#include <stdio.h>
int main() {
char str[1001];
fgets(str, sizeof(str), stdin);
int i = 0;
while (str[i] != '\0' && str[i] != '\n') {
char current = str[i];
i++;
if (str[i] >= '2' && str[i] <= '9') {
int count = str[i] - '0';
for (int j = 0; j < count; j++) {
printf("%c", current);
}
i++;
} else {
printf("%c", current);
}
}
return 0;
}
This code attempts to decode a run-length encoded string where a character followed by a digit (2-9) means that character should be repeated that many times. For example, "a3" should output "aaa". However, the logic has a flaw in how it handles the digit and moves through the string.
- The code reads the current character, then immediately increments `i` to check the next character for a digit. But if the next character is not a digit, it prints the current character and moves on. However, after printing a repeated character, it increments `i` again, which may skip over characters that should be processed normally.
- Consider what happens when a digit is found: the loop prints the repeated character, then increments `i` to skip the digit. But the outer `while` loop will then check the next character. However, if the digit is the last character in the string, the increment may cause the loop to read past the null terminator.