Char array pointer and "error: too many initializers for 'char []'"

Question

The following piece of code gives me the error: "too many initializers for 'char []'" :

int main()
{
    int input;
    char numbers[] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

    std::cin >> input;

    std::cout << ((input > 9) ? "Greater than 9" : numbers[input-1]) << std::endl;
}

What's needed for this to work is for numbers to be a pointer variable, i.e:

char * numbers[] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

I'm new to c++ and I'm trying to understand why this array needs to be a pointer and what exactly is happening in memory which requires this to be a pointer?

In other languages such as Java you can do the following:

import java.util.Scanner;

public class Playground
{
    public static void main(String[] args)
    {
        Scanner scanner = new Scanner(System.in);
        int input = scanner.nextInt();

        String[] numbers = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
        System.out.println((input > 9) ? "Greater than 9" : numbers[input-1]);
    }
}

No Pointers are required here, and from my understanding there's no need for them in this kind of scenario either.

Answer 1

As @Max Langhof said in the comment char is a single character. You are trying to initialize an array of characters with an array of strings

You mention using C++, so instead of using C style arrays [] use the std::array available in the array header file. Along with std::string

std::array<std::string,SIZE> = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

Declaring it as a pointer it essentially creates an array of char pointers which in plain english translates to an array of char arrays.

If you want to learn more about C style char arrays The Definitive C++ Book Guide and List

Take the following code
const char* a [] = {"sadas", "dasdas"}; std::cout << a[0][0]; // will output 's'

a[0] will return sadas

Having a pointer to a char array will make the value const (read-only) a[0][0] = 'b' // illegal

Answer 2

Since the strings are literals, i.e. constants, you should declare the array like this:

const char* numbers[] = { "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };

This is the C way of things with the array defined at compile time. So the memory required for the array is calculated by the compiler. Hence you have to declare it as an array of const char*, i.e. the contents can't be changed as it is defined at compile time.

You could declare the array like this:

char numbers[9][6] = { "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };

What this does is declare a 2D array of 9 elements, each element containing 6 chars. So now you could change the value of each word at run time if you want to. The square brackets [] make all the difference. When you use char*[] the compiler treats elements as literals.

You would have to allocate them on the heap if you don't know the size of each word in the array beforehand.

The C++ way is to use std:string (allocated on the heap internally for you) and these can be changed at run time:

std::string numbers[] = { "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };

Answer 3

What's needed for this to work is for numbers to be a pointer variable.

Your confusion lies here (and possibly also in what char signifies).

numbers is not a pointer variable, with or without the *.
The * is about the elements, not the array.

This:

char numbers[] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

is an array of chars. It doesn't compile because a char is a single character, and none of the expressions "one", "two" etc fit that description.

Instead, they are string literals. These are each implemented as arrays of char (ironically, the sort of array you just tried to create!). We typically access them via pointers, of type const char* (in the olden days you could use char*, but that's no longer true; perhaps you're using an ancient compiler).

So, to fix your array, it needs to be not an array of chars, but an array of const char*s.

And that's what you're doing here (const added by me for modern correctness):

const char* numbers[] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

… it's an array of const char*s.

And, yes, since C declaration syntax is confusing, you're arguably better off using modern C++ tools as shown in other answers.

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No Pointers are required here, and from my understanding there's no need for them in this kind of scenario either.

That's not really true either, but it looks like it is because Java is a completely different language with different syntax and abstractions.

Somewhere underneath the bonnet, String is a pointer (because Java has managed objects), and furthermore it'll have a pointer inside it that points to dynamically-allocated data (so, in fact, there's additional indirection in the Java example!).

The only difference is that it's done for you, transparently, rather than you having to spell it out as you do in C.

It's best not to directly compare two unrelated languages; there's usually very little value in doing so.