I'm trying to add 2 to a class variable using a function, but it gives me this undefined reference to addTwo(int) even though I already have it declared.
#include <stdio.h>
#include <iostream>
using namespace std;
class Test {
public:
int addTwo(int test);
int test = 1;
};
int addTwo(int test);
int main() {
Test test;
cout << test.test << "\n";
addTwo(test.test);
cout << test.test;
}
int Test::addTwo(int test) {
test = test + 2;
return test;
}
The defined member function int Test::addTwo(int test) do differ from the declared global function int addTwo(int test);, which the compiler searches for.
To eliminate the error, define the global function or change the call of the global function to call of the member function.
In order to "add 2 to a class variable using a function", you should stop shadowing the member variable by the argument. (You can use this->test for using member variable, but this won't be needed in this case)
Try this:
#include <iostream>
using namespace std;
class Test {
public:
int addTwo();
int test = 1;
};
int main() {
Test test;
cout << test.test << "\n";
test.addTwo();
cout << test.test;
}
int Test::addTwo() {
test = test + 2;
return test;
}
Since it is a member function of the instance test you have to call it as
test.addTwo(test.test);
Instead, you're calling it as
addTwo(test.test);
and it doesn't know what that function is. As far as the compiler is concerned, addTest(int) doesn't exist because you haven't defined it outside of the class definition.
Related
I know what "forward function declaration" means, but I want get the same with variables.
I have this code snippet:
#include <iostream>
int x;
int main()
{
std::cout << x << std::endl; // I want get printed "2" but I get compile error
return 0;
}
**x = 2;**
In the std::cout I want print "2" value, but trying to compile this I get this compile error: error: 'x' does not name a type.
While this doesn't appear somthing of programmatically impossible, I can't compile successfully.
So what is the right form to write this and obtain a forward variable declaration?
Variable declarations need extern. Variable definitions need the type, like declarations. Example:
#include <iostream>
extern int x;
int main()
{
std::cout << x << '\n';
}
int x = 2;
Normally you'd use extern to access a variable from a different translation unit (i.e. from a different .cpp file), so this is mostly an artifical example.
You can declare
extern int x;
in this file, and in some other file
int x = 2;
You could use Class and declare variable inside. Also, anonymous namespace is needed as Vlad said. Example:
#include<iostream>
namespace
{
class MyClass
{
public:
static int x;
};
}
int main()
{
std::cout << MyClass::x;
}
int MyClass::x = 2;
I tried the following code:
#include <iostream>
using namespace std;
int main()
{
char string[4]='xyz';
return 0;
}
Since string is a keyword the compiler should give error but it runs fine. Can anyone explain why it compiles successfully.
string is not a keyword.
It's the name of a type declared in the standard library.
When you give it a name, you're doing something called shadowing. This is more clear in the following example:
{
int x = 0;
{
int x = 5;
std::cout << x << std::endl;
}
std::cout << x << std::endl;
}
What gets printed?
Well, 5 first then 0.
This is because the x in the second scope overrides the x from the first. It "shadows" the first declaration.
This works with typenames as well:
struct MyStruct {
int x;
};
...
{
...
int MyStruct = 10;
...
}
Here, MyStruct gets overridden within that scope.
That same thing happens in your example with std::string
Can anybody tell me what is wrong in the following code when I initialize a global array and want to print its value outside main() function
#include <iostream>
using namespace std;
int global_array[5] = {10,20,30,40,50};
cout << global_array[2];
int main()
{
cout << "Hello World!" ;
}
The error keep popping is
error: 'cout' does not name a type|
The statement cout << global_array[2]; is not a declaration (it is an expression). Only declarations are allowed outside of functions.
So, if you want to print anything outside of main function, you can only do so by having the expression within another function.
I think the problem is that the code you have that does the printing is outside of any function. Statements in C++ need to be inside a function. For example:
#include <iostream>
using namespace std;
void hello();
int global_array[5] = {10,20,30,40,50};
void hello()
{
cout << global_array[2];
}
int main()
{
hello();
cout << "Hello World!" ;
}
Before asking a question, you can search: ‘cout’ does not name a type
Thanks you.
if you want to call it from outside the main it should be in a function something like this
#include <iostream>
using namespace std;
int global_array[5] = {10,20,30,40,50};
int pre()
{
cout << global_array[2];
return 0;
}
int x = pre();
int main()
{
cout<<"Hello World";
return 0;
}
as i mentioned it on comment it can be done via c++ classes.
