使用 pycuda (lerp) 进行线性插值

发布于 2024-12-25 07:27:55 字数 635 浏览 3 评论 0原文

我是一个刚刚接触 pyCUDA 的休闲 Python 爱好者。我试图弄清楚如何使用 pyCUDA 实现线性插值（lerp）。 CUDA CG函数为： http://http.developer.nvidia.com/Cg/lerp .html

我的最终目标是在 pycuda 中根据一组加权随机点进行双线性插值。我从来没有编写过 C 语言或 CUDA 程序，并且正在边学习边学习。

这就是我已经取得的进展：

import pycuda.autoinit
import pycuda.driver as drv
import pycuda.compiler as comp

lerpFunction = """__global__ float lerp(float a, float b, float w)
{
    return a + w*(b-a);
}"""

mod = comp.SourceModule(lerpFunction) # This returns an error telling me a global must return a void. :(

对此的任何帮助都非常棒！

原文

I am a recreational pythonista who just got into pyCUDA. I am trying to figure out how to implement a linear interpolation (lerp) using pyCUDA. The CUDA CG function is: http://http.developer.nvidia.com/Cg/lerp.html

My ultimate goal is a bilinear interpolation in pycuda from a set of weighted random points. I've never programmed C, or CUDA for that matter, and am learning as I go.

This is how far I've gotten:

import pycuda.autoinit
import pycuda.driver as drv
import pycuda.compiler as comp

lerpFunction = """__global__ float lerp(float a, float b, float w)
{
    return a + w*(b-a);
}"""

mod = comp.SourceModule(lerpFunction) # This returns an error telling me a global must return a void. :(

Any help on this would be fantastic!

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把时间冻结 2025-01-01 07:27:55

错误消息非常明确 - CUDA 内核无法返回值，它们必须声明为 void，并且可修改的参数作为指针传递。将 lerp 实现声明为这样的设备函数会更有意义：

__device__ float lerp(float a, float b, float w)
{
    return a + w*(b-a);
}

然后从内核内部调用每个需要插值的值。您的 lerp 函数缺乏很多“基础设施”来成为有用的 CUDA 内核。

编辑：一个真正基本的内核沿着同样的路线可能看起来像这样：

__global__ void lerp_kernel(const float *a, const float *b, const float w, float *y)
{
    int tid = threadIdx.x + blockIdx.x*blockDim.x; // unique thread number in the grid
    y[tid] = a[tid] + w*(b[tid]-a[tid]);
}

The error message is pretty explicit - CUDA kernels cannot return values, they must be declared void, and modifiable arguments passed as pointers. It would make more sense for your lerp implementation to be declared as a device function like this:

__device__ float lerp(float a, float b, float w)
{
    return a + w*(b-a);
}

and then called from inside a kernel for each value that requires interpolation. Your lerp function lacks a lot of "infrastructure" to be a useful CUDA kernel.

EDIT: A really basic kernel along the same lines might look something like this:

__global__ void lerp_kernel(const float *a, const float *b, const float w, float *y)
{
    int tid = threadIdx.x + blockIdx.x*blockDim.x; // unique thread number in the grid
    y[tid] = a[tid] + w*(b[tid]-a[tid]);
}

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