include <iostream>
using namespace std;
class demo_class
{
int x;
float y;
public:
void fun(void);
};
void fun2(void)
{
cout<<"i am fun2\n";
}
void demo_class::fun(void)
{
cout<<"i am fun\n";
cout<<"i can call fun2\n";
fun2();
}
int main()
{
demo_class ob1;
ob1.fun();
return 0;
}
I am not understanding that as the scope of fun function is only in demo_class
then how can it call fun2 function, should not it show error as the access of fun function only within the demo_class?
Name lookup would try to examine all the possible scopes, until it finds at least one at any scope, then the name lookup stops.
In this case, the name fun2 can't be found at class scope, then the further scope, i.e. globle scope is examined and ::fun2 is found.
There is no reason to disallow calling a free function from within a member function. If this was the case classes would be rather useless (they would be a way to prevent code reuse instead of supporting it).
As mentioned in a comment, in cout<<"i am fun2\n"; you are calling a non-member function and calling fun is not much different from that.
Further, with a grain of salt your example is not much different from
#include <iostream>
using namespace std;
class demo_class
{
};
void fun2(void)
{
cout<<"i am fun2\n";
}
void fun3(demo_class& dc)
{
cout<<"i am fun\n";
cout<<"i can call fun2\n";
fun2();
}
int main()
{
demo_class ob1;
fun3(ob1);
return 0;
}
A member function can always be transformed into a free function. If fun would access private members we would have to declare it as friend to make the above work, but otherwise no problem here.
You can also do the reverse and call member functions in free functions as in
struct foo {
void bar(){}
};
void func(foo& f) {
f.bar();
}
int main() {
foo f;
func(f);
}
This is my A.h file
class A
{
public:
void menuChoice();
void displaystartingMenu(); //EDIT
};
This is my A.cpp file
#include "A.h"
void displaystartingMenu()
{
cout<<"Please enter your choice:";
}
void A::menuChoice()
{
displaystartingMenu();
cout<<"Hello"<<endl;
}
int main()
{
A a;
a.menuChoice();
}
i tried to put
void menuChoice();
on the top of my cpp file but it still won't compile . it gives me error
In function ‘int main()’: A.cpp:112:13: error: ‘menuChoice’ was not declared in this scope menuChoice();
How do I compile : g++ A.cpp A.h
By right I don't need to even declare the function of top of the cpp because I have already declare it in my header file and I have included in my .cpp. What went wrong here?
EDIT:
Error :
In function `A::menuChoice()':
A.cpp:(.text+0x229): undefined reference to `A::displaystartingMenu()'
Option 1: Make an A instance and call that instance's menuChoice method:
#include <iostream>
class A {
public:
void menuChoice();
};
void A::menuChoice() {
std::cout << "Hello" << std::endl;
}
int main() {
A a;
a.menuChoice();
return 0;
}
Option 2: Make menuChoice a static method and call it as A::menuChoice:
#include <iostream>
class A {
public:
static void menuChoice();
};
void A::menuChoice() {
std::cout << "Hello" << std::endl;
}
int main() {
A::menuChoice();
return 0;
}
Edit: Addressing the new problem, when you tried to define A::displaystartingMenu you wrote:
void displaystartingMenu() {
// ...
}
It must be defined like this:
void A::displaystartingMenu() {
// ...
}
menuChoice is a non static member function of class A. In order to call it you need to instantiate an object A, as follows
int main()
{
A a;
a.menuChoice();
return 0;
}
Before I present the code which is found at the bottom of this post I would like to talk about the issue and the fix's that I do not desire. Okay basically I've created a GUI from scratch sort of and one requirement I wanted for this was allow components to have their own click executions so if i click a button or tab etc.. It would call Component->Execute(); Well normally you would do something like a switch statement of ids and if that components ID equaled n number then it would perform this action. Well that seemed kinda dumb to me and I thought there has to be a better way. I eventually tried to incorporate a feature in JAVA where you would do like Component.AddActionListener(new ActionListener( public void execute(ActionEvent ae) { })); or something like that and I thought that this feature has to be possible in C++. I eventually came across storing void functions into a variable in which could be executed at any time and modified at any time. However I hadn't noticed an issue and that was this only worked with static functions. So below you'll see my problem. I've patched the problem by using a pointer to SomeClass however this would mean having an individual function call for every class type is there no way to store a function callback to a non-static class member without doing the below strategy? and instead doing a strategy like the commented out code?
//Main.cpp
#include <iostream> //system requires this.
#include "SomeClass.h"
void DoSomething1(void)
{
std::cout << "We Called Static DoSomething1\n";
}
void DoSomething2(void)
{
std::cout << "We Called Static DoSomething2\n";
}
int main()
{
void (*function_call2)(SomeClass*);
void (*function_call)() = DoSomething1; //This works No Problems!
function_call(); //Will Call the DoSomething1(void);
function_call = DoSomething2; //This works No Problems!
function_call(); //Will Call the DoSomething2(void);
SomeClass *some = new SomeClass(); //Create a SomeClass pointer;
function_call = SomeClass::DoSomething3; //Static SomeClass::DoSomething3();
function_call(); //Will Call the SomeClass::DoSomething3(void);
//function_call = some->DoSomething4; //Non-Static SomeClass::DoSomething4 gives an error.
