将double转换为struct tm [英] Convert double to struct tm

问题描述

我有一个 double 包含秒。我想将它转换为 struct tm 。

I have a double containing seconds. I would like to convert this into a struct tm.

我找不到一个标准函数， 。我必须手动填写 struct tm 吗？

I can't find a standard function which accomplishes this. Do I have to fill out the struct tm by hand?

我只是有意转换为 time_t 和 http://www.StackOverflow.com 不会让我发布，除非我链接。

I just accidentally asked this about converting to a time_t and http://www.StackOverflow.com will not let me post unless I link it.

推荐答案

MSalters回答是正确的，但我想我会添加一些细节。

MSalters answer is correct, but I thought I'd add a bit of detail.

您需要将 double 转换为r值 time_t ，然后使用 localtime 转换为 struct tm 。例如，如果 double foo 是包含秒的变量：

You'll need to cast your double to an r-value time_t then use localtime to convert to a struct tm. For example if your double foo is your variable containing seconds:

const time_t temp = foo;
tm* bar = localtime(&temp);

有关 localtime 的回报：

结构可以在 std :: gmtime ， std :: localtime 和 std :: ctime ，并且可能会在每次调用时覆盖。

The structure may be shared between std::gmtime, std::localtime, and std::ctime, and may be overwritten on each invocation.

意味着对这些函数的任何调用可能会覆盖 bar 中的值。一个更好的选择是复制 localtime 的返回：

Meaning subsequent calls to any of these functions may overwrite the value in bar. A preferable option would be copying localtime's return:

const time_t temp = foo;
tm bar = *localtime(&temp);

另一个参考点是你不能 code> localtime（dynamic_cast< const time_t *>（& foo）） 或 localtime（（const time_t *）& foo） ，因为 foo 是 double 和 time_t 是一个整数值。该值需要隐式转换。只需输入 double * 即可产生 nullptr 。

One more point of reference is that you cannot simply localtime(dynamic_cast<const time_t*>(&foo)) or localtime((const time_t*)&foo) because foo is a double and time_t is an integral value. The value needs to be implicitly converted. Simply casting the double* results in a nullptr.

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