c - 迭代 C 字符串不起作用

Question

首先，我对这段代码的目标是:将一个句子放入一个 C 字符串中。遍历句子并查看特定字母出现了多少次。

此代码有些工作但没有给出正确的数字？不知道为什么:

#include <stdio.h>
#include <string.h>

int tracker=0;
int letterCount (char *sentence)
{
    int s=strlen(sentence);
    int i=0;


    for (i=0; i<s; i++){
        if (sentence[i]=='h') {
            tracker++;
        }
    }
    return tracker;
}

int main(int argc, const char * argv[])
{

    char *string="Hi there, what's going on? How's it going?";


    letterCount(string);

    printf("this sentensce has %i H's", tracker);


    return 0;
}

我得到的输出:

this sentensce has 2 H's

不太对。有什么想法吗？

Answer 1

如果你的意思是不区分大小写，这是正确的代码 H:

#include <stdio.h>
#include <string.h>

int tracker=0;
int letterCount (char *sentence)
{
    int s=strlen(sentence);
    int i=0;


    for (i=0; i<s; i++){
        if (sentence[i]=='h' || sentence[i]=='H') {  //'h' is not the same as 'H'
            tracker++;
        }
    }
    return tracker;
}

int main(int argc, const char * argv[])
{

    char *string="Hi there, what's going on? How's it going?";


    letterCount(string);

    printf("this sentensce has %i H's", tracker);


    return 0;
}

您刚刚在代码中拼错了小写字母和大写字母。 记住，C 语言是区分大小写的!