在C中（更具体地说，C表示CUDA），计算大量浮点数（比如两万个值）的校验和的最佳方法是什么，这很容易用printf打印，而不使用任何库？ / p>
我可以用浮动精度中的所有值求和，但我担心舍入误差或饱和度，或者nan / inf值会使某些变化无法检测到。
这用于在同一个gpu硬件上比较同一个二进制文件的运行之间的变量值，这仅用于调试，而不是用于安全性。
更清楚的是，如果数组中的任何浮点值发生变化，校验和的所有数字都会发生变化（很有可能）会很好，这样校验和很容易在视觉上进行比较。
答案 0 :(得分：1)
这正是Cyclic Redundancy Checks的用途。 Boost有a CRC library，网上有很多源代码实现。可能16位CRC最适合您，因为眼球效果很容易。但是如果你对假阳性偏执，你可能需要一个32位CRC。
答案 1 :(得分：1)
如果使用的是IEEE-754浮点数，则可以将浮点数转换为指针，然后将其重新解释为无符号整数指针，并将其求和，以避免任何浮点舍入问题。您基本上就是在这一点上创建一个代表浮点数的实际位的校验和，而不是浮点值本身。
例如：
float array[20] = { /* initialized to some values */ };
unsigned int total = 0;
for (int i=0; i < 20; i++)
{
float* temp_float_ptr = &array[i];
unsigned int* temp_uint_ptr = (unsigned int*)temp_float_ptr;
total += (*temp_uint_ptr);
}
编辑：正如评论中所提到的，这不会以任何方式创建安全校验和......这是一种非常简单的校验和形式，但希望它可以用于您的调试目的
答案 2 :(得分：1)
对于stackoverflow响应，这可能太大了，但这是我从pycrc的输出一起攻击的crc.cu文件。它包括其他回复中已提到的几种技术。我最信任crc版本，但是当数组应该充满零时，add和xor版本很方便。
/* The MIT License
Copyright (c) <year> <copyright holders>
Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in
all copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
THE SOFTWARE. */
/*
* (formerly) file crc.h
* Functions and types for CRC checks.
*
* Generated on Sun May 15 16:28:36 2011,
* by pycrc v0.7.7, http://www.tty1.net/pycrc/
* using the configuration:
* Width = 32
* Poly = 0x04c11db7
* XorIn = 0xffffffff
* ReflectIn = True
* XorOut = 0xffffffff
* ReflectOut = True
* Algorithm = table-driven
*
* , and then hacked by Drew Wagner to work in CUDA.
* NOTE: Note, most of this code was generated by the MIT license
* version of the pycrc. Accordingly, this derivative work is also
* licensed under the MIT license. This license applies ONLY to this file!
*
*****************************************************************************/
#ifndef __CRC_CU__
#define __CRC_CU__
/**
* The definition of the used algorithm.
*****************************************************************************/
#define CRC_ALGO_TABLE_DRIVEN 1
/**
* The type of the CRC values.
*
* This type must be big enough to contain at least 32 bits.
*****************************************************************************/
typedef uint32_t crc_t;
/**
* Calculate the initial crc value.
*
* \return The initial crc value.
*****************************************************************************/
__device__ crc_t crc_init(void)
{
return 0xffffffff;
}
/**
* Calculate the final crc value.
*
* \param crc The current crc value.
* \return The final crc value.
*****************************************************************************/
__device__ crc_t crc_finalize(crc_t crc)
{
return crc ^ 0xffffffff;
}
/**
* (formally) file crc.c
* Functions and types for CRC checks.
*
* Generated on Sun May 15 16:28:42 2011,
* by pycrc v0.7.7, http://www.tty1.net/pycrc/
* using the configuration:
* Width = 32
* Poly = 0x04c11db7
* XorIn = 0xffffffff
* ReflectIn = True
* XorOut = 0xffffffff
* ReflectOut = True
* Algorithm = table-driven
*****************************************************************************/
/**
* Static table used for the table_driven implementation.
