Use VC++2010 to open the solution in the candidate folder blank1. The project contains one source file blank1.c. The function fun() judges whether the string pointed to by parameter s is a palindrome . If it is a palindrome, the function returns 1. If not, it returns 0. A palindrome reads exactly the same forward and backward, case‑insensitive. For example: "LEVEL" and "Level" are palindromes, while "LEVLEV" is not.
Complete the program by filling in the blanks. Do not add or delete lines, do not change program structure.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int fun(char * s)
{
char * lp, * rp;
/**********found**********/
lp = 【1】 ;
rp = s + strlen(s) - 1;
while( (toupper(*lp)==toupper(*rp)) && (lp < rp) )
{
/**********found**********/
lp ++; rp 【2】 ;
}
/**********found**********/
if (lp < rp) 【3】 ;
else return 1;
}
main()
{
char s[81];
printf("Enter a string: ");
scanf("%s",s);
if(fun(s))
printf("\n\"%s\" is a Palindrome.\n\n",s);
else
printf("\n\"%s\" isn't a Palindrome.\n\n",s);
}Answers
【1】s 【2】-- 【3】return 0
Explanation
Return value rule
return 1: It is a palindrome.return 0: It is NOT a palindrome.
The program uses two pointers:
lp: left pointer, starts at the first character of string.rp: right pointer, starts at the last character of string.
while( (toupper(*lp)==toupper(*rp)) && (lp < rp) ) Loop runs only when two conditions are both true:
- The uppercase of left‑side character equals uppercase of right‑side character.
- Left pointer is still before right pointer (
lp < rp).
Inside loop: move left pointer forward, move right pointer backward toward the middle.
When the while loop stops:
- If
lp < rpis still true → we met different characters. Not palindrome → return 0. - If pointers meet or cross each other → all matched. Is palindrome → return 1.
Test case 1: Even‑length palindrome ABCCBA（偶数长度）
Characters: A B C C B A
- lp points to A, rp points to A → equal, lp++, rp--
- lp points to B, rp points to B → equal, lp++, rp--
- lp points to C, rp points to C → equal, lp++, rp-- Now
lpis at index 3,rpis at index 2.lp < rp→3 < 2→ false. While loop exits. Skipif(lp<rp), executeelse return 1. Result: it is a palindrome.
Test case 2: Odd‑length palindrome ABCBA（奇数长度）
Characters: A B C B A
- lp=0(A), rp=4(A): equal → lp++, rp--
- lp=1(B), rp=3(B): equal → lp++, rp-- Now lp and rp both point to middle character
C(index 2). Enter while‑loop: characters equal,lp<rp(2<2) is false. Loop condition fails. Loop stops directly. Executeelse return 1. Result: it is a palindrome.
One non‑palindrome example ABCDE
First compare A vs E: not equal. The while loop stops immediately. lp < rp holds true. Run return 0. Output: not a palindrome.
Why return 0?
0stands for logical false in C language, meaning the proposition "this string is palindrome" is false.1stands for logical true, meaning the proposition holds true.
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