Programming Output Evaluation

The left shift operator (<<) requires integer operands. Option B compiles because both i and j are int types, so i<

In Java, String's toString() method returns the same String object (this). The expression "String".toString() returns the literal "String" itself. Since it's the same literal, the == comparison returns true, so the code prints "Equal". The == operator compares object references for String, not content.

The Hibernate mapping configuration with automatically generates incrementing numeric IDs for new entities. When you save a new object, Hibernate reads the next ID value from the database and assigns it. This is a built-in ID generation strategy that requires no manual intervention.

In Java, integer division or modulo operations by zero using the / or % operators will throw an ArithmeticException. Floating-point division by zero does not throw an exception but results in Infinity or NaN.

The program declares four integer variables (a=1, b=2, c=3, d=4) and prints them using two printf statements. The first printf outputs a and b, the second outputs c and d with a leading space. The output is 1 2 3 4. Option A and B are incorrect because the code compiles and runs without errors. Option D is incorrect because option C is valid.

The expression uses the comma operator which evaluates left operand and returns right operand. --a is prefix decrement so a becomes 0. b++ is postfix increment so b returns 2 then becomes 3. The comma expression (--a, b++) returns 2. Then c = 2 - 3 = -1. Output: a=0, b=3, c=-1. Option A is wrong (c=-3 incorrect). Option D is wrong (c=0 incorrect).

Evaluate comma expressions left to right, returning rightmost value. (a,a) returns a=1. (b,c) returns c=3. (c,d) returns d=4. (d,b) returns b=2. So e = 1 + 3 + 4 - 2 = 6. Option A is wrong - no compile error. Option B would require different arithmetic. Option D is incorrect.

The format string "%.2" is incomplete - it lacks a type specifier like 'f' for float. In C, this is undefined behavior, but many implementations will simply print the format string literally when it's invalid, resulting in "%.2" as output. This is a common trick question testing knowledge of proper printf format syntax.

The semicolon ';' is a statement terminator in C, not an operator. Having two semicolons 'int b=6;;' is valid - the second semicolon just creates an empty statement. The program correctly computes c = 5 + 6 = 11 and prints it.

The nested loops iterate i from 1-3 and j from 1-2. When i==j (i=1,j=1), continue skips to next j. For i=2, j=2: j%3=2 > 1 is false (2 is not > 1), so no break. printf prints '2'. For i=3, j=1: j%3=1 is not > 1, printf prints '3'. Output: 2,3.

The swap macro expands to multiple statements separated by semicolons, not a single compound statement. Without braces, only 'temp=a' executes conditionally. Since 5 > 6 is false, the entire block is skipped. temp is static and initialized to 0 by default. printf outputs 'a=6 b=0' because 'a=b; b=temp;' execute unconditionally after the if block.

val=5 is unsigned int (typically 16-bit range 0-65535). val-11 = 5-11 = -6, which wraps around to 65530 (65535-6+1). The %u format prints unsigned values, so output is 5 65530. Option A is wrong - no compile error. Option B shows signed behavior which %u won't display.

*&z is a pointer dereference of z's address, which just gives z. *&z *= *&x * *&y becomes z *= x * y, or z = 2 * 4 * 3 = 24. The *& operators cancel out - & takes address, * dereferences it. This is a pointer trick question testing understanding of address and dereference operators.