segmentation fault error in C using strcpy

Question

char** key;

strcpy(*key, "Hello");
strcpy(*(key+1), "World");

printf("%s", *key);

The second strcpy has no error, while the first strcpy has a segmentation fault. How should I modify to achieve the original purpose?

Uh. Is this the whole of your code? You haven't initialized key to point to anything. So its an uninitialized pointer that you are dereferencing to char* to pass to strcpy. — kfsone, Oct 09 '13 at 05:50

score 1 · Answer 1 · edited Oct 16 '13 at 13:17

1

What you are doing is Undefined Behavior.

char * strcpy ( char * destination, const char * source )

strcpy expects a destination which can be modified. You are passing a char ** which is causing the problem since you have not allocated any memory for it to write to.

This is what (perhaps) you were trying:

  char* key = malloc(sizeof(char)*7); // 7 because it can Hold World with a Nul character

  strcpy(key, "Hello");
  strcpy((key+1), "World");

  printf("%s", key);

edited Oct 16 '13 at 13:17

answered Oct 09 '13 at 05:45

Sadique

22,181
7
62
90

would you please explain a little bit more? – Song Zhibo Oct 09 '13 at 05:47
`sizeof(char)` is always redundant when used like this, it is *defined by the C standard* to be 1. – hyde Mar 01 '14 at 14:50

score 1 · Answer 2 · answered Oct 09 '13 at 05:48

1

It is not clear in your code whether you are allocating any buffer for key. I believe that's why you are getting SEG fault.

answered Oct 09 '13 at 05:48

Santanu C

1,302
3
19
36

score 0 · Answer 3 · answered Oct 09 '13 at 05:48

You must allocate memory to before you do strcpy(), if you want to skip allocating memory try strdup().

To simplify, not sure you really want char **key, doing with char *key

char* key = malloc(sizeof(char) * 100); //allocate to store 100 chars
strcpy(key, "Hello");
strcpy(key + strlen(key), "World");
//or strcat(key, "World");

the second strcpy has no error, while the first strcpy has a segmentation fault

How do you know 2nd one does not have error when it never executed because of segmentation fault?

Steve · Answer 4 · 2013-10-09T06:01:02.613

0

No memory has been allocated for key. You can allocate memory using malloc before the first strcpy.

*key = malloc(32 * sizeof(char));     // pick a size of buffer

I suspect the second call will also cause a segfault. It's (probably) going to be writing into unallocated space.

edited Oct 09 '13 at 06:01

answered Oct 09 '13 at 05:48

Steve

7,051
2
28
52

@DennisMeng Good point, I read the code incorrectly. – Steve Oct 09 '13 at 06:01

segmentation fault error in C using strcpy

4 Answers4