当且仅当我删除Foo的自定义析构函数时,以下代码才会编译.
struct Foo {
std::unique_ptr <int> bar;
~Foo (void) {} // This Line
};
std::vector <Foo> foos;
foos.push_back (Foo ());
以下是我认为我对这种情况的理解:
它失败,因为unique_ptrs无法复制,并std::vector::push_back (thing)调用thing's复制构造函数.如果我编写Foo一个明确移动的自定义复制构造函数bar,那么一切都会好的.
但是,禁用This Line将导致代码编译.
我想,这应该无法编译,即使没有This Line,因为我仍然在尝试push_back一个unique_ptr.
为什么没有自定义析构函数就能成功,为什么添加自定义析构函数会导致它失败？
编辑:gcc -std=gnu++11在Debian Linux 64位上使用
int main(){
vector<Customer*> newcustomer;
newcustomer.push_back(new Customer("III", 123333, 555));
newcustomer.push_back(new Customer("LOL", 122222, 444));
newcustomer.push_back(new Customer("PPL", 121111, 333));
for (int i = 0; i < 3; i++){
cout << newcustomer[i]->getName() << endl;
cout << newcustomer[i]->getPhone() << endl;
cout << newcustomer[i]->getID() << endl;
cout << endl;
}
system("pause");
return 0;
}
所以我创建了一个名为customer的类,你可以插入新客户,getName返回名称,getPhone返回电话号码,GetID返回ID,现在我想从该向量中删除所有内容,不知道该怎么做.
我刚开始使用C++,现在我有一个非常基本的问题.
我写了两节课:
坐标:
#include <stdio.h>
class Coordinate {
private:
int x;
int y;
public:
Coordinate(int a, int b) {
x = a;
y = b;
};
void printTest() {
printf("%d %d\n", x, y);
};
};
测试:
class Test {
private:
int q;
Coordinate *point;
public:
Test(int a, int b, int c) {
q = a;
point = new Coordinate(b, c);
};
virtual ~Test() {
delete point;
}
};
主功能:
int main() {
Test *test = new Test(1, 2, 3);
// …在本文中http://cpp-next.com/archive/2009/08/want-speed-pass-by-value/comment-page-1/#comment-1877:
T& T::operator=(T const& x) // x is a reference to the source
{
T tmp(x); // copy construction of tmp does the hard work
swap(*this, tmp); // trade our resources for tmp's
return *this; // our (old) resources get destroyed with tmp
}
但鉴于副本的缺失,这种表述显然效率低下!现在"显而易见"的是,编写复制和交换分配的正确方法是:
T& operator=(T x) // x is a copy of the source; hard work already done
{
swap(*this, x); // trade our resources for x's
return *this; // our (old) resources get destroyed with …我有以下课程:
class Patient {
public:
Patient(int x);
~Patient();
private:
int* RP;
};
Patient::Patient(int x) { RP = new int [x]; }
Patient::~Patient() { delete [] RP; }
我在堆栈上创建了这个类的实例,如下所示:
void f() { Patient p(10); }
现在,当f()返回时,我得到一个"双重免费或损坏"错误,它向我发出信号,表示尝试删除的内容不止一次.但我不明白为什么会这样.数组的空间是在堆上创建的,只是因为分配了空间的函数返回,我不希望回收空间.
我认为如果我在堆上分配空间(使用new关键字),那么回收该空间的唯一方法是使用delete关键字.救命!
根据要求,这是实际的代码(为简洁起见略有删节)
这是完整的类定义(拆分.cpp和.h文件,但一起显示):
class Patient {
public:
Patient(int numDecisionEpochs);
~Patient();
void recordRP(const int& age, const bool& t);
void recordBiopsy(const int& age, const int& result);
void recordPSA(const int& age, const double& level);
void recordPSA(const int& age);
private:
int* …我做了一个程序来评估这样做之间的性能差异:
func3(func2(func1()));
对此:
retval1 = func1();
retval2 = func2(retval1);
func3(retval2);
我更喜欢后者的可读性和易于调试,我想知道编译器(MSVC 12.0)是否会优化发布版本中的中间对象.我的测试程序是这样的:
#include <iostream>
using namespace std;
struct Indicator {
Indicator() {cout << "Default constructor" << endl;}
Indicator(const Indicator& other) {cout << "Copy constructor" << endl;}
const Indicator& operator=(const Indicator& other) {cout << "Assignment operator" << endl;}
~Indicator() {cout << "Destructor" << endl;}
};
Indicator func1()
{return Indicator();}
Indicator func2(Indicator&& i)
{return std::move(i);}
Indicator func3(Indicator&& i)
{return std::move(i);}
int main() {
Indicator i = func3(func2(func1()));
cout << &i << endl; …我正在使用GCC 5.2和clang 3.6进行测试,两者都在C++ 14模式下,并且它们提供相同的输出.
对于以下代码
#include <iostream>
#include <type_traits>
struct S {
// S& operator= (S&&) noexcept { return *this; }
};
int main() {
std::cout << std::is_nothrow_move_constructible<S>::value
<< std::is_nothrow_move_assignable<S>::value;
}
11得到的结果.但是如果取消注释移动赋值运算符,则输出变为01.如何noexcept对移动赋值运算符的显式规范可能会影响移动构造函数的规范？
我想做的是编译由 CMake 构建的项目。在我的代码中我有下一个方法：
/** "in-place" version of TriangularView::solve() where the result is written in \a other
*
* \warning The parameter is only marked 'const' to make the C++ compiler accept a temporary expression here.
* This function will const_cast it, so constness isn't honored here.
*
* See TriangularView:solve() for the details.
*/
template<typename MatrixType, unsigned int Mode>
template<int Side, typename OtherDerived>
void TriangularView<MatrixType,Mode>::solveInPlace(const MatrixBase<OtherDerived>& _other) const
{
OtherDerived& other = _other.const_cast_derived();
eigen_assert( cols() == rows() && ((Side==OnTheLeft && …在C++ 11中测试线程时,我创建了以下示例:
#include <iostream>
#include <thread>
class Foo {
public:
Foo(void) {
std::cout << "Constructor called: " << this << std::endl;
}
~Foo(void) {
std::cout << "Destructor called: " << this << std::endl;
}
void operator()() const {
std::cout << "Operatior called: " << this << std::endl;
}
};
void test_normal(void) {
std::cout << "====> Standard example:" << std::endl;
Foo f;
}
void test_thread(void) {
std::cout << "====> Thread example:" << std::endl;
Foo f;
std::thread t(f);
t.detach();
}
int main(int …