[英]Why is there no emplace or emplace_back for std::string?
[英]Why is emplace_back() behaving like this?
#include <iostream>
#include <vector>
class Base
{
private:
static int m_count;
public:
Base()
{
std::cout << " Base created. Count = " << ++m_count << std::endl;
}
~Base()
{
std::cout << " Base destroyed. Count = " << --m_count << std::endl;
}
void sayHello() const
{
std::cout << " Base says hello" << std::endl;
}
};
int Base::m_count = 0;
int main()
{
{
std::vector< Base > vBase;
vBase.emplace_back ( Base() ); // <- Why does ~Base() get called here
vBase[0].sayHello(); // <- Why is this function accessible after call to dtor
}
return 0;
}
程序輸出...
Base created. Count = 1
Base destroyed. Count = 0
Base says hello
Base destroyed. Count = -1
在調用vBase.emplace_back ( Base() );
您首先創建一個Base
對象。 該向量在原地創建另一個Base
,然后將第一個Base
擁有的資源移動到新的Base
中。 然后刪除第一個鹼基。 在您的向量中,您現在有一個移動的構造Base
,這就是調用sayHello()
起作用的原因。
您可能想要做的是讓emplace_back
實際構造對象,而無需手動創建臨時對象。 您只需提供構建Base
所需的參數即可做到這一點。 像這樣:
vBase.emplace_back();
你錯過了安位點。 Emplace 函數從給定的參數就地構造對象,而不是例如push_back
從預先存在的對象復制構造它。 您應該已經編寫了vBase.emplace_back()
,它在沒有構造函數參數的情況下就地構造向量內部的對象(即默認構造)。
就目前而言,您實際上默認通過Base()
構造一個Base
對象,將其傳遞給 emplace,它調用構造函數獲取Base
對象(即移動構造函數),復制它,然后復制原始對象,即Base()
對象被摧毀。
它的副本仍在向量中,這就是您仍然可以訪問它的原因。 被破壞的是臨時的。 第二個析構函數調用是當向量超出范圍時被銷毀的副本。
所以你基本上只是在做與push_back
相同的事情。
您沒有在計數器中包含由移動和復制構造函數創建的對象,也沒有記錄調用。 如果您對日志進行更改以修復違反三/五規則的規則,您將看到以下內容:
#include <typeinfo>
#include <iostream>
/// noisy
///
/// A class logs all of the calls to Big Five and the default constructor
/// The name of the template parameter, received as if by call
/// to `typeid(T).name()`, is displayed in the logs.
template<typename T>
struct noisy
{
noisy& operator=(noisy&& other) noexcept { std::cout << "MOVE ASSIGNMENT<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; return *this; }
noisy& operator=(const noisy& other) { std::cout << "COPY ASSIGNMENT<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; return *this; }
noisy(const noisy& other) { std::cout << "COPY CONSTRUCTOR<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; }
noisy(noisy&& other) noexcept { std::cout << "MOVE CONSTRUCTOR<" << typeid(T).name() << ">(this = " << this << ", other = " << &other << ")\n"; }
~noisy() { std::cout << "DESTRUCTOR<" << typeid(T).name() << ">(this = " << this << ")\n"; }
noisy() { std::cout << "CONSTRUCTOR<" << typeid(T).name() << ">(this = " << this << ")\n"; }
};
#include <iostream>
#include <vector>
class Base : public noisy<Base>
{
public:
void sayHello() const
{
std::cout << "Base says hello" << "(this = " << this << ")" << std::endl;
}
};
int main()
{
{
std::vector< Base > vBase;
vBase.emplace_back ( Base() ); // <- Why does ~Base() get called here
vBase[0].sayHello(); // <- Why is this function accessible after call to dtor
}
return 0;
}
輸出:
CONSTRUCTOR<4Base>(this = 0x7fff300b580f)
MOVE CONSTRUCTOR<4Base>(this = 0x18a6c30, other = 0x7fff300b580f)
DESTRUCTOR<4Base>(this = 0x7fff300b580f)
Base says hello(this = 0x18a6c30)
DESTRUCTOR<4Base>(this = 0x18a6c30)
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