C中结束char * [关闭](End of char* in C [closed])

C中结束char * [关闭](End of char* in C [closed])

我有这样的代码

int main() { char* word = "kotok"; char* word1 = word; while (word1 != '\0'){ printf("%s ", word1); word1++; } return 0; }

但是在打印“kotok，otok，tok，ok，k”之后它并没有结束。 此代码继续打印一些垃圾，然后停止。 怎么克服它？

我试着添加word1[strlen(word1)] = '\0'; 但它给出了分段错误。

I have such piece of code

int main() { char* word = "kotok"; char* word1 = word; while (word1 != '\0'){ printf("%s ", word1); word1++; } return 0; }

But it doesn't end after printing "kotok, otok, tok, ok, k". This code continues prints some garbage and then stops. How to overcome it?

I tried to add word1[strlen(word1)] = '\0'; but it gives a segmentation fault.

最满意答案

word1是一个指针。 '\0'是一个角色。 这就是为什么word1 != '\0'没有达到预期的效果。

您必须取消引用指针以获取它指向的字符：

while (*word1 != '\0') {

word1 is a pointer. '\0' is a character. This is why word1 != '\0' doesn't do what you expect.

You have to dereference the pointer to get the character it points to:

while (*word1 != '\0') {