Question

我刚刚从C ++ 14标准（我的重点）中读到了这个：

4.9浮动积分转换[conv.fpint]

1 浮点类型的prvalue可以转换为整数类型的prvalue。转换截断;也就是说，丢弃小数部分。 如果截断值不可以，则行为未定义   以目的地类型表示。 [...]

让我想到了

截断后哪个float值无法表示为int？（这取决于实施吗？）
如果有，这是否意味着auto x = static_cast<int>(float)不安全？
将float转换为int然后（假设您想要截断）的正确/安全方式是什么？

Answer 1

我们回过头来，我手动制作了一些表格，这些表格在各种转换的边缘具有各种大小的整数的浮点数。注意，这假设iee754 4字节floats和8字节doubles和2＆＃39;补码有符号整数（int32_t 4字节和int64_t 8字节）。

如果你需要将位模式转换为浮点数或双精度数，你需要输入它们（技术上是UB）或memcpy它们。

要回答你的问题，任何太大而不适合目标整数的东西都是转换时的UB，而截断到零的唯一时间是double - ＆gt; int32_t。因此，使用以下值，您可以将浮点数与相关的最小值/最大值进行比较，只有在它们处于范围内时才进行投射。

请注意，使用INT_MIN / INT_MAX（或其现代限制对应物）进行交叉转换然后进行比较并不总是有效，因为这些大小值的浮点数的准确性非常低。

Inf / NaN在转换时也是UB。

// float->int64 edgecases
static const uint32_t FloatbitsMaxFitInt64 = 0x5effffff; // [9223371487098961920] Largest float which still fits int an signed int64
static const uint32_t FloatbitsMinNofitInt64 = 0x5f000000; // [9223372036854775808] the bit pattern of the smallest float which is too big for a signed int64
static const uint32_t FloatbitsMinFitInt64 = 0xdf000000; // [-9223372036854775808] Smallest float which still fits int an signed int64
static const uint32_t FloatbitsMaxNotfitInt64 = 0xdf000001; // [-9223373136366403584] Largest float which to small for a signed int64

// float->int32 edgecases
static const uint32_t FloatbitsMaxFitInt32 = 0x4effffff; // [2147483520] the bit pattern of the largest float which still fits int an signed int32
static const uint32_t FloatbitsMinNofitInt32 = 0x4f000000; // [2147483648] the bit pattern of the smallest float which is too big for a signed int32
static const uint32_t FloatbitsMinFitInt32 = 0xcf000000; // [-2147483648] the bit pattern of the smallest float which still fits int an signed int32
static const uint32_t FloatbitsMaxNotfitInt32 = 0xcf000001; // [-2147483904] the bit pattern of the largest float which to small for a signed int32

// double->int64 edgecases
static const uint64_t DoubleBitsMaxFitInt64 = 0x43dfffffffffffff; // [9223372036854774784] Largest double which fits into an int64
static const uint64_t DoubleBitsMinNofitInt64 = 0x43e0000000000000; // [9223372036854775808] Smallest double which is too big for an int64
static const uint64_t DoubleBitsMinFitInt64 = 0xc3e0000000000000; // [-9223372036854775808] Smallest double which fits into an int64
static const uint64_t DoubleBitsMaxNotfitInt64 = 0xc3e0000000000001; // [-9223372036854777856] largest double which is too small to fit into an int64

// double->int32 edgecases[when truncating(round towards zero)]
static const uint64_t DoubleBitsMaxTruncFitInt32 = 0x41dfffffffffffff; // [~2147483647.9999998] Largest double that when truncated will fit into an int32
static const uint64_t DoubleBitsMinTruncNofitInt32 = 0x41e0000000000000; // [2147483648.0000000] Smallest double that when truncated wont fit into an int32
static const uint64_t DoubleBitsMinTruncFitInt32 = 0xc1e00000001fffff; // [~2147483648.9999995] Smallest double that when truncated will fit into an int32
static const uint64_t DoubleBitsMaxTruncNofitInt32 = 0xc1e0000000200000; // [2147483649.0000000] Largest double that when truncated wont fit into an int32

// double->int32 edgecases [when rounding via bankers method(round to nearest, round to even on half)]
static const uint64_t DoubleBitsMaxRoundFitInt32 = 0x41dfffffffdfffff; // [2147483647.5000000] Largest double that when rounded will fit into an int32
static const uint64_t DoubleBitsMinRoundNofitInt32 = 0x41dfffffffe00000; // [~2147483647.5000002] Smallest double that when rounded wont fit into an int32
static const uint64_t DoubleBitsMinRoundFitInt32 = 0xc1e0000000100000; // [-2147483648.5000000] Smallest double that when rounded will fit into an int32
static const uint64_t DoubleBitsMaxRoundNofitInt32 = 0xc1e0000000100001; // [~2147483648.5000005] Largest double that when rounded wont fit into an int32

所以你想要的例子是：

if( f >= B2F(FloatbitsMinFitInt32) && f <= B2F(FloatbitsMaxFitInt32))
    // cast is valid.

B2F类似于：

float B2F(uint32_t bits)
{
    static_assert(sizeof(float) == sizeof(uint32_t), "Weird arch");
    float f;
    memcpy(&f, &bits, sizeof(float));
    return f;
}

请注意，此转换正确选择nans / inf（与它们的比较为false）除非您使用编译器的非iee754模式（例如gcc上的ffast-math）或/ fp：快速在msvc上）

Answer 2

float的值超出int范围的情况完全不足为奇。发明浮点值可以充分代表非常大（也非常小）的值。

INT_MAX + 1（通常等于2147483648）无法由int表示，但可以由float表示。
是的，static_cast<int>(float)与未定义的行为一样不安全。但是，对于足够大的整数x + y和x而言，y这样简单的事情也是UB，所以这里也没有什么大惊喜。
正确的做法取决于应用程序，就像在C ++中一样。 Boost有numeric_cast在溢出时抛出异常;这可能对你有好处。要做饱和度（将过大的值转换为INT_MIN和INT_MAX），请编写一些这样的代码
```
float f;
int i;
...
if (static_cast<double>(INT_MIN) <= f && f < static_cast<double>(INT_MAX))
    i = static_cast<int>(f);
else if (f < 0)
    i = INT_MIN;
else
    i = INT_MAX;
```
然而，这并不理想。您的系统是否有double类型可以代表int的最大值？如果是的话，它会起作用。另外，您希望如何舍入接近最小值或最大值int的值？如果您不想考虑此类问题，请按照here所述使用boost::numeric_cast。

没有未定义的行为[c ++]，哪些浮点值无法转换为int？

2 个答案: