Question

The Output of the statement $int\ x = 5;\ printf("\%d, x++");$ is:

• A. 5
• B. 6
• C. 0
• D. Error

Hint:

Contrast post-increment with pre-increment:
Post-increment ($x++$): Use the current value first, then increment. Output of ‘printf("%d", x++)‘ is 5.
Pre-increment ($++x$): Increment the value first, then use it. Output of ‘printf("%d", ++x)‘ would be 6.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks for the printed output of a C code segment involving variable initialization and the use of the post-increment operator inside a print function.

Step 2: Key Formula or Approach:
We must analyze the execution order of the post-increment operator ($x++$).
The post-increment operator uses the current value of the variable in the active expression first, and then increments the variable's value by 1.

Step 3: Detailed Explanation:

• First, an integer variable $x$ is declared and initialized to 5:
\[ \text{int } x = 5; \]

• Next, the print statement is executed:
\[ \text{printf("\%d", x++);} \]

• In this statement, the expression being evaluated is $x++$.

• Since $x++$ uses the post-increment operator:

• 1. The current value of $x$, which is 5, is passed as the argument to the ‘printf‘ function.

• 2. The ‘printf‘ function prints this value (5) to the console output.

• 3. After the value is retrieved for the print operation, the value of $x$ is incremented in memory by 1, making $x = 6$.

• If we were to print the value of $x$ in a subsequent line of code, the output would be 6.

• However, during the execution of the given statement, the printed output is 5.

Step 4: Final Answer
The output of the given statement is 5, which corresponds to option (A).