How to use multidimensional char or string arrays in a loop - c++

Question

I'm so new to C++ and I just can't figure out how to use any multidimesional arrays. I want to do something like that:
input number of product: number; //the products' name can be 7 with NULL char. (max 6)
char arr[number][7];
That works. But when I want to do that in a for loop(i):
cin>>arr[i][7];
and I don't know what the hell is compiler doing?
I just want that:
arr[0][7]=apple;
arr[1][7]=orange;
So please how can I do that?

Answer 1

#include <string>
#include <vector>
Since everybody is recommending it, I thought I'd sketch the options for you.
Note that you would have gotten this kind of answer in 10 milli-seconds by 3 different persons, if you had supplied a short, working sample code snippet (translating code 1:1 is more efficient than 'thinking up' examples that you might recognize)
Here you go:
std::vector<std::string> strings
strings.push_back("apple");
strings.push_back("banana");
// or
std::string s;
std::cin >> s; // a word
strings.push_back(s);
// or
std::getline(std::cin, s); // a whole line
strings.push_back(s);
// or:
// add #include <iterator>
// add #include <algorithm>
std::copy(std::istream_iterator<std::string>(std::cin),
std::istream_iterator<std::string>(),
std::back_inserter(strings));
Direct addressing is also possible:
std::vector<std::string> strings(10); // 10 empty strings
strings[7] = "seventh";
Edit in response to comments:
const char* eighth = "eighth";
if (strings[7] != eighth)
{
// not equal
}
// If you really **must** (read, no you don't) you can get a const char* for the string:
const char* sz = strings[7].c_str(); // warning:
// invalidated when `strings[7]` is modified/destructed

Answer 2

Unless you have a real reason (which you mustn't hide from us), make as Björn says and use a vector of strings. You can even do away with the initial request for the total size:
#include <string>
#include <vector>
#include <iostream>
std::vector<std::string> fruits;
std::string line;
while (std::getline(std::cin, line))
{
fruits.push_back(line);
}
Let's test:
std::cout << "You entered the following items:\n";
for (auto const & f : fruits) std::cout << "* " << f << "\n";

Answer 3

Because arr[i][7] is a char, and in fact one past the last element, which means you may get memory access error.
What you want to do maybe cin>>arr[i];.
How ever, this is not a very good idea, as you cannot control how many characters are read from input, which will easily cause memory overrun.
The easy way would be using std::vector<std::string> as others have suggested.

Answer 4

strcpy(&arr[0], "apple");
strcpy(&arr[1], "orange");
but for C++ is better to use std::vector<std::string> for array of strings

Answer 5

You have a two dimensional array of char
char arr[number][7];
And then trying to assign a string (char* or const char*) to them which will not work. What you can do here is assign a character, for example:
arr[0][1] = 'a';
If you can I would recommend using std::vector and std::string it would make things much clearer. In your case you could do
cin>>arr[i];
But I would not recommend it as you could only store up to 6 character char* strings (plus the null terminator). You can also have an array of char*
char* arr[number];
then dynamically allocate memory to store the strings.

Answer 6

Using std::vector and std::string will usually save you headaches once you understand them. Since you are brand new to C++, it might be useful to understand what is going on with two-dimensional arrays anyhow.
When you say
char array[N][M];
With N and M being constants, not variables, you are telling the compiler to allocate N*M items of type char. There will be a block of memory dedicated to that array of size N*M*sizeof(char). (You can declare an array of anything, not just char. Since sizeof(char) is 1, the memory will be N*M bytes long.) If you looked at raw memory, the first byte in the memory would be where array[0][0] is. The second byte would be where array[0][1] is, an so on, for M bytes. Then you would see array[1][0]. This is called row-major order.
As #jbat100 mentioned, when you say array[i][j] you are referring to a single character. When you say array[i], you are referring to the address of row i in the array. There is no pointer actually stored in memory, but when you say array[i] the compiler knows that you mean that you want the address of row i in the array:
char* row_i = array[i];
Now if i>0, then row_i points to somewhere in the middle of that block of memory dedicated to the array. This would do the same thing:
char* row_i = &array[i][0];
If you have a string, "orange" and you know that the length of it is less than M, you can store it in a given row in the array, like this:
strcpy(array[i], "orange"); // or
array[i][0] = 'o'; array[i][1] = 'a'; ... array[i][6] = 0;
Or you could have said row_i instead of array[i]. This copies 7 bytes into the array in the location of row_i. The strcpy() also copies an extra byte which is a 0, and this is the convention for terminating a character string in C and C++. So the 7 bytes are six bytes, 'o', 'r', 'a', 'n', 'g', and 'e', plus a 0 byte. Now strcmp(row_i, "orange") == 0.
Beware that if your string is longer than M, the strcpy and the simple char assignments will not (probably) produce a compile error, but you will end up copying part of your string into the next row.
Read about pointers and arrays in a good C/C++ book.