正如您在后面的示例中看到的那样,我正在比较歌词变量中的单词,并且我想获得count的唯一单词,但结果返回0。尽管您可以看到3个相似的词,但是if块不会进入。什么会导致此问题?

您可以快速通过this应用程序进行测试。

资源

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

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

    char lyric[] ="lorem ipsum is printing\nlorem ipsum is letters";
    int i = 0, j = 0, k = 0, y = 0, l = 0, s = 0;

    s = sizeof(lyric)/sizeof(char);

    for(k; k<s; ++k)
    {
        if(lyric[k] == ' ')
            ++l;

        if(lyric[k] == '\n')
            ++l;
    }

    char *row;
    char *temp[l];

    row = strtok (lyric," \n");

    while (row != NULL)
    {
        temp[i++] = row;
        row = strtok (NULL, " \n");
    }

    for (i=0;i<l; ++i){
        for (j=(i+1);j<l; ++j){

            printf("%s == %s\n", temp[i], temp[j]);
            if(temp[i] == temp[j]){
                y++;
            }
        }
    }

    printf("Unique word count : %d",y);

    return 0;
}


结果

lorem == ipsum
lorem == is
lorem == printing
lorem == lorem
lorem == ipsum
lorem == is
ipsum == is
ipsum == printing
ipsum == lorem
ipsum == ipsum
ipsum == is
is == printing
is == lorem
is == ipsum
is == is
printing == lorem
printing == ipsum
printing == is
lorem == ipsum
lorem == is
ipsum == is

Unique word count : 0

最佳答案

在您的代码中

temp[i] == temp[j]


因为操作数是指针,所以不能正常工作,而您想要的是比较指针指向的内存的内容,而不是指针本身。

您需要使用strcmp()进行比较。

关于c - 平等检查未按预期进行,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33778538/

10-14 18:43
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