In C, how would you print the two literal characters backslash and n (i.e., the text ' ') on the screen using printf?

Introduction / Context:
Escape sequences in C string literals can be confusing. The backslash introduces special sequences like \n (newline). To print a backslash character itself, you must escape it as \\ in the source to produce a single backslash at runtime. This question asks for the exact printf form to display the text \n literally, not to create a newline.

Given Data / Assumptions:

• You want the output characters backslash followed by the letter n.
• Using the standard C library function printf.
• No extra whitespace or newlines are required.

Concept / Approach:
In C string literals, "\" represents a single backslash character. Therefore, "\n" in a literal results in two output characters: backslash and n. Passing that literal to printf prints \n exactly as text. Using '\n' (single quotes) forms a character literal representing newline, not the two-character sequence.

Step-by-Step Solution:
Write a string literal that encodes backslash: "\".Append the character n to form "\n".Call printf("\n"); to print those two characters.Do not use character literals or unescaped backslashes.

Verification / Alternative check:
Running printf("\n"); produces backslash-n on the console. Running printf("\\n"); would print two backslashes followed by n (i.e., \\n).

Why Other Options Are Wrong:

• printf("\\n"); prints \\n (two backslashes), not \n.
• printf('\n'); uses a character literal for newline, not the two characters.
• The combined comment form is unnecessary and potentially misleading.
• puts("\n"); also prints \n but appends a newline automatically; the question asks specifically for printf and minimal form.

Common Pitfalls:
Forgetting that backslash must be escaped in string literals. Confusing character and string literals leads to different behavior.

Final Answer:
printf("\n");.