I am interested if creating a new std::vector
(or calling its assign
method) creates a copy of the data?
我感兴趣的是,如果创建一个新的std :: vector(或调用其assign方法)创建数据的副本?
For example,
void fun(char *input) {
std::vector<char> v(input, input+strlen(input));
// is it safe to assume that the data input points to was COPIED into v?
}
5 个解决方案
#1
11
Yes. Elements are always copied into or out of STL containers. (At least until move semantics are added in C++0x)
是。元素始终复制到STL容器中或从STL容器中复制出来。 (至少在C ++ 0x中添加移动语义之前)
EDIT: Here's how you can test for copying yourself:
编辑:这是你如何测试自己复制:
#include <vector>
#include <iostream>
class CopyChecker
{
public:
CopyChecker()
{
std::cout << "Hey, look! A new copy checker!" << std::endl;
}
CopyChecker(const CopyChecker& other)
{
std::cout << "I'm the copy checker! No, I am! Wait, the"
" two of us are the same!" << std::endl;
}
~CopyChecker()
{
std::cout << "Erroap=02-0304-231~No Carrier" << std::endl;
}
};
int main()
{
std::vector<CopyChecker> doICopy;
doICopy.push_back(CopyChecker());
}
The output should be:
输出应该是:
Hey, look! A new copy checker!
I'm the copy checker! No, I am! Wait, the two of us are the same!
Erroap=02-0304-231~No Carrier
Erroap=02-0304-231~No Carrier你看!一个新的复制检查器!我是复制检查员!不,我是!等等,我们两个是一样的! Erroap = 02-0304-231~无载流子错误= 02-0304-231~无载波
#2
9
Elements are always copied into or out of STL containers.
元素始终复制到STL容器中或从STL容器中复制出来。
Although the element may just be a pointer, in which case the pointer is copied but not the underlying data
虽然元素可能只是一个指针,但在这种情况下指针被复制但不复制基础数据
#3
1
About the move semantics, here is how you could move the contents in C++0x if you wanted to:
关于移动语义,如果你想要,你可以在这里移动C ++ 0x中的内容:
void fun_move(char *input)
{
std::vector<char> v;
auto len = strlen(input);
v.reserve(len);
std::move(input, input+len, std::back_inserter(v));
}
#4
0
If you want your data to be moved, use std::swap_ranges
, but you have to allocate for memory first :
如果要移动数据,请使用std :: swap_ranges,但必须先为内存分配:
vector<T> v;
v.reserve(std::distance(beg, end));
std::swap_ranges(beg, end, v.begin());
#5
0
If you do not want the object copy semantics, then you can create a vector of pointers-to-objects instead so that only the pointer is copied. However you then have to ensure that the pointers then remain valid for the lifetime of the container.
如果您不想要对象复制语义,则可以创建指向对象的向量,以便仅复制指针。但是,您必须确保指针在容器的生命周期内保持有效。
#1
11
Yes. Elements are always copied into or out of STL containers. (At least until move semantics are added in C++0x)
是。元素始终复制到STL容器中或从STL容器中复制出来。 (至少在C ++ 0x中添加移动语义之前)
EDIT: Here's how you can test for copying yourself:
编辑:这是你如何测试自己复制:
#include <vector>
#include <iostream>
class CopyChecker
{
public:
CopyChecker()
{
std::cout << "Hey, look! A new copy checker!" << std::endl;
}
CopyChecker(const CopyChecker& other)
{
std::cout << "I'm the copy checker! No, I am! Wait, the"
" two of us are the same!" << std::endl;
}
~CopyChecker()
{
std::cout << "Erroap=02-0304-231~No Carrier" << std::endl;
}
};
int main()
{
std::vector<CopyChecker> doICopy;
doICopy.push_back(CopyChecker());
}
The output should be:
输出应该是:
Hey, look! A new copy checker!
I'm the copy checker! No, I am! Wait, the two of us are the same!
Erroap=02-0304-231~No Carrier
Erroap=02-0304-231~No Carrier你看!一个新的复制检查器!我是复制检查员!不,我是!等等,我们两个是一样的! Erroap = 02-0304-231~无载流子错误= 02-0304-231~无载波
#2
9
Elements are always copied into or out of STL containers.
元素始终复制到STL容器中或从STL容器中复制出来。
Although the element may just be a pointer, in which case the pointer is copied but not the underlying data
虽然元素可能只是一个指针,但在这种情况下指针被复制但不复制基础数据
#3
1
About the move semantics, here is how you could move the contents in C++0x if you wanted to:
关于移动语义,如果你想要,你可以在这里移动C ++ 0x中的内容:
void fun_move(char *input)
{
std::vector<char> v;
auto len = strlen(input);
v.reserve(len);
std::move(input, input+len, std::back_inserter(v));
}
#4
0
If you want your data to be moved, use std::swap_ranges
, but you have to allocate for memory first :
如果要移动数据,请使用std :: swap_ranges,但必须先为内存分配:
vector<T> v;
v.reserve(std::distance(beg, end));
std::swap_ranges(beg, end, v.begin());
#5
0
If you do not want the object copy semantics, then you can create a vector of pointers-to-objects instead so that only the pointer is copied. However you then have to ensure that the pointers then remain valid for the lifetime of the container.
如果您不想要对象复制语义,则可以创建指向对象的向量,以便仅复制指针。但是,您必须确保指针在容器的生命周期内保持有效。