我们不久前遇到了这个问题，我手动制作了一些表，这些表在各种转换为各种大小的整数的边缘具有精确的浮点数模式。请注意，这假设 iee754 4 字节floats和8个字节doubles和 2 的补码有符号整数 (int32_t4 个字节和int64_t8 字节）。
如果您需要将位模式转换为浮点型或双精度型，您需要输入双关语（技术上为 UB）或memcpy them.
为了回答你的问题，任何太大而无法放入目标整数的东西都是 UB 转换时，并且截断为零的唯一时间是double -> 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，并且仅在它们在范围内时才进行转换。
请注意，使用INT_MIN/INT_MAX（或其现代极限对应物）交叉转换然后比较并不总是有效，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
所以对于你的例子你想要：
if( f >= B2F(FloatbitsMinFitInt32) && f <= B2F(FloatbitsMaxFitInt32))
// cast is valid.
其中 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，此转换正确拾取 nans/inf （因为与它们的比较是错误的）unless您正在使用编译器的非 iee754 模式（例如 gcc 上的 ffast-math 或 msvc 上的 /fp:fast）
