Understanding printf escapes: how many % signs are printed here? #include<stdio.h> int main() { printf("%%%% "); return 0; }

Introduction / Context:
printf uses % to introduce format specifiers. To print a literal percent sign, you must escape it by doubling: "%%". This problem asks how many percent signs appear when four percent characters are used in sequence.

Given Data / Assumptions:

• The format string is "%%%%\n".
• printf interprets each "%%" pair as one literal % in the output.
• No additional format arguments are supplied.

Concept / Approach:
Parse from left to right. The first pair "%%" becomes one %. The second pair "%%" becomes another %. The newline prints after them. Therefore the output consists of exactly two percent signs, then a newline.

Step-by-Step Solution:
"%%%%" → treat as "%%" + "%%"."%%" → prints "%".next "%%" → prints another "%"."%\n" → newline follows.Final console text: %% then newline.

Verification / Alternative check:
Replace "%%%%" with "%%" to see a single percent sign. Replace with "%%%%%%" (three pairs) to see three percent signs. The general rule is that 2n consecutive % characters print n percent signs.

Why Other Options Are Wrong:
Five percent signs would require ten % characters in the format. “No output” or “error” misunderstands printf’s escape semantics. A single % would require only "%%".

Common Pitfalls:
Forgetting that a lone % begins a format specifier and must be paired; mixing up backslash escapes like "\n" with % escapes like "%%".

Final Answer:
%%