我有一个程序可以完成很多单精度数学运算。如果我指定1.0体系结构但是针对2.X和3.X体系结构，它会产生正确的结果。会导致这种情况的原因是什么？
包括以下内容： 很长的代码示例。 编译命令和良好的输出。 编译命令和输出错误。
如果我使用gcc在CPU中运行相同的路由，我会得到与1.0体系结构相匹配的结果。
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <math.h>
#include <cuda_runtime.h>
/*
* svdcomp - SVD decomposition routine.
* Takes an mxn matrix a and decomposes it into udv, where u,v are
* left and right orthogonal transformation matrices, and d is a
* diagonal matrix of singular values.
*
* This routine is adapted from svdecomp.c in XLISP-STAT 2.1 which is
* code from Numerical Recipes adapted by Luke Tierney and David Betz.
* Originally from: "Numerical Recipes in C: The Art of Scientific Computing",
* Press, Flannery, Teukolosky, Vetterling. 1992.
*
* Input to dsvd is as follows:
* a = mxn matrix to be decomposed, gets overwritten with u
* m = row dimension of a
* n = column dimension of a
* w = returns the vector of singular values of a
* v = returns the right orthogonal transformation matrix
*/
#define SIGN(a, b) ((b) >= 0.0f ? fabsf(a) : -fabsf(a))
#define MIN(x,y) ( (x) < (y) ? (x) : (y) )
#define MAX(x,y) ((x)>(y)?(x):(y))
#define PERR(call) \
if (call) {\
fprintf(stderr, "%s:%d Error [%s] on "#call"\n", __FILE__, __LINE__,\
cudaGetErrorString(cudaGetLastError()));\
exit(1);\
}
#define ERRCHECK \
if (cudaPeekAtLastError()) { \
fprintf(stderr, "%s:%d Error [%s]\n", __FILE__, __LINE__,\
cudaGetErrorString(cudaGetLastError()));\
exit(1);\
}
__device__ int
svd(float *a, int m, int n, float *w, float *v, int skip_u)
{
int flag, i, its, j, jj, k, l, nm;
float c, f, h, s, x, y, z;
float anorm = 0.0f, g = 0.0f, scale = 0.0f;
float rv1[3];
/* Householder reduction to bidiagonal form */
for (i = 0; i < n; i++)
{
/* left-hand reduction */
l = i + 1;
rv1[i] = scale * g;
g = s = scale = 0.0f;
if (i < m)
{
for (k = i; k < m; k++)
scale += fabsf(a[k*n+i]);
if (scale)
{
for (k = i; k < m; k++)
{
a[k*n+i] /= scale;
s += powf(a[k*n+i], 2);
}
f = a[i*n+i];
g = -SIGN(sqrtf(s), f);
h = f * g - s;
a[i*n+i] = f - g;
if (i != n - 1)
{
for (j = l; j < n; j++)
{
for (s = 0.0f, k = i; k < m; k++)
s += a[k*n+i] * a[k*n+j];
f = s / h;
for (k = i; k < m; k++)
a[k*n+j] += f * a[k*n+i];
}
}
for (k = i; k < m; k++)
a[k*n+i] *= scale;
}
}
w[i] = scale * g;
/* right-hand reduction */
g = s = scale = 0.0f;
if (i < m && i != n - 1)
{
for (k = l; k < n; k++)
scale += fabsf(a[i*n+k]);
if (scale)
{
for (k = l; k < n; k++)
{
a[i*n+k] /= scale;
s += powf(a[i*n+k], 2);
}
f = a[i*n+l];
g = -SIGN(sqrtf(s), f);
h = f * g - s;
a[i*n+l] = f - g;
for (k = l; k < n; k++)
rv1[k] = a[i*n+k] / h;
if (i != m - 1)
{
for (j = l; j < m; j++)
{
for (s = 0.0f, k = l; k < n; k++)
s += a[j*n+k] * a[i*n+k];
for (k = l; k < n; k++)
a[j*n+k] += s * rv1[k];
}
}
for (k = l; k < n; k++)
a[i*n+k] *= scale;
}
}
anorm = MAX(anorm, fabsf(w[i]) + fabsf(rv1[i]));
}
/* accumulate the right-hand transformation */
for (i = n - 1; i >= 0; i--)
{
if (i < n - 1)
{
if (g)
{
for (j = l; j < n; j++)
v[j*n+i] = (a[i*n+j] / a[i*n+l]) / g;
/* float division to avoid underflow */
for (j = l; j < n; j++)
{
for (s = 0.0f, k = l; k < n; k++)
s += a[i*n+k] * v[k*n+j];
for (k = l; k < n; k++)
