Pointers and string literals in C

Question

Suppose, in C language I have the following code, where pointer variable "word" points to a string literal.

const char * word = "HELLO";

Why does this works -

printf("\nword[1]:  '%c'", word[1]);
printf("\nword[2]:  '%c'", word[2]);

and this doesn't ?

printf("\nAddress of word[1]:  %p", word[1]);
printf("\nAddress of word[2]:  %p", word[2]);

Well, word[x] has a character type and not pointer... So your format specifiers are not matching. Why do you think it should work? — Eugene Sh.
– Eugene Sh., Commented Mar 3, 2020 at 20:38
unrelated: prefer printf("........\n"); as opposed to printf("\n...");. — pmg
– pmg, Commented Mar 3, 2020 at 20:39
You would have to add the & to get the pointer as using [] for arrays automatically dereferences the value — Warpstar22
– Warpstar22, Commented Mar 3, 2020 at 20:39
When you say printf("\nAddress of word[1]: %p", word[1]); doesn't work, do you mean doesn't work as expected? You're not having a compilation error, are you? — Fiddling Bits
– Fiddling Bits, Commented Mar 3, 2020 at 20:40

0___________ · Accepted Answer · 2020-03-03 20:41:42Z

3

because the latter is char not pointer. word[1] is the same as *(word + 1) and you just dereference the char pointer. The result is char

you need to:

printf("\nAddress of word[1]:  %p", (void *)&word[1]);
printf("\nAddress of word[2]:  %p", (void *)&word[2]);

or

printf("\nAddress of word[1]:  %p", (void *)(word + 1));
printf("\nAddress of word[2]:  %p", (void *)(word + 2));

answered Mar 3, 2020 at 20:41

0___________

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