Question

char *a = "apple";
printf("%s\n", a);  // fine
printf("%s\n", a[1]);  // compiler complains an int is being passed

Why does indexing a string pointer give me an int? I was expecting it to just print the string starting at position one (which is actually what happens when i use &a[1] instead). why do i need to get the address?

Answer 1

That’s just how the [] operator is defined – a[1], when a is a char *, fetches the next char after the one pointed to by a (a[0] is the first one).

The second part of the puzzle is that char values are always promoted to int (or rarely, unsigned int) when passed as part of a function’s variable-length argument list.

a is equivalent to &a[0], and it prints from the first character – so it makes sense that &a[1] would print starting from the second character. You can also just use a + 1 – that’s completely equivalent.

If you use the %c conversion specifier, which prints a single character, you can use a[1] to print just the second character:

printf("%c\n", a[1]);