Why getting a "bus error" when playing with pointers?

Question

so I was playing with pointers because I didn't know what else to do and usually, I imagine what's going on under the hood after each instruction. But I recently came against an error that I don't really understand.

char *str = "test";
printf("%c", ++*str);

Output :

zsh: bus error

Expected output was a 'u' because as far as I know, it first dereference the first address of the variable 'str' wich is a 't' than increment it right ? Or am I missing something ?

Changing the code like so is not giving me any error but why ?

printf("%c", *++str);

Thank you !

`char *str = "test";` is not correct in C++, so I assume this is C. Please only tag the language you are actually using — 463035818_is_not_an_ai, Aug 19 '21 at 13:56
`++*str` attempts to increment the read-only data in the string literal. — William Pursell, Aug 19 '21 at 13:56
Don't write code like this: `printf("%c", ++*str);`. Break out your increment into a separate statement. As @WilliamPursell explains, the dereference occurs *first,* before the increment. — Robert Harvey, Aug 19 '21 at 13:57
String literals are not modifiable. You could use `char str_data[] = "test"; char *str = str_data;` or `char *str = (char[]){ "test" };` instead, — Ian Abbott, Aug 19 '21 at 13:59

score 3 · Accepted Answer · answered Aug 19 '21 at 14:00

3

You cannot modify the data in a string literal. What you expect will work if you do:

char buf[] = "test"; 
char *str = buf;
putchar(++*str);

because the content of buf is writeable.

answered Aug 19 '21 at 14:00

William Pursell

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Why getting a "bus error" when playing with pointers?

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