Programming Output Evaluation

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.

The printf function expects a matching argument for its format specifier, but none is provided. This leads to undefined behavior, meaning the output is unpredictable.

Since printf has a format specifier but no arguments supplied, it reads from the stack or registers in an undefined manner, yielding unpredictable output rather than a compile error.

With three format specifiers and zero arguments passed to printf, the function attempts to read missing values from the stack, resulting in unpredictable output.

In C++, modifying a variable twice without an intervening sequence point causes undefined behavior. Here c++ both modifies c and yields its original value, but there's no sequence point between the two c evaluations in the cout statement. This violates sequencing rules, making the output undefined rather than predictably 01, 00, or 10.

In main, v is initialized to 10. Its address is passed to afunction. Inside afunction, the local pointer parameter x is reassigned to point to a newly allocated memory address, leaving the original variable v in main completely unmodified. Thus, v remains 10.