Why is this C code involving constant pointer crashing on me?

Question

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getting segmentation fault in a small c program

Here's my code:

char *const p1 = "john";

p1[2] = 'z';  //crashes

printf("%s\n", p1);

I know p1 is a "read-only" variable, but I thought I could still modify the string ("john" ). I appreciate any tips or advice.

Hmm, on this webpage - http://www.cplusplus.com/forum/beginner/10602/ - it says you could change John though — Caffeinated, Feb 12 '12 at 18:42
@Adel: it's wrong. Whether or not you can change "john" is implementation dependent. You can't rely on it, and therefore shouldn't to it. — Mat, Feb 12 '12 at 18:44
`p1` does **not** have to be a read-only variable (you just added that restriction for no reason), but `*p1` **is** read-only. A small asterisk can make a big difference in C! — Kerrek SB, Feb 12 '12 at 18:56
Modifying string literals is UB. Seg fault therefore perfectly plausible. — David Heffernan, Feb 12 '12 at 19:19

score 4 · Accepted Answer · answered Feb 12 '12 at 18:39

You cannot safely modify string literals, even if the pointer doesn't look const. They will often be allocated in read-only memory, hence your crashes - and when they're not in read-only memory modifying them can have unexpected consequences.

If you copy to an array this should work:

char tmp[] = "john";
char *const p1 = tmp;
p1[2] = 'z'; // ok

score 3 · Answer 2 · answered Feb 12 '12 at 18:40

Keyword const means that variable of that type should stay constant. You shouldn't change it.

Also even if you declare this string as char *p1 = "john"; it would be constant string literal and changing it would cause undefined behaviour. You should declare it as char p1[] = "john"; in order to achieve behaviour you are looking for.

score 3 · Answer 3 · edited May 23 '17 at 12:03

1) Forget the "const" qualifier. This is WRONG, in C and C++, on ANY platform:

  char *p1 = "john";
  p1[2] = 'z';  //crashes

Here's why:

2) So how do you get around the access violation?

Simple: you allocate writable memory (instead of writing to a string constant, which is probably allocated in read-only memory):

  #define BUFSIZE 80
  ...
  char p1[BUFSIZE];
  strcpy (p1, "john");
  p1[2] = 'z';  //no problem

3) OK: so then what's the deal with "const"?

You were right about that part. Here's (one of many) discussions about the (subtle) difference between a "const pointer" and a "pointer to a const":

http://www.codeguru.com/cpp/cpp/cpp_mfc/general/article.php/c6967

score 2 · Answer 4 · answered Feb 12 '12 at 18:39

2

your pointer points to memory that isn't allowed to be changed ( the constant string )

do char p1[100] = "john";

answered Feb 12 '12 at 18:39

Keith Nicholas

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Sergey Kalinichenko · Answer 5 · 2012-02-12T18:45:54.160

1

This is because you are not supposed to modify constants: after all, they are called constants for a reason. Modifying a constant is undefined behavior according to the C standard, which often means that your program is going to crash.

Note that it has nothing to do with your pointer being constant: the crash is because what your pointer points to is a string constant.

Here is how to do what you are trying to do legally:

char p1[] = "john";
p1[2] = 'z';  //no longer crashes :)
printf("%s\n", p1);

edited Feb 12 '12 at 18:45

answered Feb 12 '12 at 18:39

Sergey Kalinichenko

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That should not crash there - fix your comment :) – bdonlan Feb 12 '12 at 18:40
@bdonlan I modified the answer to point out that the crash is due to modifying a constant pointed to by the pointer, thanks for a good point. – Sergey Kalinichenko Feb 12 '12 at 18:44
What I mean is, you have a `//crashes` at a line that should not crash. – bdonlan Feb 12 '12 at 18:45
oh that's partly my fault! I should make it clearer like **CRASHEZ!** – Caffeinated Feb 12 '12 at 18:49
@downvoter it is generally a good idea to explain your downvotes, especially for simple, correct answers. Once sentence is usually enough. – Sergey Kalinichenko Feb 12 '12 at 22:13

score 1 · Answer 6 · answered Feb 12 '12 at 18:49

There are two, unrelated, problems in that code. First, the const is not in the right place.

char * const p = "john";
char const * p = "john";
const char * p = "john";

The latter two are pointers to unmodifiable strings. If you had done this, then the code would not have compiled.

The first option char * const is not really a read-only variable. It means a pointer which points to modifiable data, and that the pointer cannot be changed such that it points to another string. But that's not relevant to your problem.

Your constis not relevant here. You have attempted to modify a string that shouldn't be modified. Literal strings should never be modified, and this is the cause of your crash.

score 0 · Answer 7 · answered Feb 12 '12 at 19:17

p1 is a pointer to constant data, not a constant pointer. Moreover it points to a literal constant, which typically resides in the code space, and modern operating systems and processors usually protect against code trying to modify code space.

You should not be surprised that it crashes, but in general the behaviour is undefined.

Why is this C code involving constant pointer crashing on me?

7 Answers7