Thread: C++ simple program errors in declaring variables

Hey, i tried to write a simple C++ porgram about classes but i have some problems...take a look at my code:

Code:

#include <iostream>

using namespace std;

class Data{
    private:
        int height;
        int age;
        char name;
    public:
        int printAll();
};

void Data::printAll(info.height){
    cout << "info.height: " << info.height << "\n";
}

int main(void){
    Data info;
        cout << "Height:";
        cin >> info.height;
        info.printAll(info.height);

return 0;
}

output errors:

Code:

class_program.cpp:14: error: variable or field ‘printAll’ declared void
class_program.cpp:14: error: ‘info’ was not declared in this scope
class_program.cpp: In function ‘int main()’:
class_program.cpp:7: error: ‘int Data::height’ is private
class_program.cpp:21: error: within this context
class_program.cpp:7: error: ‘int Data::height’ is private
class_program.cpp:22: error: within this context
class_program.cpp:22: error: no matching function for call to ‘Data::printAll(int&)’
class_program.cpp:11: note: candidates are: int Data::printAll()

Can anyone help me...? Thank you very much!

int printAll();
void Data::printAll(info.height)

See the difference?

can i declare the printAll (inside class) as void ?

Code:

int printAll();

void Data::printAll(info.height){
    cout << "info.height: " << info.height << "\n";
}

The debugger has actually told you everything you would need to solve these problems. If you don't know how to read debugger messages, that's pretty much necessary for becoming a programmer.

Notice how printAll() has been given two different return types (<int> and <void>), and two different sets of parameters. In order to define a function that has already been prototyped, the return type and parameters have to be the same. Otherwise the compiler considers them two different functions.

The parameter you've given it in the definition is unusable. Parameters need to have the type specified. As well, when you call a class method using the dot operator (ie, info.printAll(args)), the function automatically knows everything about the calling object (you'll get this better when you learn about keyword this). For the moment, here is a silly example to help visualize:

Code:

You create a class PERSON
This class contains variables like numberOfLegs, height, and walkingSpeed
The class also contains a method: void walk()

You then create an object of class PERSON, namely Bob.

When you tell Bob to walk ( ie: bob.walk() ), you don't need to tell Bob how many legs he has, how tall he is,
or how fast he normally walks;
All PERSONs already know these things!

it is the basic concept of class that u can not access data member, declared in private access specifier, from other than public. only member function can access the data member
as clear from the error that "int height is private".
u r using the arguement "info.height" why? if u have nothing to do with the arguement then why u use them actually u r missing the concept of "public" and private access specifier i think u should have a look at them
always use a member function to assign some value to data member

>>void Data::printAll(info.height){
>>info.printAll(info.height);
What do you think this does? Why do you think it's right?

I, too, am curious about this. Elysia's post alone sums up the issue nicely.

Stop posting answers. Prefer to steer the OP in the correct direction rather than post answers. This is more than likely for homework and we do not post outright answers to homework b/c we are not a cheat site nor do we wish to become one.

I have removed the answer.

ok bubba i will be carefull next time