Why do i get a Segmentation fault

perfect_circle · 12-09-2004, 06:25 PM

I think i found it.

this works fine:

Code:

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

int main()
{
	char s[]="bla";
	strcpy(s,"bla");
	printf("%s\n",s);
}

But this causes segmentation fault:

Code:

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

int main()
{
	char *s="bla";
	strcpy(s,"bla");
	printf("%s\n",s);
}

change your

Code:

 char *dev="tap";

to

Code:

char dev[]="tap";

Quote:

Code:

char s[10];
s = NULL;

sorry.. this is not allowed.

itsme86 · 12-09-2004, 07:13 PM

Quote:

Originally posted by perfect_circle
Any array

Code:

char s[10];

is a pointer pointing to a statically allocated memory.
i think if u do :

Code:

char s[10]; s = NULL;

will make you loose the allocated memory.
I don't remember how memset works
but s is still just a pointer.

Code:

itsme@dreams:~/C$ cat -n bzzt.c
     1  #include <stdio.h>
     2
     3  int main(void)
     4  {
     5    char s[10];
     6
     7    s = NULL;
     8    return 0;
     9  }
itsme@dreams:~/C$ gcc -Wall bzzt.c -o bzzt
bzzt.c: In function `main':
bzzt.c:7: incompatible types in assignment
itsme@dreams:~/C$

However, you can do:

Code:

char *s = malloc(10);
s = NULL;

In that event you lose track of your allocated memory. That's why the distinction between ifr_name being a pointer or an array was important.

perfect_circle · 12-09-2004, 07:37 PM

I know, i figured that by my self.

why is this:

Code:

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

int main()
{
	char *s="bla";
	strcpy(s,"bla");
	printf("%s\n",s);
}

causing a segmentation fault?

Code:

#include <stdio.h>

int main()
{
	char *s="bla";
	printf("%s\n",s);
}

this works file.

is char *s="bla"; allowed in ANSI C?

itsme86 · 12-09-2004, 09:20 PM

Compilers are not guaranteed to store string literals in writeable memory. The reason your example is failing is because you're trying to modify read-only memory (the memory that s is pointing to).

ta0kira · 12-10-2004, 07:14 AM

Quote:

Code:

char s[10];
s = NULL;

This should not work since s converts to a 'char * const &' lvalue for the assignment which cannot be modified.
ta0kira

[EDIT]
RE perfect_circle: string literals generally use the same memory, but the address is not guaranteed past the expression it is used in. This means that although 'char *s="bla";' leaves s with a pointer to internally used memory, beyond that statement, the memory may not be there anymore. That is why you should copy string literals into static arrays with []. At the point where "bla" is used in the function call in the following line, the memory location for storing a string literal may be changed, making s become invalid. "bla" is loaded into memory just long enough for it to initialize something, such as a local variable or a static array; the pointer to this temporary location should never be used. Your second block is the equivalent of 'printf("%s\n","bla");', which is why it works because you do not have any new string literals before the function call.
ta0kira

[EDIT again]
RE itsme86: it isn't read-only memory; if it was, the pointer could not be converted to a char*; you would get a compiler error in that case. The pointer cannot be changed, which is what I think you mean (try doing 'char *&s = "bla";' in C++ to see what I mean). Or maybe you mean that the mem used for the literal does not belong to the user's process, thereby causing the segfault, just like trying to use a bad pointer, which also points to non-user mem?
ta0kira

itsme86 · 12-10-2004, 09:33 AM

Please don't correct me with bad information:

This is from: http://www.faqs.org/faqs/C-faq/faq/

There are other references at the bottom if you need them.

Code:

1.32:	What is the difference between these initializations?

		char a[] = "string literal";
		char *p  = "string literal";

	My program crashes if I try to assign a new value to p[ i ].

A:	A string literal can be used in two slightly different ways.  As
	an array initializer (as in the declaration of char a[]), it
	specifies the initial values of the characters in that array.
	Anywhere else, it turns into an unnamed, static array of
	characters, which may be stored in read-only memory, which is
	why you can't safely modify it.  In an expression context, the
	array is converted at once to a pointer, as usual (see section
	6), so the second declaration initializes p to point to the
	unnamed array's first element.

	(For compiling old code, some compilers have a switch
	controlling whether strings are writable or not.)

	See also questions 1.31, 6.1, 6.2, and 6.8.

	References: K&R2 Sec. 5.5 p. 104; ISO Sec. 6.1.4, Sec. 6.5.7;
	Rationale Sec. 3.1.4; H&S Sec. 2.7.4 pp. 31-2.

Anyway, dereferencing a string literal through a pointer is not asking for trouble. It's perfectly valid. Just don't try to modify the string literal.

For instance:

Code:

{
  char *s = "some string";
  char *s2 = "some other string";

  while(*s)
    putchar(*s++);
  putchar('\n');
}

...is perfectly valid code. The memory for the string literals is guaranteed to be there and stay intact.