获取 C 中字符串的最后一个标记

问题描述：

我想要做的是给出一个输入字符串，我将不知道它的大小或标记的数量，能够打印它的最后一个标记.

what I want to do is given an input string, which I will not know it's size or the number of tokens, be able to print it's last token.

例如:

char* s = "some/very/big/string";
char* token;

const char delimiter[2] = "/";

token = strtok(s, delimiter);

while (token != NULL) {
    printf("%s\n", token);
    token = strtok(NULL, delimiter);
}

return token;

我希望我的回归是

字符串

但我得到的是(空).任何解决方法?我在网上搜索过，似乎无法找到答案.至少对于 C 编程语言来说是这样.

but I what I get is (null). Any workarounds? I've searched the web and can't seem to find an answer to this. At least for C programming language.

答

如果您对特定字符进行标记，即在您的示例中是 '/'，则根本不需要对字符串进行标记: 调用strrchr 找到最后一个的位置'/'，并将 1 添加到结果指针以跳过分隔符，如下所示:

If you tokenize on a specific character, i.e. '/' in your example, you do not need to tokenize the string at all: call strrchr to find the position of the last '/', and add 1 to the resultant pointer to skip the delimiter, like this:

char *s = "some/very/big/string";
char *last = strrchr(s, '/');
if (last != NULL) {
    printf("Last token: '%s'\n", last+1);
}

演示.

获取 C 中字符串的最后一个标记

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