main() { char *p1="Name"; char *p2; p2=(ch

main()
{
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
while(*p2++=*p1++);
printf("%s\n",p2);
}

Question Posted / guest

4 Answers
9810 Views
CitiGroup, I also Faced
E-Mail Answers

Answers were Sorted based on User's Feedback

main() { char *p1="Name"; char *p2; p2=(char *)malloc(20); ..

Answer / guest

error

Is This Answer Correct ?

2 Yes

0 No

main() { char *p1="Name"; char *p2; p2=(char *)malloc(20); ..

Answer / subbu[iit kgp]

the given program gives some meaningless output, with some
modification to the given program as

#include<stdio.h>
#include<stdlib.h>
main()
{
char a[]="ramesh";
char *p1="Name";
char *p2=a;

while(*p2++=*p1++);/*copies contents of p1 to
p2*//* here it is not possible to use while(*a++=*p1++)
because a can not change its value*/
*p2='\0';
printf("%s\n",a);

}

The output will be Name

Is This Answer Correct ?

2 Yes

1 No

main() { char *p1="Name"; char *p2; p2=(char *)malloc(20); ..

Answer / friend

i think above program ans ramesh

Is This Answer Correct ?

0 Yes

0 No

main() { char *p1="Name"; char *p2; p2=(char *)malloc(20); ..

Answer / vignesh1988i

actually in this problem, the p2 will take characters after '\0' too from p1 upto the size of p2 come to an end.

so it prints
output :

Name #^$&dhd

thank u

Is This Answer Correct ?

1 Yes

1 No

Post New Answer