Respected Sir,
I am using gcc version 4.5.2 (Ubuntu/Linaro 4.5.2-8ubuntu4)
I have one doubt regarding malloc () in gcc.
When I tried to execute the program given below, it got executed.
But since 256 = 1 0000 0000 (9 bits) , How malloc can store it at 'p'
since I have allocate only 1 byte of memory (8 bits of memory) ?
Please guide me.
Program:
# include <stdio.h>
# include <stdlib.h>
int main(void)
{
int *p;
p=(int *) malloc(1);
*p=256;
printf("%d \t %ld \n",*p,p);
free(p);
return 0;
}
Terminal:
avinash@titanic:/Remastersys/Documents/Programs/c$ gcc malloc.c -o malloc
malloc.c: In function ‘main’:
malloc.c:12:3: warning: format ‘%ld’ expects type ‘long int’, but
argument 3 has type ‘int *’
avinash@titanic:/Remastersys/Documents/Programs/c$ ./malloc
256 8044560
--
Avinash Sonawane
PICT Pune
India
