Why are strings declared as an array "char word[5] = "hello" modifiable but not strings declared as pointer (char *word = "hello")?

Question

I was under the impression that

    char word[5] = "hello";

is the same as

    char *word = "hello";

because arrays always decay into a pointer?

`char word[5] = "hello";` can't be treated as a string, it's a normal character array. `word` needs to have space for 6 characters to make it a proper C string (The trailing 0 is critical). — Shawn, Dec 07 '19 at 00:13
`char* word = "hello"` gets compiled into a different read-only section of your program - here's an answer that goes into more detail: https://stackoverflow.com/a/2589963/1158478 — ArtHare, Dec 07 '19 at 00:15

foreverska · Answer 1 · 2019-12-07T00:13:53.690

0

char *word = "hello"; Takes the address of a string literal. These are often held in read-only memory.

edited Dec 07 '19 at 00:13

answered Dec 07 '19 at 00:06

foreverska

1 Answers1