加载内联 ptx 中的函数参数 [英] load function parameters in inlined ptx

问题描述

我有以下带有内联程序集的函数，可以在 32 位 Visual Studio 2008 的调试模式下正常工作:

I have the following function with inline assembly that works fine on debug mode in 32 bit Visual Studio 2008:

__device__ void add(int* pa, int* pb)
{
  asm(".reg .u32   s<3>;"::);
  asm(".reg .u32   r<14>;"::);

  asm("ld.global.b32    s0, [%0];"::"r"(&pa));      //load addresses of pa, pb
  printf(...);
  asm("ld.global.b32    s1, [%0];"::"r"(&pb));
  printf(...);
  asm("ld.global.b32    r1, [s0+8];"::);
  printf(...);  
  asm("ld.global.b32    r2, [s1+8];"::);
  printf(...);

  ...// perform some operations
}

pa和pb在设备上全局分配如

pa and pb are globally allocated on the device such as

__device__ int pa[3] = {0, 0x927c0000, 0x20000011};  
__device__ int pb[3] = {0, 0xbb900000, 0x2000000b};

但是，此代码在发布模式下失败，在行 asm("ld.global.b32 r1, [s0+8];"::);如何在发布模式下使用内联 ptx 正确加载函数参数?

However, this code fails on release mode, on line asm("ld.global.b32 r1, [s0+8];"::); How can I load function parameters correctly with inline ptx on release mode?

附言使用 -G 标志(生成 GPU 调试信息)构建发布模式会导致代码在发布模式下正确运行.谢谢，

P.S. building the release mode with -G flag (Generates GPU debug info) causes the code to run correctly on release mode. Thank you,

推荐答案

希望这段代码能有所帮助.我仍然在猜测你到底想做什么，但我从你的代码开始并决定在 pa 和 pb 数组中添加一些值并将它们存储回来转化为 pa[0] 和 pb[0].

Hopefully this code will help. I'm still guessing at what you are trying to do exactly, but I started with your code and decided to add some values in the pa and pb arrays and store them back into pa[0] and pb[0].

此代码是为 64 位机器编写的，但将其转换为 32 位指针应该不难.我用注释标记了需要为 32 位指针更改的行.希望这能回答您关于如何使用指向设备内存的函数参数的问题:

This code is written for a 64 bit machine but converting it to 32 bit pointers should not be difficult. I have marked the lines that need to be changed for 32 bit pointers with a comment. Hopefully this will answer your question about how to use function parameters that are pointers to device memory:

#include <stdio.h>

__device__ int pa[3] = {0, 0x927c0000, 0x20000011};
__device__ int pb[3] = {0, 0xbb900000, 0x2000000b};

__device__ void add(int* mpa, int* mpb)
{
  asm(".reg .u64   s<2>;"::);  // change to .u32 for 32 bit pointers
  asm(".reg .u32   r<6>;"::);

  asm("mov.u64    s0, %0;"::"l"(mpa));      //change to .u32 and "r" for 32 bit
  asm("mov.u64    s1, %0;"::"l"(mpb));      //change to .u32 and "r" for 32 bit
  asm("ld.global.u32    r0, [s0+4];"::);
  asm("ld.global.u32    r1, [s1+4];"::);
  asm("ld.global.u32    r2, [s0+8];"::);
  asm("ld.global.u32    r3, [s1+8];"::);
  asm("add.u32    r4, r0, r2;"::);
  asm("add.u32    r5, r1, r3;"::);
  asm("st.global.u32    [s0], r4;"::);
  asm("st.global.u32   [s1], r5;"::);
}

__global__ void mykernel(){
  printf("pa[0] = %x, pb[0] = %x
", pa[0], pb[0]);
  add(pa, pb);
  printf("pa[0] = %x, pb[0] = %x
", pa[0], pb[0]);
}

int  main() {
  mykernel<<<1,1>>>();
  cudaDeviceSynchronize();
  return 0;
}

当我运行此代码时，我得到:

When I run this code I get:

$ ./t128
pa[0] = 0, pb[0] = 0
pa[0] = b27c0011, pb[0] = db90000b
$

我认为这是正确的输出.

which I believe is correct output.

我编译它:

nvcc -O3 -arch=sm_20 -o t128 t128.cu

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