Programming Output Evaluation

The while loop copies 'name' to p2 but both p1 and p2 are incremented after each assignment. When the loop exits, p2 points to the null terminator, so printf(%s, p2) prints an empty string. The key is understanding post-increment behavior.

This is identical to question 82605. The while loop copies 'name' to p2 but both p1 and p2 are incremented after each assignment. When the loop exits, p2 points to the null terminator, so printf(%s, p2) prints an empty string.

x=y++ + x++: x = 35 + 20 = 55, then y becomes 36, x becomes 56. y= ++y + ++x: y increments to 37 first, x increments to 57 first, so y = 37 + 57 = 94. Final output: 57 94. Requires understanding pre and post increment.

x = 5. x<<2 (left shift by 2) = 5 * 4 = 20. x>>2 (right shift by 2) = 5 / 4 = 1 (integer division). Output: 5,20,1. Left shift multiplies by 2^n, right shift divides by 2^n.

Tracing: x=20, y=35. x=y+++x++ uses original values: x=35+20=55, then y++ makes y=36, x++ makes x=56. y=++y+++x: y becomes 37, x becomes 57, then y=37+57=94. Output: 57 and 94.

++i is pre-increment (increments first, then returns new value), so i becomes 2 and prints 2. i++ is post-increment (returns current value, then increments), so it prints 2 again (current value) then increments to 3. Output: 2,2. Note: void main() is non-standard but works in some compilers.

The statement 'System.out.println("i="+++i);' contains a syntax error. The '++' is the pre-increment operator, but the expression '+++i' (which could be parsed as '++ + +i' or similar) is invalid Java syntax. In Java, you cannot have three plus signs in a row like this - it's ambiguous and the compiler will reject it.

In Java, when you write 'int x=010;', the leading zero makes it an octal literal. Octal 010 equals decimal 8 (1*8^1 + 0*8^0 = 8). So the output is 'X=8'.

The code int x=y=10; attempts to declare variables x and y and assign them values in a single statement. In Java, this syntax is invalid because y is being used before it has been declared - you cannot chain variable declarations like this when the first variable doesn't exist yet. The correct way would be int y=10, x=y; or declare them separately. Therefore, a compile-time error occurs.

The bitwise OR operation i|j compares each bit of i (binary 11 = decimal 3) and j (binary 10 = decimal 2). Bitwise OR produces 1 if at least one operand has a 1 in that position. So: 11 | 10 = 11 (binary) which equals 3 in decimal. The output is i|j=3.

The prefix 0x indicates a hexadecimal literal in Java. The value 0x10 in hexadecimal equals 16 in decimal (1×16 + 0×1 = 16). Therefore, the output is X=16.

In Java (and several other programming languages), integer literals starting with 0 are interpreted as octal (base-8) numbers, not decimal. The literal '010' in octal equals 8 in decimal (1×8¹ + 0×8⁰ = 8). Therefore, when the code prints 'X=' + x, it outputs 'X=8'. This is a common trick question testing knowledge of octal notation in programming languages.

The bitwise OR operator (|) compares each bit position and returns 1 if either operand has a 1 at that position. Binary 3 is 011 and binary 2 is 010, so 3|2 = 011|010 = 011 = 3.

In Java, hexadecimal literals begin with 0x. The value 0x10 in hexadecimal equals decimal 16 (1 × 16^1 + 0 × 16^0 = 16). The prefix indicates number system, not value.