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

Answer 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...

Answer 2

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};

Answer 3

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.

Answer 4

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.