Consider the following C program that compares a float and a double value initialised with 1.1. What will it print when executed? main() { float me = 1.1; double you = 1.1; if (me == you) printf("yes"); else printf("No"); }

Introduction / Context:
Floating point numbers in C are represented approximately using binary fractions, and float and double types have different precision. Comparing values of these types can lead to surprising results because decimal values like 1.1 cannot be represented exactly. This question checks whether the learner understands that a float initialised with 1.1 and a double initialised with 1.1 will not usually compare equal, even if they were written with the same decimal literal in the source code.

Given Data / Assumptions:
- The variable me is declared as float and initialised with the literal 1.1.
- The variable you is declared as double and initialised with the literal 1.1.
- The program compares me and you using the equality operator ==.
- If the comparison is true, the program prints the string yes, otherwise it prints No.
- We assume a typical implementation where float has less precision than double and both follow IEEE 754 rules.

Concept / Approach:
The key concept is that 1.1 is a double literal in C, and it is not exactly representable in binary. When assigning 1.1 to a float, additional rounding occurs because float has fewer bits of precision. The double variable you stores a more precise approximation of 1.1, while the float variable me stores a less precise approximation. When the equality comparison me == you is performed, me is promoted to double, but its less precise value does not match the more precise value stored in you. Therefore the comparison evaluates to false, and the else branch executes.

Step-by-Step Solution:
1. The float variable me is initialised: me = 1.1;. The double literal 1.1 is converted to float, losing some precision. 2. The double variable you is initialised: you = 1.1;. The same literal is stored as a double with more precision bits. 3. Internally, the value stored in me is a rounded float approximation; the value stored in you is a more accurate double approximation of 1.1. 4. When evaluating if (me == you), the float me is promoted to double for the comparison, but its approximate value still does not match the value in you. 5. Because the promoted float value and the double value are slightly different, the equality comparison yields false. 6. The false branch of the if statement executes, and the program prints the string No.

Verification / Alternative check:
If you modify the program to print the values with high precision using printf("%.20f %.20f", (double) me, you); you will see that the two numbers differ slightly in their least significant digits. This confirms that me and you are not exactly equal, even though their decimal initialisers look the same. Running the original program on a typical system will produce No, reinforcing the advice that direct equality comparisons between floating point values of different types or results of arithmetic calculations should be avoided in favour of tolerance based comparisons.

Why Other Options Are Wrong:
Option A is incorrect because it assumes that me and you are exactly equal, which is not true in most implementations due to rounding differences. Option C is wrong because the program structure does not allow both yes and No to be printed; exactly one of the branches will run. Option D is incorrect because there is nothing in the code that causes a compilation error as long as stdio.h is included; the types are compatible for comparison after usual promotions. Only option B correctly states that the program prints No, reflecting the failed equality comparison.

Common Pitfalls:
Programmers often think of floating point numbers as exact representations of decimal values and assume that if two variables are initialised with the same literal, they must compare equal. This question demonstrates that even small differences in precision can cause equality comparisons to fail. Another common pitfall is mixing float and double types in expressions without understanding the implicit promotions that occur. As a best practice, it is safer to use one floating type consistently in a calculation and to compare results using an absolute or relative tolerance instead of direct equality.

Final Answer:
Because the float and double approximations of 1.1 do not match exactly, the comparison me == you is false and the program prints "No".