How to return two elements of an array

I would like to return both elements of n[].
I cant find out how to do it.

Code:

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short* GetPosition(char letter, short number)
{
    short n[2] = {0,0};

    if    (letter == 'a' && number == 1) n[] = {7,  3};
    else if(letter == 'a' && number == 2) n[] = {7, 10};
    else if(letter == 'a' && number == 3) n[] = {7, 17};
    else if(letter == 'a' && number == 4) n[] = {7, 24};
    else if(letter == 'a' && number == 5) n[] = {7, 31};
    else if(letter == 'a' && number == 6) n[] = {7, 38};
    else if(letter == 'a' && number == 7) n[] = {7, 45};
    else if(letter == 'a' && number == 8) n[] = {7, 52};

    return n[0, 1];
}

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Eh, what is n? It looks like you have incorrect syntax.

Chances are, you want to return a std::pair<int, int>, or some struct or class object.

That doesn't look like valid code.

I edited the code, changed the return value from char* to short*.

n[2] is an array of two elements.

Quote:

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Chances are, you want to return a std::pair<int, int>, or some struct or class object.

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You mean i cant return both elements of an array?

No, you would then be returning a pointer to the first element of a local array, which means that the caller would get a pointer that points to something that no longer exists, which is a Bad Thing. Plus, you still have invalid syntax, e.g.,

Code:

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n[] = {7,  3};

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One way of doing what you want is to #include <utility> and use std::pair<short, short> with std::make_pair, e.g.,

Code:

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std::pair<short, short> GetPosition(char letter, short number)
{
    using std::make_pair;
    if    (letter == 'a' && number == 1) return make_pair(7, 3);
    else if(letter == 'a' && number == 2) return make_pair(7, 10);
    // ...
}

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Thank you Laserlight!

Code:

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n[] = {7,  3};

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May i ask why would this be invalid? Im assigning values to an array.

Quote:

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Originally Posted by Ducky

May i ask why would this be invalid? Im assigning values to an array.

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... which is invalid. You cannot assign to an array.

Arrays

"When an initialization of values is provided for an array, C++ allows the possibility of leaving the square brackets empty . In this case, the compiler will assume a size for the array that matches the number of values included between braces

Sorry i dont get it. This is what i did.

Quote:

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Originally Posted by Ducky

Arrays

"When an initialization of values is provided for an array, C++ allows the possibility of leaving the square brackets empty . In this case, the compiler will assume a size for the array that matches the number of values included between braces

Sorry i dont get it. This is what i did.

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Nope, that was assignment, not initialization. Initialization is this line:

Code:

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short n[2] = {0,0};

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Quote:

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Originally Posted by Ducky

When an initialization of values is provided for an array, C++ allows the possibility of leaving the square brackets empty . In this case, the compiler will assume a size for the array that matches the number of values included between braces

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That's correct.

Quote:

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Originally Posted by Ducky

Sorry i dont get it. This is what i did.

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No, you attempted to assign to an array, not to initialise an array. You did initialise n, on line 3 of your code snippet.

Quote:

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Originally Posted by Ducky

And can i assign only one element of

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Yes, by using the first or second public member variables of the pair, e.g.,

Code:

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std::pair<short, short> position = GetPosition(c, n);
std::cout << position.first << ' ' << position.second << std::endl;

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However, you may find it better to define your own Position class instead of using std::pair.

Thanks to both of you!

So once an array is initialized you cannot change their elements?

Quote:

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Originally Posted by Ducky

So once an array is initialized you cannot change their elements?

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No, you may be able to assign to the elements of an array.

Ah ok i understand. So i should have written something like this i guess.

Code:

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if    (letter == 'a' && number == 1)
      n[0] = {7} ;
      n[1] = {3} ;

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Quote:

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Originally Posted by Ducky

So i should have written something like this i guess.

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More like:

Code:

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n[0] = 7;
n[1] = 3;

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but then you would still have the problem where you return a pointer to an object that no longer exists by the time control returns to the caller.

The thing is that without other changes it would still be wrong to return n (pointer to local array).

More likely, and with other simplifications:

Code:

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//pass array in to be modified
short* GetPosition(short* n, char letter, short number)
{
    n[0] = n[1] = 0;
    if (letter == 'a' && number >= 1 && number <= 8) {
        n[0] = 7;
        n[1] = 7 * number - 4;
    }
}

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Great code! Thank you Anon!