Latest Persistent Aptitude Question SOLUTION: #include func(char *s1,char * s2) { char *t; t=s1; s1=s2; s2=t; } int main() { char *s1="jack", *s2="jill"; func(s1,s2); printf(" %s %s ",s1,s2); return 0; }

• Placement Papers
• /
• Persistent
• /
• Category

Persistent Question Papers:

(#M40039755) PERSISTENT Category question predicate the o/p Keep an EYE

#include
func(char *s1,char * s2)
{
char *t;
t=s1;
s1=s2;
s2=t;
}
int main()
{
char *s1="jack", *s2="jill";
func(s1,s2);
printf("
%s
%s ",s1,s2);
return 0;
}

Asked In Persistent RAHUL UBHALE (9 years ago)

Unsolved Read Solution (7)

Is this Puzzle helpful?   (14)   (3) Submit Your Solution

Read Solution (7) :

Arnab Maity  (8 years ago)

The correct output is jackjill (since it is just following call by value mechanism).
To get output jill jack the following is the code :-

#include

func(char **s1,char **s2)
{
char *t;
t=*s1;
*s1=*s2;
*s2=t;
}
int main()
{
char *s1="jack", *s2="jill";
func(&s1,&s2);
printf("%s %s ",s1,s2);
return 0;
}

Is this solution Helpfull? Yes (15) | No   

Ankit Gupta  (9 years ago)

again function is not reyrning anything so the answr is jack jill,,,,,but if we print within the function then it must be jill jack :)

Is this solution Helpfull? Yes (8) | No   

Narendra Gopi  (9 years ago)

jack jill

Is this solution Helpfull? Yes (2) | No (4)  

Sourav Nandy  (9 years ago)

jack jill cause its call by value not call by reference

Is this solution Helpfull? Yes (2) | No   

aditya purohit  (9 years ago)

jackjill
it will just print both the string jack and jill. Coz int the function func we are just assining jack again in s1

Is this solution Helpfull? Yes (1) | No   

Rashi  (5 years ago)

Initially s1 =jack and s2=jill,then if func() the s1 value is stored in t and s2 value is stored in s1 so here s1=jill and again t value is stored in s2 so here s2=jack.
Ans= jill jack.

Is this solution Helpfull? Yes (1) | No   

aspirant  (9 years ago)

wont it b jill jack??

Is this solution Helpfull? Yes | No (7)  

Please Login to Read Solution.