标准大小的整数之间的转换是否总是完全由标准定义的? [英] Is conversion between exactly-sized integers always completely defined by the standard?

问题描述

考虑以下代码:

#include <cstdint>
#include <iostream>
#include <iomanip>

int main()
{
    auto x=std::uint32_t(1)<<31;
    std::cout << " x: 0x" << std::hex << x << " =  " << std::dec << x << "\n";
    int32_t sx=x;
    std::cout << "sx: 0x" << std::hex << sx << " = " << std::dec << sx << "\n";
}

我从中得到以下输出:

 x: 0x80000000 =  2147483648
sx: 0x80000000 = -2147483648

这里 x 的值不能用 int32_t 表示，C++11 标准对这种转换做了如下说明:

Here the value of x can't be represented in int32_t, and the C++11 Standard says the following about this conversion:

如果目标类型是有符号的，如果它可以在目标类型中表示(和位域宽度)；否则，该值是实现定义的.

If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined.

即使使用 intXX_t，这仍然是实现定义的，为此我们对表示有一定的保证吗?如果是，那么我如何保证结果如上图所示?我应该 memcpy 我的无符号值进行签名以获得二进制补码解释，还是有更直接的方法?

Is this still implementation-defined even with intXX_t, for which we have certain guarantees on representation? If yes, then how can I guarantee that the result will be as shown above? Should I memcpy my unsigned value to signed to get two's complement interpretation, or is there a more straightforward way?

推荐答案

int32_t sx=x; 是实现定义.

您引用的规则是，将值转换为该值超出范围的类型会导致实现定义的行为.

The rule, which you quoted, is that converting a value to a type for which that value is out-of-range causes implementation-defined behaviour.

int32_t 的范围上升到 INT32_MAX，即 2147483647，但 (uint32_t)1 <<<31 比这多 1 个.

The range for int32_t goes up to INT32_MAX which is 2147483647, but (uint32_t)1 << 31 is one more than that.

这与表示无关，而是值是否可以保持不变.

It's nothing to do with representations, it's all about whether the value can remain unchanged or not.

使用 memcpy 将生成 int32_t，它与您从中复制的其他值具有相同的表示(我认为这是明确定义的，因为 int32_t不能有填充或陷阱)

Using memcpy would generate the int32_t which has the same representation as the other value you copy from (I think this is well-defined since int32_t cannot have padding or traps)

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