What is the output of the following C program? void main() { int k, val = 1; k…

Concept

Because the same variable val is modified twice in this one expression (between two sequence points), standard C does NOT define a single guaranteed result — the output is undefined and can vary by compiler. The conventional model used by legacy compilers such as Turbo C, which this DSSSB key follows, is: each prefix ++val increments val before its value is used, so by multiplication time both factors read the same latest value of val. Under that convention the program prints 9.

Application

1. Start: val = 1.

2. First ++val raises val from 1 to 2.

3. Second ++val raises val from 2 to 3.

4. Both factors now refer to val, which holds 3, so the product is 3 × 3 = 9.

5. Hence k = 9 and printf prints 9.

Important caveat (why answers differ)

This expression modifies val twice between two sequence points, so the C standard leaves the result undefined: the compiler is free to print different values. Legacy academic compilers such as Turbo C apply both increments first and then multiply, giving 3 × 3 = 9; some modern optimising compilers may instead print 6 (2 × 3). For this DSSSB question the official key follows the Turbo C convention, so the expected answer is 9.

Cross-check

This confirms the result is self-consistent under the Turbo C / official-key convention: if both ++val operations complete before either factor of the multiplication is read, val finishes at 3 and 3 × 3 = 9. This is NOT an independent proof under the C standard — the two increments are unsequenced, so a compiler that reads one factor before applying the other operand's increment can legitimately print a different value (e.g. 6).