In C programming, what will be the output when you execute the following code that uses the logical OR operator to compute an array index? #include <stdio.h> void main() { const int SIZE = 5; int expr; double value[SIZE] = { 2.0, 4.0, 6.0, 8.0, 10.0 }; expr = 1 || 2 || 3 || 4; printf("%f", value[expr]); }

Introduction / Context:
This question checks your understanding of the logical OR operator in C and how its result can be used as an array index. It also indirectly tests your knowledge of how C treats nonzero values in boolean expressions and how printf prints floating point values using the %f format specifier.

Given Data / Assumptions:

• SIZE is a constant with value 5, so the array value has indices from 0 to 4.

• value is initialized as { 2.0, 4.0, 6.0, 8.0, 10.0 }.

• expr is assigned using the logical OR operator: expr = 1 || 2 || 3 || 4;.

• printf uses "%f" to print value[expr] as a double with six digits after the decimal point.

• We assume the code compiles cleanly with proper headers.

Concept / Approach:
In C, the logical OR operator || evaluates its operands as boolean expressions. Any nonzero value is treated as true (1) and zero is false (0). The result of a logical OR expression is either 0 or 1. Also, logical operators use left to right evaluation and short circuiting. For an array, value[0] is 2.0 and value[1] is 4.0. If expr evaluates to 1, value[expr] will select the second element, 4.0.

Step-by-Step Solution:
Step 1: Evaluate 1 || 2. Since 1 is nonzero (true), the result of 1 || 2 is 1 and the right operand 2 is not even evaluated due to short circuiting. Step 2: Now the expression conceptually becomes 1 || 3. Again, 1 is true, so 1 || 3 is 1, with 3 not evaluated. Step 3: Finally we have 1 || 4. As before, the left operand is true, so the overall expression yields 1. Step 4: Therefore expr is assigned the value 1. Step 5: value[expr] is value[1], which from the initializer is 4.0. Step 6: printf("%f", value[expr]); prints 4.000000 using the default %f formatting.

Verification / Alternative check:
You can verify the behavior by changing the expression to expr = 0 || 0 || 0 || 0; which would evaluate to 0, causing the program to print value[0], which is 2.0. Similarly, if you used expr = 0 || 0 || 0 || 4;, the expression would evaluate to 1 because the last operand is nonzero, again selecting value[1]. This confirms that logical OR returns either 0 or 1 rather than combining bits as bitwise OR would do.

Why Other Options Are Wrong:
Option A (2.000000) would be correct only if expr were 0, which is not the case here. Option C (8.000000) would correspond to index 3, but expr never becomes 3 with the logical OR operator. Option D (Compilation error) is incorrect because the code is syntactically valid once stdio.h is included; using logical OR in an assignment expression is allowed and the array index is within bounds (0 to 4).

Common Pitfalls:
A common mistake is to confuse logical OR (||) with bitwise OR (|). Bitwise OR combines bits and could produce results like 7 in an expression such as 1 | 2 | 3 | 4, which might then cause an out of bounds index. However, the question specifically uses logical OR, which always yields 0 or 1. Another pitfall is to assume that all operands of || are evaluated; due to short circuit evaluation, once the result is determined, remaining operands are not evaluated, although that does not change the final numeric result in this case. Remember that array indices must be within range, and logical OR is a convenient way to reduce a chain of conditions to 0 or 1.

Final Answer:
The expression expr evaluates to 1, so value[1] is printed, and the output is 4.000000.