*****************************************************************************/
__device__ static const crc_t crc_table[16] = {
0x00000000, 0x1db71064, 0x3b6e20c8, 0x26d930ac,
0x76dc4190, 0x6b6b51f4, 0x4db26158, 0x5005713c,
0xedb88320, 0xf00f9344, 0xd6d6a3e8, 0xcb61b38c,
0x9b64c2b0, 0x86d3d2d4, 0xa00ae278, 0xbdbdf21c
};
/**
* Reflect all bits of a \a data word of \a data_len bytes.
*
* \param data The data word to be reflected.
* \param data_len The width of \a data expressed in number of bits.
* \return The reflected data.
*****************************************************************************/
__device__ crc_t crc_reflect(crc_t data, size_t data_len)
{
unsigned int i;
crc_t ret;
ret = data & 0x01;
for (i = 1; i < data_len; i++) {
data >>= 1;
ret = (ret << 1) | (data & 0x01);
}
return ret;
}
/**
* Update the crc value with new data.
*
* \param crc The current crc value.
* \param data Pointer to a buffer of \a data_len bytes.
* \param data_len Number of bytes in the \a data buffer.
* \return The updated crc value.
*****************************************************************************/
__device__ crc_t crc_update(crc_t crc, const unsigned char *data, size_t data_len)
{
unsigned int tbl_idx;
while (data_len--) {
tbl_idx = crc ^ (*data >> (0 * 4));
crc = crc_table[tbl_idx & 0x0f] ^ (crc >> 4);
tbl_idx = crc ^ (*data >> (1 * 4));
crc = crc_table[tbl_idx & 0x0f] ^ (crc >> 4);
data++;
}
return crc & 0xffffffff;
}
// Note 1: The xor and add versions below will return 0x00000000 if the vector, or array,
// is all zeros. This can be convenient, but they will NOT detect if zero values move
// around. This invariance to changes in order is especially true for the add version.
// Note 2: Calling these introduces thread synchronization! Be wary of heisenbugs!
// Note 3: The CRC version is the most principled, but is also slowest, and makes zeros arrays less obvious.
__device__ uint32_t vector_checksum_xor(const float* array, int m, uint32_t prevValue=0x00000000)
{
__syncthreads();
if(threadIdx.x==0 && blockIdx.x==0)
{
uint32_t sum = prevValue;
uint32_t * array_ptr = (uint32_t*) array;
for(int i=0; i<m; i++)
if(array_ptr[i]!=0x00000000)
sum ^= array_ptr[i];
return sum;
} else { return 0xffffffff;}
__syncthreads();
}
// Coded for m x n column major arrays with column stride lda
__device__ uint32_t array_checksum_xor(const float* A, int m, int n, int lda, uint32_t prevValue=0x00000000)
{
uint32_t sum = prevValue;
__syncthreads();
if(threadIdx.x==0 && blockIdx.x==0)
{
for(int i=0; i<n; i++)
sum = vector_checksum_xor(&A[i*lda], m, sum);
return sum;
} else { return 0xffffffff;}
__syncthreads();
}
__device__ uint32_t vector_checksum_sum(const float* array, int m, uint32_t prevValue=0x00000000)
{
__syncthreads();
if(threadIdx.x==0 && blockIdx.x==0)
{
uint32_t sum = prevValue;
uint32_t * array_ptr = (uint32_t*) array;
for(int i=0; i<m; i++)
if(array_ptr[i]!=0x00000000)
sum += array_ptr[i];
return sum;
} else { return 0xffffffff;}
__syncthreads();
}
// Coded for m x n column major arrays with column stride lda
__device__ uint32_t array_checksum_sum(const float* A, int m, int n, int lda, uint32_t prevValue=0x00000000)
{
uint32_t sum = prevValue;
__syncthreads();
if(threadIdx.x==0 && blockIdx.x==0)
