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How can char pointer be initialized with a string (Array of characters) but an int pointer not with a array of integer?

When I tried this

int* a={1,2,3,4,5};

It gives an error saying

error: scalar object ‘a’ requires one element in initializer

But, 

char* name="mikhil"

works perfectly.

Sourav Ghosh
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Mikhil Singh
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    You can do this: int a[] = {1,2,3,4,5}; – kvr Mar 12 '16 at 06:38
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    _"But char* `name="mikhil"` works perfectly"_ No, it does not - it's wrong in C++ and **won't compile**. So, are we talking about C or C++? – edmz Mar 12 '16 at 15:27

4 Answers4

9

Because these are the rules (standards) of the language. There's nothing telling you it's possible to initialize a pointer with an array like this, for example, the following is also illegal:

char* name={'m', 'i', 'k', 'h', 'i', 'l', 0};

A literal string gets its own treatment and is not defined just as an array of characters.

Amit
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  • You're right, but this does not _answer_ this particular question. There's a basic difference in the use of the initializer here. – Sourav Ghosh Mar 12 '16 at 06:56
7

In case of you're trying to initialize an int * with

  {1,2,3,4,5};

it is wrong because {1,2,3,4,5};, as-is, is not an array of integers. It is a brace enclosed list of initializer. This can be used to initialize an int array (individual elements of the array, to be specific), but not a pointer. An array is not a pointer and vice-versa.

However, you can make use of a compound literal to initialize an int *, like

int * a = (int []){1,2,3,4,5};
Sourav Ghosh
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2

For convenience, the language allows char * (or, preferably, const char *) to be initialized with a string. This is because it is a very common usage, and it has been possible since C. You shouldn't really modify the string value (of course, that's a separate debate).

With an int, if you initialize an int a[5], it is expected you might change the values. The initialized values will be copied into your array (compiler optimizations not withstanding), and you can change them how you see fit. If int *a was allowed to initialize this way, what would it be pointing to? It isn't really clear, so the standard doesn't allow it.

Rob L
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1

the language dont support null terminator to int arrays , which is since compiler would have ambiguity that is that null terminator or 0 in int array...

it is the same reason due to which

cout<<anystring; give complete array of character i.e. complete string and this dont

cout<<a; for a 1-d int array a...infact it gives an error message and dont prints the complete 1-d array...

upvote if i somewhere satisfied / solved your query...

Shreyan Mehta
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  • `'\0'` is just a character with value `0` that can be promoted to `int`. That aside, I don't see how null terminators apply to this. The issue is that `{ 1, 2, 3}` is not an array. – Weak to Enuma Elish Mar 12 '16 at 07:04