使用 use_fast_math 标志编译时,更改代码的一部分会影响另一部分
Changing one part of code affects the other part when compiling with use_fast_math flag
我有以下内核:
__global__ void kernel()
{
float loc{269.0f};
float s{(356.0f - loc) / 13.05f};
float a{pow(1.0f - 0.15f * s, 1.0f)};
float b{pow(1.0f - 0.15f * (356.0f - loc) / 13.05f, 1.0f)};
printf("%f\n", a);
printf("%f\n", b);
}
它打印
0.000000
0.000000
但是如果我改变 b
的计算方式:
__global__ void kernel()
{
float loc{269.0f};
float s{(356.0f - loc) / 13.05f};
float a{pow(1.0f - 0.15f * s, 1.0f)};
float b{pow(1.0f - 0.15f * ((356.0f - loc) / 13.05f), 1.0f)}; // notice the added braces
printf("%f\n", a);
printf("%f\n", b);
}
它打印:
nan
nan
为什么为 b
添加大括号也会更改 a
?为什么牙套有效果?查看 godbolt.org 上的代码,我发现生成的程序集不同,但我没有足够的知识来理解究竟是什么导致了这种行为。
这是我的项目配置:
set_target_properties(test PROPERTIES
CXX_STANDARD 14
CXX_STANDARD_REQUIRED YES
CXX_EXTENSIONS NO
CUDA_STANDARD 14
CUDA_STANDARD_REQUIRED YES
CUDA_EXTENSIONS NO
CUDA_SEPARABLE_COMPILATION ON
CUDA_ARCHITECTURES "61"
)
set(CUDA_FLAGS --use_fast_math)
set(CXX_FLAGS -O0)
请注意标志 --use_fast_math
- 没有它,一切正常。
我的 GPU 是 Quadro P1000。
Cuda 编译工具,版本 11.2,V11.2.152.
你的代码数值不稳定:它包含一个catastrophic cancellation. Indeed, 1.0f - 0.15f * s
and 1.0f - 0.15f * ((356.0f - loc) / 13.05f)
are nearly equal to 0. It can be equal to few unit in the last place (ULP). Regarding the rounding, the value can be positive, negative or 0. When the base value is negative, the result of pow is undefined. The thing is the rounding is dependent of the compiler heuristic since you explicitly enabled fast-math that disable some IEEE-754 rules like taking care of the floating-point associativity and the possible presence of NaN values. With fast-math, the compiler is free to use approximations like a reciprocal (less accurate) instead of the basic division. In fact it actually do that in your case. Consequently, results are undefined and the compiler is free to set the output to 0 or NaN regarding its heuristic. I think that having a pow exponent set to 1.0f (which is well represented) does not save you from this behaviour with pow
(注意你应该使用powf
作为float
值而不是pow
用于 double
个值。
解决这个问题的正确方法是使用稳定的数值计算,它不包含任何像您的代码中那样的灾难性取消。实际解决方案非常依赖于整体计算方法。有时,需要使用完全不同的算法(例如,QR 分解而不是 LU 分解,2 步方差计算而不是 1 步)。
我有以下内核:
__global__ void kernel()
{
float loc{269.0f};
float s{(356.0f - loc) / 13.05f};
float a{pow(1.0f - 0.15f * s, 1.0f)};
float b{pow(1.0f - 0.15f * (356.0f - loc) / 13.05f, 1.0f)};
printf("%f\n", a);
printf("%f\n", b);
}
它打印
0.000000
0.000000
但是如果我改变 b
的计算方式:
__global__ void kernel()
{
float loc{269.0f};
float s{(356.0f - loc) / 13.05f};
float a{pow(1.0f - 0.15f * s, 1.0f)};
float b{pow(1.0f - 0.15f * ((356.0f - loc) / 13.05f), 1.0f)}; // notice the added braces
printf("%f\n", a);
printf("%f\n", b);
}
它打印:
nan
nan
为什么为 b
添加大括号也会更改 a
?为什么牙套有效果?查看 godbolt.org 上的代码,我发现生成的程序集不同,但我没有足够的知识来理解究竟是什么导致了这种行为。
这是我的项目配置:
set_target_properties(test PROPERTIES
CXX_STANDARD 14
CXX_STANDARD_REQUIRED YES
CXX_EXTENSIONS NO
CUDA_STANDARD 14
CUDA_STANDARD_REQUIRED YES
CUDA_EXTENSIONS NO
CUDA_SEPARABLE_COMPILATION ON
CUDA_ARCHITECTURES "61"
)
set(CUDA_FLAGS --use_fast_math)
set(CXX_FLAGS -O0)
请注意标志 --use_fast_math
- 没有它,一切正常。
我的 GPU 是 Quadro P1000。
Cuda 编译工具,版本 11.2,V11.2.152.
你的代码数值不稳定:它包含一个catastrophic cancellation. Indeed, 1.0f - 0.15f * s
and 1.0f - 0.15f * ((356.0f - loc) / 13.05f)
are nearly equal to 0. It can be equal to few unit in the last place (ULP). Regarding the rounding, the value can be positive, negative or 0. When the base value is negative, the result of pow is undefined. The thing is the rounding is dependent of the compiler heuristic since you explicitly enabled fast-math that disable some IEEE-754 rules like taking care of the floating-point associativity and the possible presence of NaN values. With fast-math, the compiler is free to use approximations like a reciprocal (less accurate) instead of the basic division. In fact it actually do that in your case. Consequently, results are undefined and the compiler is free to set the output to 0 or NaN regarding its heuristic. I think that having a pow exponent set to 1.0f (which is well represented) does not save you from this behaviour with pow
(注意你应该使用powf
作为float
值而不是pow
用于 double
个值。
解决这个问题的正确方法是使用稳定的数值计算,它不包含任何像您的代码中那样的灾难性取消。实际解决方案非常依赖于整体计算方法。有时,需要使用完全不同的算法(例如,QR 分解而不是 LU 分解,2 步方差计算而不是 1 步)。