c++ - 限制(放大器)功能的默认参数

Question

以下代码编译失败。错误信息是:

错误一:

error C3930: 'foo' : no overloaded function has restriction specifiers that are compatible with the ambient context ''

错误 2:

error C2660: 'f1' : function does not take 0 arguments

错误 3:

IntelliSense: amp-restricted function "int foo() restrict(amp)" (declared at line 5) must be called from an amp-restricted function

程序:

#include <amp.h>
#include <iostream>
using namespace std;

int foo() restrict(amp) { return 5; }

int f1(int x = foo()) restrict(amp) {
  return x;
}

int main()
{
  using namespace concurrency;

  int a[10] = {0};
  array_view<int> av(10, a);

  parallel_for_each(av.extent, [=](index<1> i) restrict(amp) {
    av[i] = f1();
  });

  for(unsigned i=0; i<10; ++i) {
    cout << av[i] << "\n";
  }
  return 0;
}

奇怪的是，当我删除 foo() 上的 restrict(amp) 并替换 lambda 中的 f1() 调用时例如，5，程序可以正常编译。那么 amp 函数的默认参数中的函数调用规则是什么？

Answer 1

MSDN Forum answer问题。

The semantics of the default arguments we have chosen are aligned with the overarching premise of C++ that parsing of a program is done in a one left-to-right pass (notwithstanding few significant exceptions to this rule, most notably member functions) - therefore since the restriction specifier is read after the function parameter declaration, any function calls located in default argument expressions are bound according to the "outer" restriction specification, for better or for worse. In other words, you read the program from the beginning with the cpu-restriction "active" (because it's the default one) and switch to restriction X for everything between "restrict(X)" and "}" closing the relevant scope.