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.

Avoiding std::string is not idiotic. There are quite a few cases where one would want to use char arrays of set sizes instead, especially when dealing with limited memory such as programming for microcontrollers or IoT devices. — OpticalMagician
– OpticalMagician, Commented Jun 22, 2020 at 3:40

Mysticial · Accepted Answer · 2011-11-07 02:54:14Z

12

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*.

edited Nov 7, 2011 at 2:54

answered Nov 7, 2011 at 2:43

Mysticial

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Comments

Loki Astari · Accepted Answer · 2011-11-07 02:49:16Z

7

Arrays devolve into pointers when passed as parameters.
So the simple way that you want is:

char* myfunc(char* myname1)
{
    return myname1;
}

If you were going to show off you can pass the array by reference.
But if you can't read ahead you will not be able to use this.

char* myfunc(char (&myname1)[12]) // Note you can only pass arrays of size 12
{                                // to this function so it is limited in use.
    return myname1;
}

TO make it more useful though you could template it:

template<int SIZE>
char* myfunc(char (&myname1)[SIZE])
{
    return myname1;
}

edited Nov 7, 2011 at 2:49

answered Nov 7, 2011 at 2:44

Loki Astari

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NullUserException · Accepted Answer · 2011-11-12 15:27:49Z

7

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;

edited Nov 12, 2011 at 15:27

NullUserException

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answered Nov 7, 2011 at 2:48

jmmartinez

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Comments

Martin Beckett · Accepted Answer · 2011-11-07 02:44:56Z

0

myname1 = "Billy" doesn't copy a string it copies a pointer to the constant local memory containing "Billy"

Take a look at strncpy() or memcpy()

answered Nov 7, 2011 at 2:44

Martin Beckett

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Comments

John Humphreys · Accepted Answer · 2011-11-07 02:45:44Z

0

Pass it as a char* instead of a char&. You're passing a reference to a single character instead of a pointer to a character array in this code.

Also use strncpy (google it) to set the value of tr char* once you're in the function.

answered Nov 7, 2011 at 2:45

John Humphreys

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Comments

Alex Force · Accepted Answer · 2011-11-07 02:46:18Z

0

void myfunc(char& ); is the problem it should take in a char * and not a char reference which is what you did.

and in the function use strcpy(char * destination, char *source);

answered Nov 7, 2011 at 2:46

Alex Force

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