segmentation fault in strcpy
 

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

int main()
{
char *dest;
char *in="String1";

dest=(char*) malloc(20);
dest="data1";
strcpy(dest,in);
printf("%s",dest);
free(dest);
return 0;
}

output:
Segmentation fault


thanks & regards
dayalan

By assignment "data1" to dest, dest points to a buffer with 6 bytes, not 20 bytes.

Well, the problem is here:
The buffer addr value dest gets from malloc() is trashed by the string assignment, so you might as well not have used malloc() at all.

This is the kind of thing you need to do:

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

int main()
{
char *dest;
char *in="String1";

dest=(char*) malloc(20);
dest="data1 data1 data1 da"; --> now it is 20 bytes now also i am getting segmentation fault
strcpy(dest,in);
printf("%s",dest);
free(dest);
return 0;
}

output:
Segmentation fault

please give me some more information.

i have doubt in your answers can you please explain
 

please send me your answers.

Here's my explanation:
Tiil now, all is correct. You've just made pointer dest to point to location where the string "data1 data1 data1 da" is (statically) stored.
Now, you made the mistake. You're trying to modify the string "data1 data1 data1 da", by using a pointer. This is WRONG! To alter the string content, dest must be an array.
, where n is a value, works.

:-/ The OP doesn't contain any questions.. just a piece of code, the output and

I fail to see what the author wants.

You don't initialize a buffer you've created with malloc(), you write to it, for the simple reason that only writing to it will work. (Sorry, mistake here: 'Your first try at assignment overwrote the pointer to your buffer because you needed to try '*dest' instead of 'dest' ... but even using the pointer properly in this case will still fail. Try it and see.')

Also you should try turning on warnings (gcc -Wall -Wextra). You will get a warning if you try the assignment above.

All that matters with malloc() is the size of the block. Of course it returns a void pointer ... that means you don't have to cast it. The practice of casting malloc() is from the time before 'void' was implemented.

Put another way, once you assign the pointer from malloc to your variable that's been declared as a char pointer, what's left to worry about? Your char pointer is no less a char pointer.

More to the point, pigsa already gave you the answer to your original question, in the very first response:Here is a corrected version of your original program:

That's not really the answer. The answer is that the assignment in question creates a string literal, which will cause a segfault when you try to overwrite it no matter how many bytes are involved. (String literals are stored in protected memory.)

I misexplained what was wrong in my remarks above, as I corrected.

Of course, you also get a segfault from 'free (dest);' after dest has been overwritten.

The result of a buffer overrun as mentioned would be "undefined behavior" ... it might work, crash, or segfault. I'm pretty sure the code didn't execute to that point, though.

When there's more than one thing wrong, it's sometimes hard to nail the pertinent one.

Therein lies the nub of the error.

use :
This I think, should work fine. Following that , the free will also work correctly.
Some teacher taught me to always use strcpy , use something like what you have done only when variable has been declared.

Someone posted dest should be an array. Thats not true IMHO; cos when you did malloc you did as much.

Sorry , my english is not that good and I dont have access to GCC right now, so all this is from top of my head.

I can think whatever I want :P

Seriously though, why can't I write into src?
from what I read, static is something which doesn't lose value between function calls..

hai duryodhan,

you are correct static is only preserve the values for the variable. but one thing i just want to remember you


" Someone posted dest should be an array. Thats not true IMHO; cos when you did malloc you did as much "

in the case of array the same code will goes through without segmentation fault. i have reason for you because when compiler does convert to object file that time the array is becomes read and write permissions but in the case of pointer i am assigning "data1" constant data this might has read permission but not for write so it causes segmentation fault occur. But still i need more information why pointer gaves me segmentation fault when i assinged "data1".

you want me to give program for the array implemented code?


Thanks & Regards
dayalan

I meant to say that:
won't cause any segmentation fault.

PS: I really don't understand why some members keep saying stupidities, when I already explained what was the real problem.

When you do something like this:
...the compiler makes a static string "hello world" in the object file. This is in a read-only data segment (called .data). The variable named c, of type pointer to a char, is set to be the address of the first character of the string in the .data segment.

If you try to modify the address of the static string in the data segment, you get a segfault because that page of memory is marked as read-only.

So, in answer to the question "why does this make a segfault?", see comments here: