Transforming from char * to char[]

Question

I am getting segmentation fault in the following code

static char * result;

char s[31];

int i;

random_string(s, 10);



 // for (i = 0; i < 15; i++){
 //     result[i] = s[i];
 // }

strcpy(result, s);

printf("el result es %s\n", result);

where the function random_string is:

void random_string(char * string, int length)
 {
  /* Seed number for rand() */

 int i;

for (i = 0; i < length -1; ++i){

    string[i] = rand() % 90 + 65;
 }

 string[length] = '\0';
}

For some reason I am getting segmentation fault when using strcpy. Also copying byte by byte is not working. What is the problem? I am out of ideas.

You should allocate the memory for the array... Look up malloc... — ppeterka, Mar 14 '13 at 12:21
This very same beginner bug question has been asked 2 times earlier this hour only. Kindly put some minimum effort into researching before posting questions. — Lundin, Mar 14 '13 at 12:34

score 4 · Answer 1 · answered Mar 14 '13 at 12:20

4

The problem is that result has not been initialized. It results in undefined behavior. Before it can be used like that, you need to make sure it points to a valid buffer. For example:

result = malloc( strlen( s ) + 1 );

answered Mar 14 '13 at 12:20

Mark Wilkins

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Dmitry Sazonov · Answer 2 · 2016-06-15T08:07:29.300

2

You forgot to allocate memory to "result" pointer. Try next:

result = malloc( strlen( s ) + 1 );
strcpy(result, s);

edited Jun 15 '16 at 08:07

answered Mar 14 '13 at 12:21

Dmitry Sazonov

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3

Please [don't cast the return value of `malloc()`](http://stackoverflow.com/a/605858/28169) in C. – unwind Mar 14 '13 at 12:26
Thanks, I don't have deep knowledge of C. – Dmitry Sazonov Mar 14 '13 at 14:09

Leo Chapiro · Accepted Answer · 2013-03-14T12:36:57.940

2

static char * result; is just an address without any allocated memory!

Try this: [EDITED]

char * result = (char*)malloc(strlen(s) + 1);

edited Mar 14 '13 at 12:36

answered Mar 14 '13 at 12:22

Leo Chapiro

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yeah that was it. Thanks. And yeah I will accept your answer as soon as it lets me :) – Alessandroempire Mar 14 '13 at 12:26
5

This code snippet is kind of scary, of course you can't call `strlen(s)` before writing to `s`. I assume that the OP's code is implied to go between the two lines, somehow. – unwind Mar 14 '13 at 12:27
Please do NOT cast malloc return value - it can hide certain subtle errors. – paxdiablo Jun 14 '16 at 08:43

score 0 · Answer 4 · answered Mar 14 '13 at 12:21

0

result is just a uninitialised pointer. You must assign result to a character buffer before using strcpy.

answered Mar 14 '13 at 12:21

suspectus

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score 0 · Answer 5 · answered Mar 14 '13 at 12:22

You need to assign a valid memory region to your result pointer using malloc or using static memory, as you are doing with your s string. Otherwise your pointer has just a random location assigned and your program receives a segfault by accessing it as it lies out of the boundaries of your program.

score 0 · Answer 6 · answered Mar 14 '13 at 12:38

0

Declare result as a char array like static char result[10];

or assign some memory to result pointer.

answered Mar 14 '13 at 12:38

µtex

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Transforming from char * to char[]

6 Answers6