Understanding the below behaviour

Question

I am getting a seg fault for the below code, is there something wrong? Here I am trying to shift the bits in a. Also, I know char * is a read only memory.

So, we have to copy into char a[] and then modify it???

    char *str = "abc";

    *str = *str << 1;

You declared a literal string (in read only memory) and tried to modify it. To your update: Yes, `char a[] = "abc";` would allow it to work. — Kevin Richardson, Jun 03 '16 at 06:43
You know that is read-only and you still try to modify it, ask your self why. char *str = "abc"; should be **const char *str** — Michi, Jun 03 '16 at 08:31

score 2 · Answer 1 · edited May 23 '17 at 11:52

2

char* string literal is pointing to read-only memory. You need to use a char array:

char str[] = "Hello";
*str = *str << 1;

See: What is the difference between char s[] and char *s?

edited May 23 '17 at 11:52

Community

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answered Jun 03 '16 at 06:52

Isaac Turner

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score 0 · Answer 2 · answered Jun 03 '16 at 06:41

What are you trying to do exactly ?

<< is a bitwise operator and *str will not refer to the whole string but only to the first character.

I also recommend you to put parenthesis around your pointer when doing any operations on it to be sure that the bit shift is not done on the address instead of on what is pointed at this address.

Understanding the below behaviour

2 Answers2