Consider the following C code: #include <stdio.h> void main() { signed int a = -1; unsigned int b = -1u; if (a == b) printf("The Lord of the Rings"); else printf("American Beauty"); } What will be the output when this program is executed on a typical 32 bit system?

Introduction / Context:
This question explores how C handles comparisons between signed and unsigned integers. The values involved are both set to represent minus one in different ways, and the comparison uses the C rules for the usual arithmetic conversions. Understanding how signed values are converted when compared to unsigned values explains why the if condition evaluates as it does.

Given Data / Assumptions:

• The type of a is signed int and it is initialized to minus one.
• The type of b is unsigned int and it is initialized to minus one with the literal -1u.
• We assume a typical 32 bit implementation where unsigned int is a 32 bit type with values from zero to 2^32 minus one.

Concept / Approach:
In two complement representation, minus one is stored as all bits set to one. For a, this bit pattern represents the signed value minus one. For b, the literal -1u is computed in unsigned arithmetic, so it also results in all bits set to one, which as an unsigned value is 2^32 minus one. When C compares a signed int and an unsigned int, it performs the usual arithmetic conversions. If the range of the unsigned type can represent all values of the signed type, the signed value is converted to unsigned before the comparison. In this case, a is converted to an unsigned value by interpreting its two complement bit pattern as an unsigned integer, which yields 2^32 minus one, the same as b.

Step-by-Step Solution:
Step 1: Represent a in binary. As a signed int with value minus one, its bits are all ones.Step 2: Represent b in binary. As an unsigned int with value -1u, it is computed in unsigned arithmetic and also results in all bits set to one.Step 3: Apply the usual arithmetic conversions for the comparison a == b. The signed value a is converted to unsigned int.Step 4: Interpreting all bits set to one as unsigned yields the value 2^32 minus one, which is exactly the value already stored in b.Step 5: Therefore, after conversion, both sides of the comparison have the same unsigned value, so the expression a == b is true and the if branch executes.

Verification / Alternative check:
Compiling and running this program on a standard 32 bit or 64 bit system that uses 32 bit ints produces the output The Lord of the Rings. Compilers may emit a warning about comparing signed and unsigned values, but they still perform the conversion and comparison as described. Changing the literal to a different value, such as -2 or another number, demonstrates how the rules affect equality in other cases.

Why Other Options Are Wrong:
Option B suggests that the else branch runs, which would require the values to differ, but they are equal after conversion.Option C claims a compilation error, but the C standard allows such comparisons, possibly with warnings.Option D mentions unpredictable output, but the behaviour is defined by the usual arithmetic conversions and is not undefined in this case.

Common Pitfalls:
Many programmers assume that a signed minus one and an unsigned maximum value can never compare equal, but due to the conversion rules, they can. Another pitfall is to ignore compiler warnings about signed and unsigned comparisons, which can hide subtle bugs when values are not carefully chosen. Always pay attention to type promotions and conversions when mixing signed and unsigned arithmetic.

Final Answer:
The correct answer is The Lord of the Rings.