Question

如果我写下面的CUDA代码：

a.cu(8): warning: statement is unreachable
          detected during instantiation of "void foo<N>() [with N=5U]" 
(12): here

我收到编译器警告：

#include <cstdio>

template <unsigned N>
void foo()
{
    std::printf("In kernel foo() with N = %u\n", N);
    if (N < 10) { return; }
    std::printf("Wow, N is really high!\n");
    /* a whole lot of code here which I don't want to indent */
}

int main() {
    foo<5>();
    foo<20>();
    return 0;
}

我感觉＆＃34;我不应该收到此警告，因为无法访问的代码仅对模板参数的某些值无法访问。如果我写＆＃34; CPU等价物＆＃34;，可以这么说：

-Wall

并使用gcc（5.4.0）构建它 - 即使我使用if (not (N < 10)) { printf("Wow, N is really high!\n"); /* a whole lot of code here which I don't want to indent */ }进行编译，我也不会收到任何警告。

现在，我可以通过编写

来规避这一点

if (not (N < 10)) { 
    return;
}
else {
    printf("Wow, N is really high!\n");
    /* a whole lot of code here which I don't want to indent */
}

但我宁愿避免不得不改变我的逻辑来跳过nvcc＆＃34; hoop＆＃34;。我也可以写

[super touchesEnded:touches withEvent:event];

但是 - 我不想缩进所有代码（同样的问题可能会再次出现，在else块中需要更多的缩进。

我能做些什么吗？此外，这不是一个错误＆＃34;或者我应该报告错误的错误？

Answer 1

怎么样：

template<unsigned N, bool>
struct FooImpl
{
  static void foo()
  {
    std::printf("In kernel foo() with N = %u\n", N);
  }
};

template<unsigned N>
struct FooImpl<N, false>
{
  static void foo()
  {
    std::printf("In kernel foo() with N = %u\n", N);
    std::printf("Wow, N is really high!\n");
    /* a whole lot of code here which I don't want to indent */
  }
};

template <unsigned N>
__global__ void foo()
{
  FooImpl<N, N < 10>::foo();
}

如何避免＆＃34;无法访问的声明＆＃34;由于编译时计算的值？

1 个答案: