Comments on: problem

By: jatinder.kumar

jatinder.kumar — Thu, 03 Oct 2013 17:56:35 +0000

*p get memory as read only. So when you printf(“%c\n”,–*p–); mesn you are goinh to change the contents of read only memory so that why you are getting segmantion fault

By: jasmeen.kaur

jasmeen.kaur — Mon, 30 Sep 2013 04:49:42 +0000

if it is — then –*p means u are changing the value at adress p(which is readonly) which will give you segmentation fault .
*p– means you are decrementing address i.e now it will point to ‘e’

By: jasmeen.kaur

jasmeen.kaur — Sun, 29 Sep 2013 17:25:40 +0000

Can u pls tell what is written before *p in printf statement

By: Ram D

Ram D — Wed, 11 Sep 2013 16:58:25 +0000

run following code in ur c99 compiler and compaire with the above one

#include
#include

void main()
{
char *p=”abcdef”;
for(int i=0;i<=6;i++)
{
p = p + (strlen(p)-i);
// printf("%c\n",–*p–);
// printf("%c\n",–*–p);
// printf("%c\n",–*(p–));
printf("%p\n",p);
printf("%c\n",*p);
}
}