问题描述
我使用Linux x86_64和clang 3.3。
I use Linux x86_64 and clang 3.3.
这在理论上是可能的吗?
Is this even possible in theory?
std :: atomic< __ int128_t>
不工作(未定义的对某些函数的引用)。
std::atomic<__int128_t>
doesn't work (undefined references to some functions).
也不工作('错误:无法编译此原子库调用')。
__atomic_add_fetch
also doesn't work ('error: cannot compile this atomic library call yet').
两个 std :: atomic
和 __ atomic_add_fetch
使用64位数字。
Both std::atomic
and __atomic_add_fetch
work with 64-bit numbers.
推荐答案
不可能用单个指令这样做,但你可以模拟它,仍然是无锁。除了最早的AMD64 CPU,x64支持 CMPXCHG16B
指令。有了一个多精度的数学,你可以很容易做到。
It's not possible to do this with a single instruction, but you can emulate it and still be lock-free. Except for the very earliest AMD64 CPUs, x64 supports the CMPXCHG16B
instruction. With a little multi-precision math, you can do this pretty easily.
恐怕我不知道instrinsic for CMPXCHG16B
在GCC,但希望你得到一个旋转循环 CMPXCHG16B
的想法。这里是一些未经测试的VC ++代码:
I'm afraid I don't know the instrinsic for CMPXCHG16B
in GCC, but hopefully you get the idea of having a spin loop of CMPXCHG16B
. Here's some untested code for VC++:
// atomically adds 128-bit src to dst, with src getting the old dst.
void fetch_add_128b(uint64_t *dst, uint64_t* src)
{
uint64_t srclo, srchi, olddst[2], exchlo, exchhi;
srchi = src[0];
srclo = src[1];
olddst[0] = dst[0];
olddst[1] = dst[1];
do
{
exchlo = srclo + olddst[1];
exchhi = srchi + olddst[0] + (exchlo < srclo); // add and carry
}
while(!_InterlockedCompareExchange128((long long*)dst,
exchhi, exchlo,
(long long*)olddst));
src[0] = olddst[0];
src[1] = olddst[1];
}
编辑:这里有一些未经测试的代码我可以找到对于GCC内在函数:
here's some untested code going off of what I could find for the GCC intrinsics:
// atomically adds 128-bit src to dst, returning the old dst.
__uint128_t fetch_add_128b(__uint128_t *dst, __uint128_t src)
{
__uint128_t dstval, olddst;
dstval = *dst;
do
{
olddst = dstval;
dstval = __sync_val_compare_and_swap(dst, dstval, dstval + src);
}
while(dstval != olddst);
return dstval;
}
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