Question 1

C file I/O with fseek and fgets: if source.txt contains the single line "Be my friend", what does this program print?  #include<stdio.h>  int main() { FILE *fs; char c[10]; fs = fopen("source.txt", "r"); c[0] = getc(fs); // read 'B' fseek(fs, 0, SEE…

Accepted Answer

Introduction / Context: This problem checks your grasp of random file access with fseek and buffered line reads with fgets. You must reason about byte positions within the file containing the single line "Be my friend". Given Data / Assumptions: File contents (no trailing newline assumed): "Be my friend". Indices (0-based): B(0) e(1) space(2) m(3) y(4) space(5) f(6) r(7) i(8) e(9) n(10) d(11). fgets(c, 5, fs) reads at most 4 bytes and then appends '\0'. Concept / Approach: fseek(fs, 0, SEEK_END) places the file position one past the last character (index 12). The next call fseek(fs, -3L, SEEK_CUR) moves the position back three bytes to index 9, which is the letter 'e' of the word "friend". Then fgets reads up to 4 bytes starting there. Step-by-Step Solution: Start: after SEEK_END, position = 12.Back up 3 → position = 9 (character 'e').Call fgets(c, 5, fs): can read up to 4 characters: e(9), n(10), d(11), then EOF.c becomes "end" followed by '\0'.puts(c) prints "end" and adds a newline. Verification / Alternative check: If the file had a trailing newline, the three characters before EOF would typically still be 'e', 'n', 'd' for this exact string, so the result remains "end". Why Other Options Are Wrong: "friend"/"frien" require seeking much earlier than index 9 or using a larger read size. “Error in fseek” is incorrect because seeking from end with a negative offset is valid. “rien” would start at index 8, not 9. Common Pitfalls: Off-by-one mistakes at EOF; forgetting that fgets reads at most n-1 characters; assuming getc affects later fseek behavior (it does not here). Final Answer: end

Question 2

Self-referential format (a classic quine-style string): what exact output does this program produce?  #include<stdio.h> char *str = "char *str = %c%s%c; main(){ printf(str, 34, str, 34);}";  int main() { printf(str, 34, str, 34); return 0; }

Accepted Answer

Introduction / Context: This is a well-known technique where a format string contains a description of itself. By passing the string to printf along with character code 34 (the double quote) and the string pointer itself, the program reconstructs a readable representation of its own source. Given Data / Assumptions: str is a format string containing %c and %s placeholders. 34 is the ASCII code for the double-quote character. printf returns the number of characters printed but here we only care about the printed text. Concept / Approach: printf(str, 34, str, 34) substitutes the first %c with a double quote, the %s with the entire contents of str, and the last %c with a double quote again. The literal text outside the format specifiers is printed verbatim, thereby emitting an assignment of str followed by a call pattern that mirrors main’s printf. Step-by-Step Solution: Replace first %c → ".Replace %s → the exact bytes of str.Replace last %c → ".Remaining literal text prints unchanged.The concatenated result matches option (a) precisely. Verification / Alternative check: Running the code prints the string definition and the main call pattern on one line. Counting quotes confirms balanced quoting. Why Other Options Are Wrong: Option (b) omits the quoted replication of str's content. “No output” and “error” are false; the code compiles and runs. The last option misinterprets the substitution behavior. Common Pitfalls: Confusing %c with the character code 34; forgetting that %s consumes the whole string until '\0'; missing that the printed output contains both the definition and the function call text. Final Answer: char *str = "char *str = %c%s%c; main(){ printf(str, 34, str, 34);}"; main(){ printf(str, 34, str, 34);}

Question 3

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

Accepted Answer

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 "%%%%
". 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 "%"."%
" → 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 "
" with % escapes like "%%". Final Answer: %%

Input / Output Questions