In the following pieces of code, B and D will compile without any error. True or

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In the following pieces of code, B and D will compile without any error. True or false ? A: StringBuffer sb1 = "abcd"; B: Boolean b = new Boolean("abcd"); C: byte b = 255; D: int x = 0x1234; E: float fl = 1.2;
Options
A. TRUE
B. FALSE

Correct Answer

TRUE

Explanation

The code segments B and D will compile without any error. A is not a valid way to construct a StringBuffer, you need to create a StringBuffer object using "new". B is a valid construction of a Boolean (any string other than "true" or "false" to the Boolean constructor will result in a Boolean with a value of "false"). C will fail to compile because the valid range for a byte is -128 to +127 (i.e., 8 bits, signed). D is correct, 0x1234 is the hexadecimal representation in java. E fails to compile because the compiler interprets 1.2 as a double being assigned to a float (down-casting), which is not valid. You either need an explicit cast, as in "(float)1.2" or "1.2f", to indicate a float.

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1. What is the output after compile and run the following code ? int Output = 10; boolean b = false; if((b == true) && ((Output += 10) == 20)) { System.out.println("We are equal " + Output); } else { System.out.println("Not equal! " + Output); }
Options
A. Compilation and output of "We are equal 10"
B. Compilation and output of "Not equal! 10"
C. Compilation error, attempting to perform binary comparison on logical data type
D. Compilation and output of "Not equal! 20"

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