In C programming, we want to round off x (a float variable) to an int value. Which of the following expressions correctly rounds x to the nearest integer value and then converts it to int?

Introduction / Context:
In C programming it is common to store real numbers in float or double variables and then convert them to an int when we need a whole number. If we simply cast a float to int, the fractional part is truncated toward zero. To obtain proper rounding to the nearest integer we must adjust the value before casting. This question tests the correct expression for rounding a float x to an int in C.

Given Data / Assumptions:

• x is a positive float or double value that we want to round to the nearest integer.
• Y is an int variable that will hold the rounded result.
• We are using standard C syntax for casts and arithmetic expressions.
• The goal is to choose the option that performs rounding, not simple truncation.

Concept / Approach:
For non negative values, a simple way to round to the nearest integer is to add 0.5 and then truncate. In C, truncation is achieved by casting a floating value to int. Therefore the general pattern is Y = (int)(x + 0.5). The parentheses are important: they ensure that 0.5 is added to x while it is still a floating value, and only then is the sum converted to int. Any expression that casts only x and then adds 0.5 will not work correctly because 0.5 would be added after truncation.

Step-by-Step Solution:
Step 1: Start with the float value x that may contain a fractional part.Step 2: Add 0.5 to x while it is still a floating point value, forming x + 0.5.Step 3: Cast the entire sum (x + 0.5) to int using (int)(x + 0.5).Step 4: Because casting to int truncates toward zero, values in the interval [n - 0.5, n + 0.5) will be mapped to n for non negative x.Step 5: Store this result in the int variable Y, which will now hold the rounded value.

Verification / Alternative check:
Take x = 2.3. Using option A, Y = (int)(2.3 + 0.5) = (int)(2.8) = 2, which is the nearest integer. For x = 2.7, Y = (int)(3.2) = 3, again correctly rounded. If we tried option C, Y = (int)x + 0.5, we would get (int)2.7 + 0.5 = 2 + 0.5 = 2.5, which is not even an int and would be truncated later in unintended ways.

Why Other Options Are Wrong:
Option B is syntactically incorrect in C, because int is a type, not a function, so int(x + 0.5) is not valid. Option C casts x to int first and then adds 0.5, which does not perform rounding on the original float value. Option D first truncates x to int, adds 0.5 (still as int arithmetic), and then casts again; it effectively leaves the original truncation unchanged and does not implement proper rounding.

Common Pitfalls:
A common mistake is to forget the parentheses and write (int)x + 0.5 or int(x) + 0.5, which truncates before adding 0.5 and therefore does not round correctly. Another pitfall is assuming that a simple cast from float to int will automatically round instead of truncate. Programmers must deliberately add 0.5 (for positive values) before casting to achieve rounding.

Final Answer:
The correct way to round a positive float x to an int is Y = (int)(x + 0.5);.