CODEWITHSUNDEEP
c#arrays
elseshawe
elseshawe

Reputation: 161

Why can't I declare an implicitly typed array without an initializer?

Why do I have to write int after new when I am declaring the array num without an initializer? E.g.

int[] num = new int[5];

For another array, num1, I just declared it first and then gave it values, and I didn't type int after new like I did for array num.

int[] num1 = new [] {1, 2, 3, 4};

Why does this shortcut not work for my first example?

Upvotes: 10

Views: 5381

Answers (5)

InBetween
InBetween

Reputation: 32790

Because you seem to expect that the way the compiler infers the type of the array is somehow based on the type to the left of the assignment operator:

int[] num = new[5];

Your thought process is:

  1. num is an array of ints.
  2. new[5] is an array of something.
  3. something has to be int because that is num's type.

That is not the way the compiler works. The compiler must figure out univocally the type to the right of the assignment expression and once it does, it will check if the assignment to whatever is on the left is valid.

In the syntaxes allowed in C#, the compiler always has enough information to figure out the type on the right side of the assignment (if it has any). Your proposed syntax wouldn't provide that information.

Upvotes: 3

Blorgbeard
Blorgbeard

Reputation: 103555

The "full" syntax for declaring an array with values is new int[] {1, 2, 3, 4};.

Your new [] {1, 2, 3, 4}; line is a shorthand syntax. It's an "implicitly typed" array. Because the values are all int, the array is automatically declared an int[].

The reason you can't also write int[] x = new [5]; is simply that this feature has not been implemented by the C# designers.

Upvotes: 9

Servy
Servy

Reputation: 203850

The type of the array can be inferred (under certain circumstances, which your example fullfills) if you provide the values for the array. Since the values are all integers, the compiler is able to infer that you want an array of integers. If you don't provide any values (or if the values aren't all implicitly convertible to the type of one of the values), it has no way of inferring what the type of the array should be, and so you need to specify it explicitly.

Upvotes: 10

Patrick Hofman
Patrick Hofman

Reputation: 157136

why I have to write int after new when I am declaring array num with default values

You don't. There are a number of shorthands possible because often the compiler can infer the type of the array you are creating, like in your case.

So this is a perfectly fine declaration:

var num1 = new[] { 1, 2, 3, 4 };

See, there is no int here, but if you hover over var it will tell you the array is an int[].

It is getting harder when you are mixing types:

var num1 = new [] { "1", 1 };

Then the compiler doesn't know and forces you to specify a type.

Upvotes: 0

krillgar
krillgar

Reputation: 12815

In your first scenario, you're declaring an empty array, so you need to determine what type it is.

int[] num = new int[5]; // New empty array of integers with 5 slots.

However, in your second example, the type of the array is implied based on the values that you assign to the slots in the array.

int[] num1 = new [] {1, 2, 3, 4}; // New array of integers with 4 slots, already assigned.

The above works because every element in the array is of the same type.

If you were to do the following, a compilation error would occur because the type can't be inferred as the elements are different types.

var array = new [] {1, "something"}; // This won't compile.

But, if you declare one of the elements as a common base type, then it will store all of the elements as that base type.

var array = new [] {(object)1, "something", DateTime.Now}; // This is an array of objects.

Upvotes: 2

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