在隐式整数降落/截断时，如何使NVCC出错？

Question

考虑此缩放内核：

__global__ void foo(int *const a, const int N){
    const auto tid = threadIdx.x + blockDim.x * blockIdx.x;
    // Implicit downcast/truncation from int64 to int32 - how can we get an error?
    a[tid] = tid * 3L;
}

如何获得

a[tid] = tid * 3L

？

的错误 在GCC中会标记类似的情况：

-Wconverion

在clang中

<source>:27:9: warning: conversion from ‘int64_t’ {aka ‘long int’} to ‘int’ may change value [-Wconversion]
   27 |     int y = x;
      |         ^

将升起：

-Wshorten64-to-32最小工作示例

<source>:6:11: warning: implicit conversion loses integer precision: 'int64_t' (aka 'long') to 'int' [-Wshorten-64-to-32]
    6 |     int y=x;
      |         ~ ^

一种可能的方法：使内核代码的主体成为功能，从内核中调用该函数，然后使用

#include <cstdint>
#include <iostream>
#include <limits>
#include <vector>

__global__ void foo(int *const a, const int N){
    // Wrong data type: it would be nice to detect this
    const int tid = threadIdx.x + blockDim.x * blockIdx.x;
    // Implicit downcast/truncation from int64 to int32 - how can we get an error?
    a[tid] = tid * 3L;
}

int main(){
    constexpr int N = 10000;
    constexpr int threads = 128;
    int *a_d;
    cudaMalloc(&a_d, N * sizeof(int));
    foo<<<(N + threads - 1) / 128, threads>>>(a_d, N);

    std::vector<int> v(N);
    cudaMemcpy(v.data(), a_d, N * sizeof(int), cudaMemcpyDeviceToHost);
    for(int i=0;i<10;i++){
        std::cout<<v.at(i)<<std::endl;
    }

    return 0;
}

：

：

Answer 1