for the first o/p qs
as p1 is moving towards the NULL('\0') character so is p2...to avoid dis
save the base address and den let p2 move forward with p1
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
char *save;
save = (char *)malloc(20);
save = p2;
if(p2==NULL)
cout<<"\n NOT ENOUGH SPACE";
else
{
while(*p2++=*p1++);
printf("%s\n",save);
}
for the second qs o/p = 25
cz post incr means first use den incr thererfore 5*5
>
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