c++ - {fmt}等同于cout.rdbuf？

Question

我有一个单元测试，它通过读取函数发送的缓冲区来验证函数:

template <typename Manifold>
void print_manifold(Manifold const& manifold)
try
{
  std::cout << "Manifold has " << manifold.N0() << " vertices and "
            << manifold.N1() << " edges and " << manifold.N2() << " faces and "
            << manifold.N3() << " simplices.\n";
  // fmt::print(
  //    "Manifold has {} vertices and {} edges and {} faces and {}
  //    simplices.\n", manifold.N0(), manifold.N1(), manifold.N2(),
  //    manifold.N3());
}
catch (...)
{
  std::cerr << "print_manifold() went wrong ...\n";
  throw;
}  // print_manifold

和:

SCENARIO("Printing results", "[utility]")
{
  // redirect std::cout
  stringstream buffer;
  cout.rdbuf(buffer.rdbuf());
  GIVEN("A Manifold3")
  {
    Manifold3 const manifold(640, 4);
    WHEN("We want to print statistics on a manifold.")
    {
      THEN("Statistics are successfully printed.")
      {
        print_manifold(manifold);
        CHECK_THAT(buffer.str(), Catch::Contains("Manifold has"));
      }
    }
}

有没有办法捕获由 fmt::print生成的输出到 stdout的输出？

当我注释掉 cout代码并取消注释 fmt代码时，我得到了由 cout <<的先前实例产生的缓冲区。

Answer 1

这比{fmt}问题更像是一个C stdio问题，但是您可以按照Redirecting stdout to pipe in C的答案中所述将stdout重定向到管道并从中读取输出。尽管这不是一个好的单元测试，因为它取决于全局状态，但是您当前的测试也存在相同的问题。