#include <iostream>
int global_array[5] = { 10,20,30,40,50 };
struct foo
{
foo()
{
std::cout << global_array[2] << std::endl;
}
};
foo f;
int main()
{
}
My problem is in the following C++ code. On the line with the 'cout' I get the error:
"'number' was not declared in this scope".
.h
using namespace std;
class a{
int number();
};
.cpp
using namespace std;
#include <iostream>
#include "header.h"
int main(){
cout << "Your number is: " << number() << endl;
return 0;
}
number(){
int x = 1;
return x;
}
Note: I'm aware this isn't the cleanest code. I just wanted to get the function working and refresh my memory on how to use headers.
For minimal fix, three basic changes are necessary.
Proper implementation of the number() method
int a::number() {
int x = 1;
return x;
}
Proper invocation of the number() method
a aObject;
cout << "Your number is: " << aObject.number() << endl;
There are many other enhancements possible though.
Addition, as pointed out by #CPlusPlus, usable scope of number() method, for example declaring it public
class a{
public:
int number();
};
Try this in your cpp file
using namespace std;
#include <iostream>
#include "header.h"
void a::number()
{
int x = 1;
return x;
}
int main()
{
cout << "Your number is: " << a().number() << endl;
return 0;
}
As for your header file replace class with a struct. The reason you are getting this error is because the compiler cant find the variable number. It is actually a method of a class.The reason you are replacing the class with a struct is because by default everything in a struct is public. So your header file called header.h should look like this
using namespace std;
struct a
{
int number();
};
There are three issues with your code.
The definition of the function number().
As you declared, it is a member function of the class "a". In your .cpp, the class name should be used as a prefix to the function. I mean,
a::number(){
int x = 1;
return x;
}
As the function is a member of the class "a", there are only two ways of accessing it,
If the function is a static function in the class, you can access it with :: operator. Like a::number().
If the function is not a static function, that is true in your case, you should instantiate the object out of the class "a" and they use "." operator with the reference. I mean,
a obj;
obj.number().
Your function number() is declared in private scope. You may recall that by default the scope is a class is private unless you specify public or protected. So the private function number() cannot be used outside the declared class unless there is a friend to it.
Below the code that I fixed,
.h
using namespace std;
class a{
public:
int number();
};
.cpp
using namespace std;
#include <iostream>
#include "header.h"
a::number(){
int x = 1;
return x;
}
int main(){
a obj;
cout << "Your number is: " << obj.number() << endl;
return 0;
}
I'm new to C++ and get a beginner's mistake:
myclass.cpp: In function ‘int main()’:
myclass.cpp: 14:16: error: ‘func’ was not declared in this scope
This is the code:
#include <iostream>
using namespace std;
class MyClass{
public:
int func(int);
};
int MyClass::func(int a){
return a*2;
}
int main(){
cout << func(3);
}
I hope you can help me.
int main(){
cout << func(3);
}
func is not a global function; it is a member function of the class. You need an instance of the class to access it.
For example:
int main()
{
MyClass obj;
std::cout<< obj.func(3);
}
func is a member function, so it must be invoked through an object. For example:
int main()
{
MyClass obj;
std::cout << obj.func(3); // 6
}
In your example, you treated it as a free function, so the compiler looked for a function with that name. Since it could not find it, it issued a compiler error.
func is a member function of MyClass. To call it, you need an object of MyClass type to invoke it on:
int main(){
MyClass m; // Create a MyClass object
cout << m.func(3);
}
Alternatively, you could make func a static member function, which means that it is not associated with any particular instance of the class. However, you would still need to qualify its name as belonging to the MyClass class:
class MyClass{
public:
static int func(int);
};
int MyClass::func(int a){
return a*2;
}
int main(){
cout << MyClass::func(3);
}