Calloc causing segmentation fault

Here is my code:

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

int main(){
int n=10;
char *s= calloc(2,sizeof(char));
sprintf(s,"%d",n);
printf(s);
return 0;
}

The intent is to assing 2 digit number to a (char *). when I run the code, I get segmentation fault. Outout from valgrind is-

==18540== Command: ./test
==18540== 
==18540== Conditional jump or move depends on uninitialised value(s)
==18540==    at 0x366C06F397: _IO_str_init_static_internal (in /lib64/libc-2.5.so)
==18540==    by 0x366C063C8A: vsprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C04D677: sprintf (in /lib64/libc-2.5.so)
==18540==    b开发者_如何学Pythony 0x4004FC: main (test.c:8)
==18540== 
==18540== Conditional jump or move depends on uninitialised value(s)
==18540==    at 0x366C06E37B: _IO_default_xsputn (in /lib64/libc-2.5.so)
==18540==    by 0x366C043D38: vfprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C063C98: vsprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C04D677: sprintf (in /lib64/libc-2.5.so)
==18540==    by 0x4004FC: main (test.c:8)
==18540== 
==18540== Conditional jump or move depends on uninitialised value(s)
==18540==    at 0x366C06F20A: _IO_str_overflow (in /lib64/libc-2.5.so)
==18540==    by 0x366C06E3E3: _IO_default_xsputn (in /lib64/libc-2.5.so)
==18540==    by 0x366C043D38: vfprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C063C98: vsprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C04D677: sprintf (in /lib64/libc-2.5.so)
==18540==    by 0x4004FC: main (test.c:8)
==18540== 
==18540== Use of uninitialised value of size 8
==18540==    at 0x366C06F241: _IO_str_overflow (in /lib64/libc-2.5.so)
==18540==    by 0x366C06E3E3: _IO_default_xsputn (in /lib64/libc-2.5.so)
==18540==    by 0x366C043D38: vfprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C063C98: vsprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C04D677: sprintf (in /lib64/libc-2.5.so)
==18540==    by 0x4004FC: main (test.c:8)
==18540== 
==18540== Invalid write of size 1
==18540==    at 0x366C06F241: _IO_str_overflow (in /lib64/libc-2.5.so)
==18540==    by 0x366C06E3E3: _IO_default_xsputn (in /lib64/libc-2.5.so)
==18540==    by 0x366C043D38: vfprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C063C98: vsprintf (in /lib64/libc-2.5.so)
==18540==    by 0x366C04D677: sprintf (in /lib64/libc-2.5.so)
==18540==    by 0x4004FC: main (test.c:8)
==18540==  Address 0x0 is not stack'd, malloc'd or (recently) free'd
==18540== 

---

You allocate space for just 2 chars and then put the string '10\0' which needs another char to hold the null/0 terminating character. So you need to allocate 3 chars for this particular example to work.

Read up C Strings for full details.

---

You allocate 2 bytes, but the string size is 3:

'1', '0', '\0' (null terminator) the result is undefined in this case since you corrupt the heap

Also, when you allocate memory, don't forget to call free at the end.

---

You need to allocate room for the null character of a null terminated string. That's 3 characters total.

Also printf(s); should be: printf("%s", s);

---

For this particular piece of code the string "10" requires 3 bytes, which are '1', '0', and '\0' . So you need 3 bytes of memory to be allocated.