[SOLVED] segmentation fault during memcpy()

Aquarius_Girl · 02-20-2010, 01:12 AM

Please point out what wrong I have done here ?
Why is this giving me a segmentation fault ?

Code:

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

int main ()
{
	int  i  = 0;
	char *d = malloc ((sizeof(char)) * 50);
	
	memcpy ((void*) d, "anisha", 6);
	
	memcpy ((void*)(d + 6), (const void*)i, 4);
}

sparker · 02-20-2010, 01:19 AM

The problem lies in:

memcpy ((void*)(d + 6), (const void*)i, 4);

If you read the man page for memcpy it gives you the prototype:

void *memcpy(void *dest, const void *src, size_t n);

In this case you are trying to copy an integer into the destination, however it is not a pointer to int. The correct solution would simply be:

memcpy ((void*)(d + 6), (const void*)&i, 4);

Dan04 · 02-20-2010, 01:22 AM

You don't need casts to convert a pointer to void*. Your code gives a good reason not to do it, because without the casts, you would have gotten a useful compiler error.

Aquarius_Girl · 02-20-2010, 01:22 AM

Quote:

Originally Posted by sparker

The problem lies in:

memcpy ((void*)(d + 6), (const void*)i, 4);

If you read the man page for memcpy it gives you the prototype:

void *memcpy(void *dest, const void *src, size_t n);

In this case you are trying to copy an integer into the destination, however it is not a pointer to int. The correct solution would simply be:

memcpy ((void*)(d + 6), (const void*)&i, 4);

Many thanks for replying!!

I tried what you said it worked fine..

and then I tried the following also, it also worked fine !

Code:

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

int main ()
{
	int  *i = malloc ((sizeof(char)) * 50);
	*i = 0;
	
	char *d = malloc ((sizeof(char)) * 50);
	
	memcpy ((void*) d, "anisha", 6);
	
	memcpy ((void*)(d + 6), (const void*)i, 4);
	
	printf ("\n== %s ==", d);
}

Aquarius_Girl · 02-20-2010, 01:26 AM

Quote:

Originally Posted by Dan04

You don't need casts to convert a pointer to void*. Your code gives a good reason not to do it, because without the casts, you would have gotten a useful compiler error.

Thanks for replying!

Sorry I couldn't understand what you were trying to say, please explain it once more !

Aquarius_Girl · 02-20-2010, 01:38 AM

Ok now the segmentation fault problem has been solved, but when I am trying to print the string, it only shows the "anisha" string and not the number attached at the end ?

May I know what's the problem now ?

Code:

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

int main ()
{
	int  i  = 1;	
	char *d = malloc ((sizeof(char)) * 50);
	
	memcpy ((void*) d, "anisha", 6);
	
	memcpy ((void*)(d + 6), (const void*)&i, 4);
	
	memcpy ((void*)(d + 10), (const void*)"\0", 1);
	
	printf ("\n==%s==", d);
}

sparker · 02-20-2010, 02:06 AM

You could just use something like this instead

Code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
	char *p = malloc((sizeof(char)) * 50);
	
	snprintf(p, 50, "%s %d", "hello", 5);
	printf("%s\n", p);

       return 0;
}

Aquarius_Girl · 02-20-2010, 02:19 AM

Quote:

Originally Posted by sparker

You could just use something like this instead

Code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
char *p = malloc((sizeof(char)) * 50);

snprintf(p, 50, "%s %d", "hello", 5);
printf("%s\n", p);

return 0;
}

Many thanks for the useful inputs !!

Yes, the code shown by you worked ! I'll read about snprintf more.

Dan04 · 02-20-2010, 04:21 AM

Quote:

Originally Posted by anishakaul

Thanks for replying!

Sorry I couldn't understand what you were trying to say, please explain it once more !

If a function is declared as taking a void*, you can pass any type of pointer. You don't need a cast. And you shouldn't cast unnecessarily because casts hide compiler warnings. If you had written

Code:

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

int main ()
{
        int  i  = 0;
        char *d = malloc ((sizeof(char)) * 50);

        memcpy (d, "anisha", 6);

        memcpy ((d + 6), i, 4);
}

then gcc would have told you

Code:

foo.c:12: warning: passing argument 2 of ‘memcpy’ makes pointer from integer without a cast

With the cast, C assumes that you meant to reinterpret the integer as a pointer. If sizeof(int) == sizeof(void*) (which is generally true on 32-bit systems), you won't get a warning.

Because the integer happens to be zero, it's interpreted as a null pointer. Dereferencing it is an error.

Dan04 · 02-20-2010, 04:23 AM

BTW, if you malloc(), you have to free().

Aquarius_Girl · 02-20-2010, 04:41 AM

@Dan04

Thanks for the useful information !