我目前正在查看Keavin Beason的smallpt代码。我使用g++ -O3 -fopenmp smallpt.cpp
编写了代码,并使用g++ -O3 smallpt.cpp
编写了代码,然后我遇到了无限循环或死锁。
仅使用int main(int argc, char *argv[])
{
int w=1024, h=768, samps = argc==2 ? atoi(argv[1])/4 : 1;
Ray cam(Vec(50,52,295.6), Vec(0,-0.042612,-1).norm()); // cam pos, dir
Vec cx=Vec(w*.5135/h);
Vec cy=(cx%cam.d).norm()*.5135, r, *c=new Vec[w*h];
#pragma omp parallel for schedule(dynamic, 1) private(r) // OpenMP
for (int y=0; y<h; y++) // Loop over image rows
{
fprintf(stderr,"\rRendering (%d spp) %5.2f%%",samps*4,100.*y/(h-1));
for (unsigned short x=0, Xi[3]={0,0,y*y*y}; x<w; x++) // Loop cols
{
for (int sy=0, i=(h-y-1)*w+x; sy<2; sy++) // 2x2 subpixel rows
{
for (int sx=0; sx<2; sx++, r=Vec()) // 2x2 subpixel cols
{
for (int s=0; s<samps; s++)
{
double r1=2*erand48(Xi), dx=r1<1 ? sqrt(r1)-1: 1-sqrt(2-r1);
double r2=2*erand48(Xi), dy=r2<1 ? sqrt(r2)-1: 1-sqrt(2-r2);
Vec d = cx*( ( (sx+.5 + dx)/2 + x)/w - .5) +
cy*( ( (sy+.5 + dy)/2 + y)/h - .5) + cam.d;
r = r + radiance(Ray(cam.o+d*140,d.norm()),0,Xi)*(1./samps);
} // Camera rays are pushed ^^^^^ forward to start in interior
c[i] = c[i] + Vec(clamp(r.x),clamp(r.y),clamp(r.z))*.25;
}
}
}
}
/* PROBLEM HERE!
The code never seems to reach here
PROBLEM HERE!
*/
FILE *f = fopen("image.ppm", "w"); // Write image to PPM file.
fprintf(f, "P3\n%d %d\n%d\n", w, h, 255);
for (int i=0; i<w*h; i++)
fprintf(f,"%d %d %d ", toInt(c[i].x), toInt(c[i].y), toInt(c[i].z));
}
编译代码会生成在其页面上看到的图像,但我无法使OpenMP并行化工作。
作为参考,我正在使用Cygwin和GCC 4.5.0在Windows 7 64位机器上进行编译。作者本人已声明他运行相同的代码并且没有遇到任何问题无论如何,但是当它完成跟踪图像时,我无法让程序实际退出。
这可能是我的特定编译器和环境的问题,还是我在这里做错了什么?这是使用OpenMP并行化的特定代码片段。我只修改了一些次要的格式以使其更具可读性。
$ time ./a
Rendering (4 spp) 100.00%spp) spp) 00..0026%%
这是程序在运行完成时产生的输出:
#include <cstdio>
#include <cstdlib>
#include <cmath>
struct Vector
{
double x, y, z;
Vector() : x(0), y(0), z(0) {}
};
int toInt(double x)
{
return (int)(255 * x);
}
double clamp(double x)
{
if (x < 0) return 0;
if (x > 1) return 1;
return x;
}
int main(int argc, char *argv[])
{
int w = 1024;
int h = 768;
int samples = 1;
Vector r, *c = new Vector[w * h];
#pragma omp parallel for schedule(dynamic, 1) private(r)
for (int y = 0; y < h; y++)
{
fprintf(stderr,"\rRendering (%d spp) %5.2f%%",samples * 4, 100. * y / (h - 1));
for (unsigned short x = 0, Xi[3]= {0, 0, y*y*y}; x < w; x++)
{
for (int sy = 0, i = (h - y - 1) * w + x; sy < 2; sy++)
{
for (int sx = 0; sx < 2; sx++, r = Vector())
{
for (int s = 0; s < samples; s++)
{
double r1 = 2 * erand48(Xi), dx = r1 < 1 ? sqrt(r1) - 1 : 1 - sqrt(2 - r1);
double r2 = 2 * erand48(Xi), dy = r2 < 1 ? sqrt(r2) - 1 : 1 - sqrt(2 - r2);
r.x += r1;
r.y += r2;
}
c[i].x += clamp(r.x) / 4;
c[i].y += clamp(r.y) / 4;
}
}
}
}
FILE *f = fopen("image.ppm", "w"); // Write image to PPM file.
fprintf(f, "P3\n%d %d\n%d\n", w, h, 255);
for (int i=0; i<w*h; i++)
fprintf(f,"%d %d %d ", toInt(c[i].x), toInt(c[i].y), toInt(c[i].z));
}
以下是可以重现上述行为的最基本代码
$ g++ test.cpp
$ ./a
Rendering (4 spp) 100.00%
$ g++ test.cpp -fopenmp
$ ./a
Rendering (4 spp) 100.00%spp) spp) 00..0052%%
这是从以下示例程序获得的输出:
{{1}}
答案 0 :(得分:2)
fprintf
没有受到关键部分或#pragma omp single/master
的保护。如果在Windows上这件事搞砸了控制台,我不会感到惊讶。