C++ passing char array to function

Question

I would rather just use a string, but we aren't supposed to as the teacher hates them and wants us to figure out ways to avoid them. So I looked into using a struct, but we aren't that far in the book and she hates it when I skip ahead. So I was thinking of doing this:

#include <iomanip>
#include <iostream>
#include <stdio.h>

using namespace std;

void myfunc(char& );

int main()
{
    char myname[12];
    cout<<"enter  a name ";
    cin>>myname;
    cout<<"myname is "<<myname;

    cout<<"myname is " << myfunc(myname);

    getchar();
    getchar();
    return 0;
}

void myfunc(char &myname1)
{
    myname1 = "Billy"
}

But this doesn't work and I don't know why.

Answer 1

One way is to do it like this:

void myfunc(char *myname1)
{
    strcpy(myname1,"Billy");
}   

You will also need to change your main to:

myfunc(myname);
cout<<"myname is " << myname;

However you have to be careful not to overflow your initial buffer.

The reason why your original code doesn't work is because you can't assign strings to char pointers. Instead you must copy the string to the char*.

Answer 2

This line of code is wrong:

cout<<"myname is " << myfunc(myname);

myfunc() doesn't return anything, its return type is void.

Try using:

char* myfunc(char *myname1)
{
    strcpy(myname1,"Billy");
    return myname;
}

Or

myfunc(myname);
cout<<"myname is " << myname;