stripping end of a string in C

Question

I need a function to remove ) character from end of a string. for example, hello,world) should be converted to hello,world.

I have written this :

char *a="hello,world)";
int a_len=strlen(a);
*(a+a_len-1)='\0';
printf("%s", a);

but nothing is shown in the output.

@MichaelFoukarakis You might want to refer to [this](http://c-faq.com/decl/strlitinit.html). — devnull, Feb 11 '14 at 13:59
@devnull: i think he meant that SIGSEGV takes place during runtime not compile time — Hayri Uğur Koltuk, Feb 11 '14 at 14:00
@HayriUğurKoltuk Maybe. I thought that _executing_ was implicit. (on executing this) — devnull, Feb 11 '14 at 14:02
Might be useful : http://stackoverflow.com/questions/5457608/c-remove-character-from-string — thikan, Feb 11 '14 at 14:04
Compile with all warnings and debug info (e.g. `gcc -Wall -g`) then **use a debugger** (e.g. `gdb`) — Basile Starynkevitch, Feb 11 '14 at 14:20

Sergey L. · Accepted Answer · 2014-02-11T14:24:13.003

You should ideally be getting a segmentation violation runtime error.

You have assigned a pointer to a string literal which resides in read-only memory. Trying to modify that is bad!

Try copying it onto the stack

char a[] ="hello,world)";

If you really have to use dynamic memory (please write that in your question) then you have to manually copy your string there:

char *a = malloc(sizeof("hello,world)"));
memcpy(a, "hello,world)", sizeof("hello,world)"));
int a_len=strlen(a);
a[a_len - 1] = '\0';

Alternatively you can also have printf truncate your string:

printf("%.*s", strlen(a) - 1, a);

Also as Basile pointed out there is strdup

char * a = strndup("hello,world)", sizeof("hello,world)") -2);

Note that here we have to truncate by two characters because sizeof includes the null terminator, but strndup will always add one.

I have to keep my code clean of libc functions, so I think memcpy is a better one to use. — alireza_fn, Feb 11 '14 at 14:24

score 1 · Answer 2 · answered Feb 11 '14 at 14:04

Analysis:

Line #1: char *a="hello,world)";

Variable a points to an array of characters, located in the (read-only) code section of the program

Line #3: *(a+a_len-1)='\0';

A memory access violation occurs, when the CPU attempts to write into a read-only memory section

Solution:

Line #1: char a[]="hello,world)";

Variable a is an array of characters located in the stack of the program, which is a read/write section

score -2 · Answer 3 · answered Feb 11 '14 at 13:57

-2

I must use dynamic memory so I have to leave char[].

trying this also does not work:

char *a=malloc(4);
a="hello,world)";
int a_len=strlen(a);
*(a+a_len-2)='\0';
printf("%s", a);

answered Feb 11 '14 at 13:57

alireza_fn

You are again assigning a pointer to read-only memory. You have to copy the string into your dynamic memory. – Sergey L. Feb 11 '14 at 13:58
also it should be -1 :) – Hayri Uğur Koltuk Feb 11 '14 at 13:59
1

Look my answer, I have edited it. Please use comments for discussions instead of posting your own not working answer. – Sergey L. Feb 11 '14 at 14:00
Should `malloc` a bigger zone then use `strcpy` – Basile Starynkevitch Feb 11 '14 at 14:19

3 Answers3