v[k*n+j] += s * v[k*n+i];
}
}
for (j = l; j < n; j++)
v[i*n+j] = v[j*n+i] = 0.0f;
}
v[i*n+i] = 1.0f;
g = rv1[i];
l = i;
}
/* accumulate the left-hand transformation */
if (!skip_u) {
for (i = n - 1; i >= 0; i--)
{
l = i + 1;
g = w[i];
if (i < n - 1)
for (j = l; j < n; j++)
a[i*n+j] = 0.0f;
if (g)
{
g = 1.0f / g;
if (i != n - 1)
{
for (j = l; j < n; j++)
{
for (s = 0.0f, k = l; k < m; k++)
s += a[k*n+i] * a[k*n+j];
f = (s / a[i*n+i]) * g;
for (k = i; k < m; k++)
a[k*n+j] += f * a[k*n+i];
}
}
for (j = i; j < m; j++)
a[j*n+i] = a[j*n+i]*g;
}
else
{
for (j = i; j < m; j++)
a[j*n+i] = 0.0f;
}
++a[i*n+i];
}
}
/* diagonalize the bidiagonal form */
for (k = n - 1; k >= 0; k--)
{ /* loop over singular values */
for (its = 0; its < 30; its++)
{ /* loop over allowed iterations */
flag = 1;
for (l = k; l >= 0; l--)
{ /* test for splitting */
nm = l - 1;
if (fabsf(rv1[l]) + anorm == anorm)
{
flag = 0;
break;
}
if (fabsf(w[nm]) + anorm == anorm)
break;
}
if (flag)
{
c = 0.0f;
s = 1.0f;
for (i = l; i <= k; i++)
{
f = s * rv1[i];
if (fabsf(f) + anorm != anorm)
{
g = w[i];
h = hypotf(f, g);
w[i] = h;
h = 1.0f / h;
c = g * h;
s = (- f * h);
if (!skip_u) {
for (j = 0; j < m; j++)
{
y = a[j*n+nm];
z = a[j*n+i];
a[j*n+nm] = y * c + z * s;
a[j*n+i] = z * c - y * s;
}
}
}
}
}
z = w[k];
if (l == k)
{ /* convergence */
if (z < 0.0f)
{ /* make singular value nonnegative */
w[k] = -z;
for (j = 0; j < n; j++)
v[j*n+k] = -v[j*n+k];
}
break;
}
if (its >= 30) {
}
/* shift from bottom 2 x 2 minor */
x = w[l];
nm = k - 1;
y = w[nm];
g = rv1[nm];
h = rv1[k];
f = ((y - z) * (y + z) + (g - h) * (g + h)) / (2.0f * h * y);
g = hypotf(f, 1.0f);
f = ((x - z) * (x + z) + h * ((y / (f + SIGN(g, f))) - h)) / x;
/* next QR transformation */
c = s = 1.0f;
for (j = l; j <= nm; j++)
{
i = j + 1;
g = rv1[i];
y = w[i];
h = s * g;
g = c * g;
z = hypotf(f, h);
rv1[j] = z;
c = f / z;
s = h / z;
f = x * c + g * s;
g = g * c - x * s;
h = y * s;
y = y * c;
for (jj = 0; jj < n; jj++)
{
x = v[jj*n+j];
z = v[jj*n+i];
v[jj*n+j] = x * c + z * s;
v[jj*n+i] = z * c - x * s;
}
z = hypotf(f, h);
w[j] = z;
if (z)
{
z = 1.0f / z;
c = f * z;
s = h * z;
}
f = (c * g) + (s * y);
x = (c * y) - (s * g);
if (!skip_u) {
for (jj = 0; jj < m; jj++)
{
y = a[jj*n+j];
z = a[jj*n+i];
a[jj*n+j] = y * c + z * s;
a[jj*n+i] = z * c - y * s;
}
}
}
rv1[l] = 0.0f;
rv1[k] = f;
w[k] = x;
}
}
return(0);
}
__global__ void
svd_kernel(float *v)
{
float a[9], w[3];
a[0] = 8.0f;
a[1] = 3.0f;
a[2] = 7.0f;
a[3] = 7.0f;
a[4] = 9.0f;
a[5] = 1.0f;
a[6] = 3.0f;
a[7] = 7.0f;
a[8] = 2.0f;
svd(a, 3, 3, w, v, 1);
}
int main()
{
int i, j;
float *v_d, v[9];
PERR(cudaMalloc(&v_d, 9*sizeof(float)));
svd_kernel<<<1,1>>>(v_d);
cudaDeviceSynchronize();
ERRCHECK;
PERR(cudaMemcpy(v, v_d, 9*sizeof(float), cudaMemcpyDeviceToHost));
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%6.3f\t", v[i*3+j]);
}
printf("\n");
}
return 0;
}
正确的结果：
$ nvcc -arch=sm_10 -o svd svd.cu
$ ./svd
-0.657 -0.685 0.314
-0.668 0.337 -0.664
-0.349 0.646 0.679
结果破碎：
$ nvcc -arch=sm_20 -o svd svd.cu
$ ./svd
-0.661 -0.660 0.356
-0.642 0.253 -0.724
0.019 0.460 0.888
答案 0 :(得分：1)
似乎CUDA 6解决了这个问题。