In C programming, what is the output of this code that uses the left shift operator on an integer? #include <stdio.h> void main() { int x = 1, z = 3; int y = x << 3; printf("%d\ ", y); }

Introduction / Context:
This question checks your understanding of the bitwise left shift operator in C. Shifting an integer left by n positions is equivalent to multiplying it by 2^n, provided there is no overflow and the value remains within the valid range of the type.

Given Data / Assumptions:

• x is initialized to 1.

• z is initialized to 3 but is not used in the computation of y.

• y is assigned x << 3.

• printf prints y as a decimal integer.

• We assume a typical two's complement representation and that the result does not overflow int.

Concept / Approach:
The expression x << 3 shifts the bits of x three positions to the left. For positive integers that fit in the type, each left shift by one multiplies the value by 2. Therefore, shifting left by 3 multiplies the value by 2^3 = 8. Starting from x = 1, the result y should be 1 * 8 = 8, which is well within the range of int and causes no overflow.

Step-by-Step Solution:
Step 1: Write x in binary: 1 is 0001 in simplified 4-bit notation (and similar in larger width). Step 2: Shift left by 1 position: 0001 becomes 0010, which is 2 in decimal. Step 3: Shift left by another position: 0010 becomes 0100, which is 4 in decimal. Step 4: Shift left by a third position: 0100 becomes 1000, which is 8 in decimal. Step 5: Therefore x << 3 evaluates to 8, so y = 8 and printf prints 8 followed by a newline.

Verification / Alternative check:
Using the multiplication view, each left shift multiplies the value by 2, so x << 3 is x * 2^3 = 1 * 8 = 8. This matches the bitwise reasoning and confirms that the result is 8. Because 8 is a small positive integer, there is no risk of exceeding int's capacity on any realistic platform, so we do not need to consider overflow scenarios here.

Why Other Options Are Wrong:
Option A (-2147483648) might be associated with overflow scenarios in 32-bit two's complement int, but there is no overflow when shifting 1 by three bits. Option B (-1) could appear in questions about right shift of negative numbers, but here we are left shifting a positive number. Option C ("Run time error") is incorrect; bitwise shifts on valid operands do not cause runtime errors by themselves in this simple context.

Common Pitfalls:
One pitfall is confusing left shift << with right shift >>, which behaves differently, especially for signed negative values. Another common mistake is forgetting that shifting too far can cause undefined behavior or overflow, but in this case the shift is small and safe. Also, some learners misinterpret the presence of an unused variable (z) as relevant to the answer; in fact, it has no effect on y or the output at all.

Final Answer:
The value of y is 1 << 3 = 8, so the program prints 8.