如何消除 C++ 中 time.h 的警告？
当我使用它时，
#include<time.h>
//...
int n = time(0);
//...
我收到有关将时间转换为 int 的警告。有没有办法消除这个警告？
When I use this
#include<time.h>
//...
int n = time(0);
//...
I get a warning about converting time to int. Is there a way to remove this warning?
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是的，将 n 更改为
time_t。如果您查看大多数/所有系统上 time.h 中的签名，您会发现这就是它返回的内容。请注意，Arak 是正确的：使用 32 位 int 是一个问题，至少 由于 2038 错误。但是，您应该考虑到，对整数 n（而不是 time_t）进行的任何类型的算术只会增加您的代码提前遇到该错误的可能性。
PS：如果我在原始答案中没有说清楚，对编译器警告的最佳响应几乎总是解决您被警告的情况。例如，将较高精度的数据强制转换为较低精度的变量会丢失信息 - 编译器试图警告您，您可能刚刚创建了一个地雷错误，直到很久以后才会有人绊倒。
Yes, change n to be a
time_t. If you look at the signature in time.h on most / all systems, you'll see that that's what it returns.Note that Arak is right: using a 32 bit int is a problem, at a minimum, due to the 2038 bug. However, you should consider that any sort of arithmetic on an integer n (rather than a time_t) only increases the probability that your code will trip over that bug early.
PS: In case I didn't make it clear in the original answer, the best response to a compiler warning is almost always to address the situation that you're being warned about. For example, forcing higher precision data into a lower precision variable loses information - the compiler is trying to warn you that you might have just created a landmine bug that someone won't trip over until much later.
时间返回
time_t而不是整数。最好使用该类型，因为它可能大于int。如果您确实需要
int，则显式地对其进行类型转换，例如：Time returns
time_tand not integer. Use that type preferably because it may be larger thanint.If you really need
int, then typecast it explicitly, for example:我认为您正在使用 Visual C++。即使您是针对与
g++不同的32bit平台进行编程，time(0)的返回类型也是64bit int。要删除警告，只需将time(0)分配给64bit变量即可。I think you are using Visual C++. The return type of
time(0)is64bit inteven if you are programming for32bitplatform unlikeg++. To remove the warning, just assigntime(0)to64bitvariable.您可能想使用 time_t 类型而不是 int 类型。
请参阅 http://en.wikipedia.org/wiki/Time_t 中的示例。
You probably want to use a type of time_t instead of an int.
See the example at http://en.wikipedia.org/wiki/Time_t.
原因是 time() 函数返回 time_t 时间，因此在这种情况下您可能需要静态转换为 int 或 uint。这样写：
The reason is time() functions returns a time_t time so you might need to do a static cast to an int or uint in this case. Write in this way: