如何将size_t转换为double或int C ++ [英] how to cast the size_t to double or int C++

问题描述

我的问题是

我有一个size_t数据，但现在我想将其转换为double或int。

I have a size_t data, but now I want to convert it to double or int.

如果我做了

 size_t data = 99999999;
 int convertdata = data;

编译器将报告警告。因为它可能溢出。

the compiler will report warning. because it maybe overflow.

你有一些方法像boost或其他方法来执行转换吗？

Do you have some method like the boost or some other method to do the convert?

推荐答案

演员， Blaz Bratanic建议：

size_t data = 99999999;
int convertdata = static_cast<int>(data);

很可能会使警告静音（虽然原则上编译器可以警告任何它喜欢，有一个转换）。

is likely to silence the warning (though in principle a compiler can warn about anything it likes, even if there's a cast).

但它不能解决警告告诉你的问题，即从 size_t 到 int 真的可能溢出。

But it doesn't solve the problem that the warning was telling you about, namely that a conversion from size_t to int really could overflow.

如果可能，设计你的程序，需要将 size_t 值转换为 int 。只需将它存储在 size_t 变量（如你已经做过的那样）。

If at all possible, design your program so you don't need to convert a size_t value to int. Just store it in a size_t variable (as you've already done) and use that.

转换为 double 不会导致溢出，但对于非常大的 size_t 值，可能会导致精度损失。再次，将 size_t 转换为 double 没有什么意义;你最好把这个值保存在 size_t 变量中。

Converting to double will not cause an overflow, but it could result in a loss of precision for a very large size_t value. Again, it doesn't make a lot of sense to convert a size_t to a double; you're still better off keeping the value in a size_t variable.

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