{
for(int i=0; i<n; i++)
{
sum = vector_checksum_sum(&A[i*lda], m, sum);
}
return sum;
} else { return 0xffffffff;}
__syncthreads();
}
__device__ uint32_t vector_checksum_crc(const float* array, int m, uint32_t sum=0xffffffff)
{
__syncthreads();
if(threadIdx.x==0 && blockIdx.x==0)
{
const unsigned char * array_ptr = (const unsigned char*) array;
sum = crc_update(sum, array_ptr, m*sizeof(float));
sum = crc_finalize(sum);
return sum;
} else { return 0xffffffff;}
__syncthreads();
}
// Coded for m x n column major arrays with column stride lda
__device__ uint32_t array_checksum_crc(const float* A, int m, int n, int lda, uint32_t sum=0xffffffff)
{
__syncthreads();
if(threadIdx.x==0 && blockIdx.x==0)
{
for(int i=0; i<n; i++)
{
const unsigned char * array_ptr = (const unsigned char*) A;
sum = crc_update(sum, array_ptr, m*sizeof(float));
}
sum = crc_finalize(sum);
return sum;
} else { return 0xffffffff;}
__syncthreads();
}
#endif
答案 3 :(得分：0)
：
我相信最好的答案是TonyK和Jason的答案的结合。在将float*转换为uint32_t*之后，在缓冲区上使用32位CRC。从你的编译器提供uint32定义，或者根据你的平台自己定义它（通常在32位机器上无符号长，在64位机器上无符号int。）这是一个good CRC explanation and implementation
答案 4 :(得分：0)
一个正确的CRC库可能是你最好的选择，但只是为了潜在的兴趣：你可以使用每个字节（即*(uint8_t*)&重新解释）索引到一个表中，而不是对你的值中的位进行异或运算。 32位随机数，然后将这些表条目异或。这意味着值中的单个位变化会随机翻转输出中的位。如果不希望有尽可能多的查找表，那么就可以得到合理的结果，对已经使用过的表中的位进行循环移位。它在概念上比数学哈希算法简单得多....
答案 5 :(得分：0)
如果要计算GPU上的校验和，请使用__int_as_float（）和__float_as_int（）内在函数来处理浮点数作为整数。 CUDA 4.0附带的Thrust库使计算这个校验和变得容易 - 这是minmax Thrust示例，移植到您正在寻找的位置。
#include <thrust/device_vector.h>
#include <thrust/host_vector.h>
#include <thrust/transform_reduce.h>
#include <thrust/functional.h>
#include <thrust/extrema.h>
// sumAsInt contains a float, but implements a binary
// operator that adds them as if they were ints.
template <typename T>
struct sumAsInt
{
T val;
};
// sumAsInt_unary_op is a functor that initializes a sumAsInt
// with a given value T.
template <typename T>
struct sumAsInt_unary_op : public thrust::unary_function<T,T>
{
__host__ __device__
sumAsInt<T> operator()(const T& x) const {
sumAsInt<T> result;
result.val = x;
return result;
}
};
// sumAsInt_binary_op is a functor that accepts two sumAsInt
// structs and returns a new sumAsInt that contains the
// sum of the two floats, as if they were integers.
template <typename T>
struct sumAsInt_binary_op : public thrust::binary_function<T,T,T>
{
__host__ __device__
sumAsInt<T> operator()(const sumAsInt<T>& x, const sumAsInt<T>& y) const {
sumAsInt<T> result;
result.val = __int_as_float(__float_as_int(x.val)+__float_as_int(y.val));
return result;
}
};
int main(void)
{
// initialize host array
float x[7] = {-1, 2, 7, -3, -4, 5};
int sum = 0;
for ( int i = 0; i < sizeof(x)/sizeof(x[0]); i++ ) {
sum += *((int *) (&x[i]));
}
printf( "CPU sum: %d\n", sum );
// transfer to device
thrust::device_vector<float> d_x(x, x + 7);
// setup arguments
sumAsInt_unary_op<float> unary_op;
sumAsInt_binary_op<float> binary_op;
sumAsInt<float> init = unary_op(0.0f/*d_x[0]*/); // initialize with first element
// compute sum-as-int
sumAsInt<float> result = thrust::transform_reduce(d_x.begin(), d_x.end(), unary_op, init, binary_op);
printf( "GPU sum: %d\n", *((int *) (&result.val)) );
// std::cout << result.val << std::endl;
return 0;
}