What is the problem?

The following code does not run.
What is the problem?
It returns message () expected.

Code:

# include<iostream.h>
# include<conio.h>
# include<stdio.h>
class person
{
private:
int age;
char name[20];
public:
void getdata();
void putdata();
};
void person::getdata()
{
cout<<"Enter the name";
cin>>name;
cout<<"Enter age";
cin>>age;
}
void person:: putdata()
{
cout<<"\nName  :"<<name<<"\n";
cout<<"Age   :"<<age;
}

int main()
{
person a;
void  pline(char ch '#',int i=40);
clrscr();
a.getdata();
pline();
a.putdata();
return(0);
}
void pline(char ch,int i)
{
for(i=0;i<=40;i++)
printf("%c",ch);
}

It seems you're having confusion with your pline() function:

1) Your pline() function has no prototype.

Code:

void pline(char ch,int i); //Put this above main() !

2) Your first call to your pline() function does not have valid agruments:

Why do you have

Code:

void  pline(char ch '#',int i=40); //We don't like this

in main()? When you're calling a function, don't declare the arguments within the function call. Instead, pass it like:

Code:

pline('#',40); //We like this

3) Your second call to the pline() function has no arguments:

Code:

pline(); //invalid

The pline() function takes two arguments, so you'll need to pass to it two arguments.

By the way, I *think* you meant:

Code:

void pline(char ch,int i)
{
     for(int x=0;x<=i;x++) //loop 40 times (or whatever number gets passed to it)
     printf("%c",ch);
}

>>1) Your pline() function has no prototype.
His method is perfectly valid. You are allowed to prototype inside of a function.

void pline(char ch = '#',int i=40);

Originally posted by XSquared
>>1) Your pline() function has no prototype.
His method is perfectly valid. You are allowed to prototype inside of a function.

void pline(char ch = '#',int i=40);

Well, it may be valid, but in a large project you would have a hard time finding your first prototype again once you have to change the implementation

Yup, you just forgot the '=' sign :-)

You are allowed to prototype inside of a function.

Wow, my bad. I've just never seen that before. I thought he was just confused about functions...lol.

The for-loop will loop once more than expected.
That is, if I pass 20 it will loop 21 times, because of the =0, <= combination.