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");
}
void main()
{
char str1[] = "abcd";
char str2[] = "abcd";
if(str1==str2)
printf("Equal");
else
printf("Unequal");
}A. Equal
B. Unequal
C. Error
D. None of these.
Answer: Option B
Solution(By Examveda Team)
Strings are compared using strcmp() function defined under string.h header file.
This Question Belongs to C Program >> Arrays And Strings
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Comments ( 12 )
Related Questions on Arrays and Strings
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...................