如果我编写以下CUDA代码:
#include <stdio.h>
template <unsigned N>
__global__ void foo()
{
printf("In kernel foo() with N = %u\n", N);
if (N < 10) { return; }
printf("Wow, N is really high!\n");
/* a whole lot of code here which I don't want to indent */
}
int main() {
foo<5><<<1,1>>>();
foo<20><<<1,1>>>();
return 0;
}
我收到编译器警告:
a.cu(8): warning: statement is unreachable
detected during instantiation of "void foo<N>() [with N=5U]"
(12): here
我“觉得”我不应该收到此警告,因为只有模板参数的某些值才无法访问代码。如果我写“ CPU等效”,可以这么说:
#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;
}
并使用gcc(5.4.0)构建它-即使使用
-Wall
进行编译,我也不会收到任何警告。现在,我可以通过写来规避
if (not (N < 10)) {
printf("Wow, N is really high!\n");
/* a whole lot of code here which I don't want to indent */
}
但是我宁可避免不得不颠倒自己的逻辑以跳过nvcc的“篮球”。我也可以写
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 */
}
但是-我不想缩进所有代码(并且可能再次发生相同的问题,在else块中甚至需要更多缩进。
有什么我可以做的吗?另外,这不是“错误”,还是我应该报告为错误的错误功能?
最佳答案
关于什么:
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();
}
关于c++ - 如何避免由于编译时计算的值导致的“无法到达的语句”?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/43158760/