我初学C， 运行环境是 Mac OS 10.7.4  xcode 4.1
刚在做一个课后题，写了以下程序， 运行结果很奇怪，麻烦高手帮忙看一下。谢谢！
这个程序是判断输入的单词里，是不是出现过的letter都正好出现过两次。比如Dada就符合要求。但是我输入horseshoer，返回结果是false。

/* Exercise 8.2*/

#include <stdio.h>
#include <stdlib.h>

int isDoubloon (char letter[])
{
    int histogram [26] = {0};
    for (int i=0; i<100; i++)
    {
        letter [i]= tolower (letter [i]);
        int index = letter[i] -'a';
        histogram [index]++;
    }
    int count=0;
    for (int j=0; j<26; j++)
    {
        if (histogram [j]==0 || histogram [j]==2)
        {
            count++;
        }        
       printf("j is %c, count is %i, histogram is %i\n",j+'a', count, histogram [j]);   /* This line is for debug */
    }
    if (count==26)
    {
        return 1;
    }
    return 0;
}

int main ()
{
    char word [100];
    printf("Please enter your word: ");
    scanf("%s",word);
    if (isDoubloon(word))
    {
        printf("True\n");
    }
    else
    {
        printf("False\n");
    }
    return 0;
}


输出结果：


Please enter your word: horseshoer
j is a, count is 1, histogram is 0
j is b, count is 2, histogram is 0
j is c, count is 3, histogram is 0
j is d, count is 4, histogram is 0
j is e, count is 5, histogram is 2
j is f, count is 6, histogram is 0
j is g, count is 7, histogram is 0
j is h, count is 8, histogram is 2
j is i, count is 9, histogram is 0
j is j, count is 10, histogram is 0
j is k, count is 11, histogram is 0
j is l, count is 12, histogram is 0
j is m, count is 13, histogram is 0
j is n, count is 14, histogram is 0
j is o, count is 15, histogram is 2
j is p, count is 16, histogram is 0
j is q, count is 17, histogram is 0
j is r, count is 18, histogram is 2
j is s, count is 19, histogram is 2
j is t, count is 20, histogram is 0
j is u, count is 21, histogram is 0
j is v, count is 22, histogram is 0
j is w, count is 22, histogram is 1
j is x, count is 23, histogram is 0
j is y, count is 24, histogram is 0
j is z, count is 25, histogram is 0
False

每次运行结果都不一样，总会有几行的histogram显示是1， 麻烦高手帮忙解答一下，谢谢！

[ 本帖最后由 rixobo 于 2012-9-16 10:24 编辑 ]

我也是刚入门 ，我勉强能看懂一点
逻辑上没错
猜测 如果V只有一个那么
当J为22的时候 就是v 那么histogram [j]==1
count不会加 才会出现下一个count和V一样
程序好像出错了
建议查找错误  先检查字符串长度是否和输入的一样
再输出 litter[i]-'a';
看看是不是有错误
也可以换另一种方法做
int k=1;
for(j=0;j<26j++) {
    if(histogram [j]!=0&&histogram!=2) {
        k=0;
        return k;
    }
}
return k;

输入的单词里面没有v或者w的。。。

慢慢看

字符串是以‘\0’结束：一下是两种输出，看有何不一样

#include <stdio.h>
void main()
{    char str[100];
    int i=0;
scanf("%s",str);
while(str[i]!='\0')
{
    printf("%4c",str[i]);
    i++;
}
printf("\n");
for(i=0;i<100;i++)
{
    printf("%4c",str[i]);
   
}
printf("\n");


}

显然第二种输出，出现了乱码。

所以楼主这种循环是否不妥，，
是否改用：

while(str[i]!='\0')

这种形式，

若说的不对请高手指教

谢谢！
改了之后程序对了！

额 长见识了

或者我在main里面
char word[100]={0};
应该也可以