What will be the output of the following program?

void main()
{
      char str1[] = "abcd";
      char str2[] = "abcd";
      if(str1==str2)
            printf("Equal");
      else
            printf("Unequal");
}

How this program runs explain

String str1 = “ABCD“ ;
String str2 = str1.toLowerCase();
str2.replace(‘a’ , ‘b’ );
str2.replace(‘b’ , ‘c’ );
System.out.printIn(str2);

We can not compare string with operators so i think ans c is right .

str1 and str2 are both separate memory allocations for the array elements, when the condition says str1==str2 , we are comparing its memory allocation and not the string i.e. "abcd", and since both the strings cannot be in the same memory location , therefore the answer will always be UNEQUAL no matter what the string may be, well atleast that is what I understood.

String is not compared the way it is being compared like integers. That's why the answer is "Unequal".

i thing c answer is correct

hi, friends !!!

if(str1==str2);// we will get if(0) i.e if(false)

so the else part is executed.

ans is error , braces missing sir

ANKITA U R RIGHT !!

It's a run time error...answer is c)error

what do compare == operator. Is this match address

i am not getting.
can u tell in simple...................