passing 2d char array and assigning value

Question

I have a char array declared as

char userName[20][35];

I think that should be 20 buffers of 35 chars each.

So now I try and send it to a function:

readNames(userName[20][35])
{


}

I tried several different ways, is that the correct way to pass the array? Now I need to assign a value I read out of user input.

strcpy(userName[numberOfUsers] , userInput);

So two parts, did I send the array to the function ok? And how how do I do the strcpy? It doesn't seem to work.

Answer 1

You can do it several ways. For safety, you should add a size limit parameter so the callee (your function) knows how large the dominant dimension is:

void readNames(char names[][35], size_t n)
{
    // your code here using names[0] through names[n-1]
}

and called as:

char names[20][35];
readNames(names, sizeof(names)/sizeof(*names));

Similarly, this also works (and in fact is synonymous with the above):

void readNames(char (*names)[35], size_t n)
{
    // your code here using names[0] through names[n-1]
}

Both work, but there is a better way with C++ if you truly want to stick to C-style arrays. (shown later).

Lastly, this:

strcpy(userName[numberOfUsers] , userInput);

is correct syntax. Just make sure to remember that numberOfUsers cannot exceed or meet the boundary of the array (the size_t n you pass from the caller in the above samples). The addressing is just like C, zero based up to (n-1).

---

Fixed Array Size Determination with Template Deduction

A much more robust mechanism for doing this with C++ involves using template deduction to ensure the function knows the dimensions. It is considerably more flexible, as it can be used with different array declarations:

#include <iostream>

template<size_t N, size_t M>
void readNames(char (&names)[N][M])
{
    // use names[0]... names[N-1], where
    // each is a char [0..M-1] buffer.

    std::cout << __PRETTY_FUNCTION__ << '\n';
}

int main()
{
    char shortnames[20][30];
    readNames(shortnames);

    char longnames[50][100];
    readNames(longnames);

    return 0;
}

Output

void readNames(char (&)[N][M]) [N = 20, M = 30]
void readNames(char (&)[N][M]) [N = 50, M = 100]

---

How about that Standard Library, Eh?

If you really want to do the above, knock yourself out, but honestly, we're a decade-plus deep into the new millennium. Stop using technology from the 80's and 90's:

std::vector<std::string> readNames()
{
    std::vector<std::string> res;

    while (some-condition)
    {
        // get name here
        std::string name;

        //add to result
        res.push_back(name);
    }
    return res;
}

Invoked like this:

std::vector<std::string> names = readNames();

Answer 2

The syntax you chose is something of a function prototype, not a call. Each is written differently.

To call readnames with that array:

readNames (userName);

To declare the function:

void readNames (char names[20][35])
{
  ...  manipulate names here ...
}

Change void to another type if the function returns a value.

Answer 3

void getinput(char names[][10], int s)
{
int j;
cout << "enter value ";
for (int i = 0; i < s; i++)
{
    j = 0;
    while (**condition**)\\what will be here,this condition is not working
    {
        cin >> names[i][j];
        j++;

    }
}
}