到位危险的隐式转换 [英] Dangerous implicit conversion in emplace

问题描述

以下代码使用gcc 6.3（ https://godbolt.org/g/sVZ8OH 进行编译，不会出现任何错误/警告。 ，但是由于下面标记了无效的内存访问，它包含危险的不确定行为。根本原因是在emplace_back中执行的隐式转换。有人可以建议一种避免此类代码错误的好方法或最佳实践吗？

The following code compiles without any errors/warnings with gcc 6.3 (https://godbolt.org/g/sVZ8OH), but it contains a dangerous undefined behavior due to invalid memory access marked below. The root cause is the implicit conversion performed in emplace_back. Can anyone suggest a good way or best practices to avoid such bugs in code?

#include <iostream>
#include <vector>

struct Foo
{
  explicit Foo(const int& i) : i{i} {}

  void foo() const { std::cout << i; }  // invalid memory access, as i is an invalid ref!!
  const int& i;
};

void bar(const double& d) {
  std::vector<Foo> fv;
  fv.emplace_back(d);
}

推荐答案

如果要接受const引用，但您不想要引用临时变量，使用rvalue引用参数声明一个附加构造函数-并将其删除。

If you are going to accept a const reference, but you don't want a reference to a temporary, declare an additional constructor with an rvalue reference argument - and delete it.

struct Foo
{
  explicit Foo(const int& i) : i{i} {}
  explicit Foo(const int&& i) = delete;  // This preferentially matches a reference to a
                                         // temporary, and compilation fails.

  void foo() const { std::cout << i; }  // invalid memory access, as i is an invalid ref!!
  const int& i;
};

（我假设实际问题不仅仅是一个int复杂。对于int，持有按值算是正确的答案。）

(I am assuming that the actual problem is more complex than just an int. For an int, holding it by value is the right answer.)

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