string handling in c using pointers

Question

I have written the following program (it is given as an example in one of the best text books). When I compile it in my Ubuntu machine or at http://www.compileonline.com/compile_c_online.php, I get "segmentation fault"

The problem is with while( *p++ = *str2++)

I feel it is a perfectly legal program. Experts, please explain about this error.

PS: I searched the forum, but I found no convincing answer. Some people even answered wrong, stating that *(unary) has higher precedence than ++ (postfix).

#include <stdio.h>
#include <string.h>

int main()
{
char *str1= "Overflow";
char *str2= "Stack";

char *p = str1;
while(*p)
++p;

while( *p++ = *str2++)
;

printf("%s",str1);

return 0;
}

Thanks

score 3 · Answer 1 · answered Feb 07 '14 at 12:19

3

str1 and str2 point to string literals. You aren't allowed to modify those. Even if you could, there isn't enough memory allocated for the string to hold the characters you're trying to append. Instead, initialize a sufficiently large char array from a string literal:

char str1[14] = "Overflow";

answered Feb 07 '14 at 12:19

user2357112

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This will give "lvalue required as increment operand" error – Aravind mavnoor Feb 07 '14 at 12:24
@Aravindmavnoor: Did you try to do `str1++`? You still need the `p` variable. You can't increment an array. – user2357112 Feb 07 '14 at 12:25
@Aravindmavnoor: `str1++` is the thing you're *not* supposed to do, because it won't work. You need to declare `p` as a `char *` and initialize it to `str1`. – user2357112 Feb 07 '14 at 12:30
i haven't read ISO/IEC standard on C. Below link is useful http://stackoverflow.com/questions/2589949/c-string-literals-where-do-they-go – Aravind mavnoor Feb 07 '14 at 17:25

Lundin · Accepted Answer · 2014-02-07T13:01:12.787

1

I feel it is a perfectly legal program.

Unfortunately, it is not. You have multiple, severe bugs.

First of all, you are creating pointers to string literals char *str1= "Overflow"; and then you try to modify that memory. String literals are allocated in read-only memory and attempting to write to them results in undefined behavior (anything can happen).

Then you have while(*p) ++p; which looks for the end of the string, to find out where to append the next one. Even if you rewrite the pointers to string literals into arrays, you don't have enough free memory at that location. You must allocate enough memory to hold both "Overflow" and "Stack", together with the string null termination.

You should change your program to something like this (not tested):

#include <stdio.h>

int main()
{
  char str1[20] = "Overflow"; // allocate an array with enough memory to hold everything
  char str2[]  = "Stack"; // allocate just enough to hold the string "Stack".

  char *p1 = str1;
  char *p2 = str2;
  while(*p1)
    ++p1;

  while(*p1++ = *p2++)
    ;

  printf("%s",str1); // should print "OverflowStack"

  return 0;
}

Or of course, you could just #include <string.h> and then strcat(str1, str2).

edited Feb 07 '14 at 13:01

answered Feb 07 '14 at 12:29

Lundin

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`string.h`, with an `r` in it. – user2357112 Feb 07 '14 at 12:32
I accept your answer and it works fine. My doubt is I am reading "A Book on C by Al Kelly and Ira Pohl" in which author says that strcat can be implemented without using string.h in many ways and one way is as shown in my question which gives error. Why did author mislead ? – Aravind mavnoor Feb 07 '14 at 16:56
i haven't read ISO/IEC standard on C. Below link is useful http://stackoverflow.com/questions/2589949/c-string-literals-where-do-they-go – Aravind mavnoor Feb 07 '14 at 17:26
Author of A book on C is correct.IT was my mistake to add 2 statements containing string literals. Now my doubts is cleared – Aravind mavnoor Feb 08 '14 at 01:26

score 0 · Answer 3 · answered Feb 07 '14 at 12:18

0

Because you are crossing the string boundary of Overflow (str1) is why you are getting sigsegv.

str1 does not have enough memory allocated to accomodate beyond Overflow.

answered Feb 07 '14 at 12:18

Aniket Inge

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Why should compiler do that? – Aniket Inge Feb 07 '14 at 12:22
@Aravindmavnoor: If you expect C to do convenient things for you, you will usually be disappointed. – user2357112 Feb 07 '14 at 12:23
C compiler will do exactly what you tell it to do. – Aniket Inge Feb 07 '14 at 12:24
it is nothing to do with compiler. I cannot change pointer to string literals as in the program above – Aravind mavnoor Feb 08 '14 at 01:28

string handling in c using pointers

3 Answers3