Predict the output when using ungetc correctly on stdin. The user repeatedly inputs the character 'a':\n\n#include<stdio.h>\n\nint main()\n{\n int i;\n char c;\n for(i=1; i<=5; i++)\n {\n scanf("%c", &c); /* given input is 'a' */\n printf("%c", c);\n ungetc(c, stdin);\n }\n return 0;\n}\n\nWhat is printed?

Introduction / Context:
ungetc allows you to “push back” a character onto an input stream so that a subsequent read returns it again. This pattern can intentionally repeat input.

Given Data / Assumptions:

• The loop runs five times.
• Each iteration reads one character from stdin, prints it, then pushes it back onto stdin.
• The initial input supplied is 'a' and remains available via ungetc for the next iteration.

Concept / Approach:
After printing the character, ungetc(c, stdin) ensures that the next scanf("%c", &c) reads the same character again. This repeats for every loop iteration, producing the same output character each time.

Step-by-Step Solution:
Iteration 1: read 'a', print 'a', push back 'a'.Iteration 2..5: read the pushed-back 'a', print, push back again.Total visible output: five 'a' characters concatenated.

Verification / Alternative check:
Replacing ungetc with no pushback would consume the input and require additional characters to be provided by the user for subsequent iterations.

Why Other Options Are Wrong:
aaaa: only four characters; the loop prints five.Garbage value / Error: the code is valid and deterministic here.

Common Pitfalls:
Forgetting that ungetc only guarantees one character of pushback; excessive pushback may fail depending on the implementation.

Final Answer:
aaaaa