Java Programming - Flow Control - Discussion
Discussion Forum : Flow Control - General Questions (Q.No. 3)
3.
public void test(int x)
{
int odd = 1;
if(odd) /* Line 4 */
{
System.out.println("odd");
}
else
{
System.out.println("even");
}
}
Answer: Option
Explanation:
The compiler will complain because of incompatible types (line 4), the if expects a boolean but it gets an integer.
Discussion:
15 comments Page 2 of 2.
Manoj said:
1 decade ago
Here if (odd) expects value of odd variable.
As declared earlier value of odd is 1 then condition holds true. Odd should be printed. Didn't understand why it is compiler error.
As declared earlier value of odd is 1 then condition holds true. Odd should be printed. Didn't understand why it is compiler error.
Prashant said:
1 decade ago
Here if (odd) expects value of odd variable.
As declared earlier value of odd is 1 then condition holds true. Odd should be printed. Didn't understand why it is compiler error.
As declared earlier value of odd is 1 then condition holds true. Odd should be printed. Didn't understand why it is compiler error.
Chandra bhushan said:
1 decade ago
If condition take two value 1 and 0 ant 1=true 0=false.
Now I is int value which contain value 1 so it should be true.
Now I is int value which contain value 1 so it should be true.
Jaladi said:
1 decade ago
Using the integer 1 in the while statement, or any other looping or conditional construct for that matter, will result in a compiler error. This is old C Program syntax, not valid Java.
AKASH said:
1 decade ago
I am confused about it. At 4th Line if expects any value if it's equal to 0 than compilation will go in if condition otherwise else condition.
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