//function_call(); //Not used because of error above.
function_call2 = SomeClass::DoSomething5; //Store the SomeClass::DoSomething(SomeClass* some);
function_call2(some); //Call out SomeClass::DoSomething5 which calls on SomeClass::DoSomething4's non static member.
system("pause");
return 0;
}
//SomeClass.hpp
#pragma once
#include <iostream>
class SomeClass
{
public:
SomeClass();
~SomeClass();
public:
static void DoSomething3(void);
void DoSomething4(void);
static void DoSomething5(SomeClass* some);
};
//SomeClass.cpp
#include "SomeClass.h"
SomeClass::SomeClass(void)
{
}
SomeClass::~SomeClass(void)
{
}
void SomeClass::DoSomething3(void)
{
std::cout << "We Called Static DoSomething3\n";
}
void SomeClass::DoSomething4(void)
{
std::cout << "We Called Non-Static DoSomething4\n";
}
void SomeClass::DoSomething5(SomeClass *some)
{
some->DoSomething4();
}
Secondary Fix for what I'll do not an exact answer I wanted but it meets my needs for now along with allowing additional features which would have become overly complicate had this not existed.
//Component.hpp
#pragma once
#include <iostream>
#include <windows.h>
#include <d3dx9.h>
#include <d3d9.h>
#include "Constants.hpp"
#include "ScreenState.hpp"
#include "ComponentType.hpp"
using namespace std;
class Component
{
static void EMPTY(void) { }
static void EMPTY(int i) { }
public:
Component(void)
{
callback = EMPTY;
callback2 = EMPTY;
callback_id = -1;
}
Component* SetFunction(void (*callback)())
{
this->callback = callback;
return this;
}
Component* SetFunction(void (*callback2)(int), int id)
{
this->callback_id = id;
this->callback2 = callback2;
return this;
}
void execute(void)
{
callback();
callback2(callback_id);
}
}
The syntax for pointers-to-member-functions is as follows:
struct Foo
{
void bar(int, int);
void zip(int, int);
};
Foo x;
void (Foo::*p)(int, int) = &Foo::bar; // pointer
(x.*p)(1, 2); // invocation
p = &Foo::zip;
(x.*p)(3, 4); // invocation
Mind the additional parentheses in the function invocation, which is needed to get the correct operator precedence. The member-dereference operator is .* (and there's also ->* from an instance pointer).
I am trying to call the function hello which belongs to the class program1
#include <iostream>
using namespace std;
class program1
{
program1();
~program1();
hello();
}
program1::hello()
{
cout<<"hello";
}
int main()
{
program1.hello(); //Call it like a normal function...
cin.get();
}
Names inside a class are private by default.
class program1 {
public:
program1();
~program1();
void hello() ;
};
// ...
int main(int, char **) {
program1 myProgram;
myProgram.hello();
return 0;
}
Alternatively, you can invoke a method on a temporary:
int main(int, char **) {
program1().hello();
return 0;
}
but that's probably for later in the semester.
you forgot to create an object:
int main()
{
program1 p1;
p1.hello();
}
Class definition should end with ;
Secondly, you need to instantiate class to call members on it. ( i.e., creation of an object for the class )
In C++, methods should have return type.
program1::hello(); // method should have a return type.
class members and methods are private by default, which means you cannot access them outside the class-scope.
So, the class definition should be -
class program1
{
public: // Correction 1
program1();
~program1();
void hello(); // Correction 2
};
void program1::hello() // Need to place return type here too.
{
cout<<"hello";
}
Now on creation of object for class program1, it's method hello() can be called on it.
This version is edited. (make sure you include all the body of the methods)
#include <iostream>
using namespace std;
class program1
{
public: // To allow outer access
program1();
~program1();
void hello(); // The void
}; // Don't miss this semicolon
void program1::hello() // The void
{
cout<<"hello";
}
int main()
{
program1 prog; // Declare an obj
prog.hello(); //Call it like a normal function...
cin.get();
}
I noticed that you left out return type for your hello() function.
If you want to call hello() as a member function, then as suggested you should create an object to it.
program1 prog;
prog.hello();
If you want to call it without an object, the you should use static function.
class program1
{
public: // To allow outer access
program1();
~program1();
static void hello(); // The void
}
then you can call it this way:
program1::hello();
Therefore the working code should be this way:
#include <iostream>
using namespace std;
class program1 {
public:
void hello();
}; // Don't miss this semicolon
class program2 {
public:
void static hello(); // to demonstrate static function
}; // Don't miss this semicolon
void program1::hello() {
cout << "Hello, I'm program1." << endl;
}
void program2::hello() {
cout << "Hello, I'm program2." << endl;
}
int main(void) {
program1 prog1;
prog1.hello(); // Non-static function requires object
program2::hello(); // Static function doesn't
return 0; // Return 0
}