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/