segmentation fault run time error

Question

i have written a code . this looks as follows .

#include<stdio.h>

#include<string.h>

 int do_print2(char *q[]);
 int main()
{

  char *p[]={"pointervaule","NAM", "JAM", "CALM"};
  do_print2(p);
  return 1; 
}

int do_print2(char *p[])
{ 
    printf("this is print1 char *p \n");
    strcat(p[0],"added");
    printf("%s\n", (p[0]));
    return 1;

}

after compilation, i am trying to run it, i am getting segmentation fault. help me in learning what is the reason for that error. thank in advance.

score 4 · Accepted Answer · edited May 23 '17 at 12:21

4

In your code: strcat(p[0],"added"); try to write on read only memory that is illegal in C. Because p[0] points to a constant string.

Not p is pointer to char array, but not 2-dimensional char array.

Read: Difference between char* str[] and char str[][] and how both stores in memory? an answer with diagrams and code examples, to understand it better.

edited May 23 '17 at 12:21

Community

1
1

answered Jul 18 '13 at 20:48

Grijesh Chauhan

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hii Grijesh chauhan, firstly very thank full for helping me in fixing this error, and can u let me know a book or some material kind of thing that makes me comfortable in dealing with pointer and strings. ( any tutorial ? ) – Srinivas Thanneeru Jul 18 '13 at 20:55
@SrinivasThanneeru A huge list: [The Definitive C Book Guide and List](http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) and for [pointers](http://stackoverflow.com/questions/tagged/pointers) – Grijesh Chauhan Jul 18 '13 at 20:59
@SrinivasThanneeru If one need to write code in C he/she should read [FAQ](http://c-faq.com/) regularly – Grijesh Chauhan Jul 18 '13 at 21:03

score 3 · Answer 2 · answered Jul 18 '13 at 20:54

3

The OS says that the C strings are in the read section of the object (i.e. protected from writing).

Due to historical reasons "bla bla" is really a const char * const data type, but is allowed to get away in the C compilers eyes some teenage interdependencies. But the headmaster (OS) is less forgiving and expels the running of such code in the corridors. (how many metaphors in that statement).

answered Jul 18 '13 at 20:54

Ed Heal

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In `C` the type of string literal is `char[N]`. It is an array, not a pointer, but there's no `const` in the type. With `gcc` you can force the compiler to treat string litterals as `'const char *'`. The option `-Wwrite-strings` will do that. While it is **not** standard C, it is really useful to make the code more robust. – Grijesh Chauhan Jul 18 '13 at 21:10
I really meant it is in the read section of code i.e. not on the stack. – Ed Heal Jul 18 '13 at 21:24
Ok just wanted to inform about this, some people are not aware of this and also confuse between c and c++, (because in C++ string literals are `const char *`). Btw you have nice lines in second para. – Grijesh Chauhan Jul 18 '13 at 21:37

score 1 · Answer 3 · answered Jul 18 '13 at 21:45

You can't write to read only memory, better way to do this:

#include<stdio.h>

#include<string.h>

int do_print2(char q[][20]);
int main()
{
  char p[4][20] = {{0}, {0}, {0}, {0}};
  strcat(p[0],"pointervaule");
  strcat(p[1],"NAM");
  strcat(p[2],"JAM");
  strcat(p[3],"CALM");
  do_print2(p);
  return 1; 
}

int do_print2(char p[][20])
{ 
   printf("this is print1 char *p \n");
   strcat(p[0],"added");
   printf("%s\n", (p[0]));
   return 1;

}

segmentation fault run time error

3 Answers3