有什么情况下，用emplace_back替换push_back是不正确的？ [英] Are there any cases where it is incorrect to replace push_back with emplace_back?

问题描述

我可以用 emplace_back 替换 std :: vector :: push_back 并用C ++ 11编译器编译它？从阅读 emplace_back 引用我收集它不应该发生，但我承认我不完全得到右值引用。

push_back 被替换为 emplace_back / code>：

  #include< vector& 
 struct S {
 S（double）{} 
 private：
 explicit S（int）{} 
}; 
 int main（）{
 std :: vector< S>（）。push_back（0）; // OK 
 std :: vector< S>（）。emplace_back（0）; //错误！ 
} 
  

调用 push_back 需要将其 0 类型的类型转换为 c>。由于这是一个隐式转换，不考虑显式构造函数 S :: S（int）， S :: S（double） code>被调用。另一方面， emplace_back 执行直接初始化，因此 S :: S（double）和 S :: S（int）。后者是更好的匹配，但它 private ，所以程序是不成形的。

Can I break a valid C++03 program by replacing std::vector::push_back with emplace_back and compiling it with C++ 11 compiler? From reading emplace_back reference I gather it shouldn't happen, but I'll admit I don't fully get rvalue references.

解决方案

I constructed a short example that actually fails to compile when push_back is replaced by emplace_back:

#include <vector>
struct S {
    S(double) {}
  private:
    explicit S(int) {}
};
int main() {
    std::vector<S>().push_back(0); // OK
    std::vector<S>().emplace_back(0); // error!
}

The call to push_back needs to convert its argument 0 from type int to type S. Since this is an implicit conversion, the explicit constructor S::S(int) is not considered, and S::S(double) is called. On the other hand, emplace_back performs direct initialization, so both S::S(double) and S::S(int) are considered. The latter is a better match, but it's private, so the program is ill-formed.

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