Pointers in C - int array VS char array [duplicate]

String literals have types of character arrays. For example the string literal "John" has type char[5] (string literals include the terminating zero).

Used in expressions arrays with rare exceptions are converted to pointers to their first elements.

You can imagine this declaration

char *names[] = { "John", "Paul", "George", "Ringo" };

like

char *names[] = { &"John"[0], &"Paul"[0], &"George"[0], &"Ringo"[0] };

Thus the array names is initialized by valid addresses of first characters of the string literals.

As for this declaration

int *numbers[] = { 11, 12, 13, 14 };

then the array numbers is initialized by invalid addresses values like 11, 12 and so on that do not point to actual objects.

You could write for example

int i = 11, j = 12, k = 13, l = 14;

and then

int *numbers[] = { &i, &j, &k, &l };

In this case the array would be initialized by valid addresses.

As for the printf calls then the function is designed such a way that when the conversion specifier s is encountered the function considers the corresponding argument as an address of a zero-terminating string and tries to output this string.

While the conversion specifier d serves to output objects of the type int. So for example if you want to output the integer pointed to by numbers[0] then you should write

printf("%d\n", *numbers[0] );

"John", … are of type char *.

11, … are of type int (without indirection).

So it is:

int numbers[] = { 11, 12, 13, 14 };
printf("%d\n", numbers[0]);

This question is explained here.

In the first example each name is an array and you have an array of arrays.

char *names[] = { "John", "Paul", "George", "Ringo" };

In the second example with int you have a simple array and using *var[] = **var but you don't have such ting in memory.

int *numbers[] = { 11, 12, 13, 14 };