C++ 容器元素的初始化

//default string constructor and five string copy constructors invoked
vector<string> svec(5);
书上说,编译器首先使用string默认构造函数创建一个临时值来初始化svec，然后使用复制构造函数将临时值复制到svec的每一个元素。

1、首先这个临时值是一个string类型的临时值吧，用它来初始化svec，是不是创建了一个空的string类型的容器？

2、然后说调用复制构造函数，是一个什么样的复制构造函数，从英文解释来看，应该是string类的复制构造函数，但是怎么能用上面那个临时制调用string类的复制构造函数呢？

还是我理解的根本就不对，求高手们指点!?

程序清单：

#include <iostream>
#include <list>
#include <string>
#include <vector>
#include <deque>

using namespace std;

int main(int argc,char *argv[])
{
char *words[] = {"first","second","third","fourth","fifth"};

size_t words_size = sizeof(words)/sizeof(char *);


vector<string> words1(words,words+words_size);

list<string> words2(words,words+words_size);

cout<<"The elements of words1 are below:"<<endl;
for(vector<string>::iterator it=words1.begin();
it!=words1.end();++it)
{
cout<<*it<<" ";
}
cout<<endl;

cout<<"The elements of words2 are below:"<<endl;
for(list<string>::iterator it=words2.begin();it!=words2.end();
++it)
{
cout<<*it<<" ";
}
cout<<endl;


vector<string>::iterator mid = words1.begin() + words1.size()/2;


vector<string> vfront(words1.begin(),mid);

vector<string> vback(mid,words1.end());



deque<string> dfront(words1.begin(),mid);

deque<string> dback(mid,words1.end());

cout<<"The elements of vfront are below:"<<endl;
for(vector<string>::iterator it=vfront.begin();it!=vfront.end();
++it)
{
cout<<*it<<" ";
}
cout<<endl;

cout<<"The elements of vback are below:"<<endl;
for(vector<string>::iterator it=vback.begin();it!=vback.end();
++it)
{
cout<<*it<<" ";
}
cout<<endl;

cout<<"The elements of dfront are below:"<<endl;
for(deque<string>::iterator it=dfront.begin();it!=dfront.end();
++it)
{
cout<<*it<<" ";
}
cout<<endl;

cout<<"The elements of dback are below:"<<endl;
for(deque<string>::iterator it=dback.begin();it!=dback.end();
++it)
{
cout<<*it<<" ";
}
cout<<endl;


list<string> slist(25,"love");
int count = 0;
cout<<"The elements of slist are below:"<<endl;
for(list<string>::iterator it=slist.begin();it!=slist.end();
++it)
{
cout<<*it<<" ";
++count;
if(count%5 == 0)
{
cout<<endl;
}
}
// cout<

return 0;
}
编译并运行程序后的执行结果：
The elements of words1 are below:
first second third fourth fifth
The elements of words2 are below:
first second third fourth fifth
The elements of vfront are below:
first second
The elements of vback are below:
third fourth fifth
The elements of dfront are below:
first second
The elements of dback are below:
third fourth fifth
The elements of slist are below:
love love love love love
love love love love love
love love love love love
love love love love love
love love love love love

对于vector容器来说，声明变量后讲调用构造函数，假如你已经开始学习面向对象编程了，你将容易理解。 vector x(5);调用一个参数的构造函数，构造出具有5个int型值得对象，默认值为0 (5,5);调用2个参数的构造函数，初始化为5个相同的元素，其值为5。 而(5,5,5);没有对应的构造函数，将调用错误。

书上的意思大概是 svec[0]=string(); for (size_t i=1;i<5;i++) svec[i]=svec[0]; 之类的东西吧……

书上说的有问题，做一个实验即可验证： 这是vector的一个构造函数，const Allocator&是迭代器，不用管 explicit vector ( size_type n, const T& value= T(), const Allocator& = Allocator() ); Repetitive sequence constructor: Initializes the vector with its content set to a repetition, n times, of copies of value. 从这句翻译上是，n次调用构造函数（类型是T，value）来初始化vector对象的每一个元素，而不是仅仅调用一次T类型的构造函数，在这里的T相当于你的string，value使用默认，为空string 实验： #include<iostream> #include<vector> #include<string> using namespace std; 例： class string1{ public: string1(){cout<<"default"<<endl;}; string1(const string1& x){cout<<"copy"<<endl;} }; int main(){ vector<string1> sec(5); return 0; } 运行结果为5个default，也就是说不调用赋制构造函数的，更说明书上说的是错误的 当然你的问题也就回答了， 自己运行一下这个程序，应该明白的