C：字符串问题

发布于 2024-08-22 08:23:15 字数 745 浏览 2 评论 0原文

我是 C89 的新手，不太了解字符串是如何工作的。我正在 Windows 7 上进行开发。

这是我在 Java 中尝试做的事情：

String hostname = url.substring(7, url.indexOf('/'));

这是我在 C89 中尝试执行此操作的笨拙尝试：

// well formed url ensured
void get(char *url) {
    int hostnameLength;
    char *firstSlash;
    char *hostname;

    firstSlash = strchr(url + 7, '/');
    hostnameLength = strlen(url) - strlen(firstSlash) - 7;
    hostname = malloc(sizeof(*hostname) * (hostnameLength + 1));
    strncpy(hostname, url + 7, hostnameLength);
    hostname[hostnameLength] = 0; // null terminate
}

更新以反映答案

对于 hostnameLength 共 14 个，hostname 是 malloc()'d 31 个字符。为什么会出现这种情况？

原文

I am new to C89, and don't really understand how strings work. I am developing on Windows 7.

Here is what I am trying to do, in Java:

String hostname = url.substring(7, url.indexOf('/'));

Here is my clumsy attempt to do this in C89:

// well formed url ensured
void get(char *url) {
    int hostnameLength;
    char *firstSlash;
    char *hostname;

    firstSlash = strchr(url + 7, '/');
    hostnameLength = strlen(url) - strlen(firstSlash) - 7;
    hostname = malloc(sizeof(*hostname) * (hostnameLength + 1));
    strncpy(hostname, url + 7, hostnameLength);
    hostname[hostnameLength] = 0; // null terminate
}

Update to reflect answers

For a hostnameLength of 14, hostname is malloc()'d 31 characters. Why does this happen?

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红颜悴 2024-08-29 08:23:15

// 现在怎么办？ 是 strncpy()：

hostname = malloc(hostnameLength + 1);
strncpy(hostname, url + 7, hostnameLength);
hostname[hostnameLength] = '\0'; // don't forget to null terminate!

// now what? is strncpy():

hostname = malloc(hostnameLength + 1);
strncpy(hostname, url + 7, hostnameLength);
hostname[hostnameLength] = '\0'; // don't forget to null terminate!

回复收藏 0 原文

可是我不能没有你 2024-08-29 08:23:15

之后，您需要执行以下操作：

hostname = malloc(sizeof(char) * (hostnameLength+1));
strncpy(hostname,  url + 7, hostnameLength);
hostname[hostnameLength] = 0;

有关复制的详细信息，请参阅 strncpy。它确实需要提前分配目标指针（因此是 malloc），并且只会复制这么多字符......

After that, you need to do:

hostname = malloc(sizeof(char) * (hostnameLength+1));
strncpy(hostname,  url + 7, hostnameLength);
hostname[hostnameLength] = 0;

See strncpy for details on the copying. It does require your destination pointer to be allocated in advance (hence the malloc), and will only copy so many characters...

回复收藏 0 原文

~没有更多了~