#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;
}

• 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;
  }
  
• 9 years agoHelpfull: Yes(15) No(0)
• 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 :)
  
• 9 years agoHelpfull: Yes(8) No(0)
• jack jill
  
• 10 years agoHelpfull: Yes(2) No(4)
• jack jill cause its call by value not call by reference
• 9 years agoHelpfull: Yes(2) No(0)
• jackjill
  it will just print both the string jack and jill. Coz int the function func we are just assining jack again in s1
• 10 years agoHelpfull: Yes(1) No(0)
• 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.
• 6 years agoHelpfull: Yes(1) No(0)
• wont it b jill jack??
  
• 10 years agoHelpfull: Yes(0) No(7)