C++ 组合字符串向量
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Combining a vector of strings
提问by Bogdan
I've been reading Accelerated C++and I have to say it's an interesting book.
我一直在阅读Accelerated C++，不得不说这是一本有趣的书。
In chapter 6, I have to use a function from <algorithm> to concatenate from a vector<string> into a single string. I could use accumulate, but it doesn't help because string containers can only push_back characters.
在第 6 章中，我必须使用 <algorithm> 中的函数将 vector<string> 连接成单个字符串。我可以使用累积，但它没有帮助，因为字符串容器只能 push_back 字符。
int main () {
using namespace std;
string str = "Hello, world!";
vector<string> vec (10, str);
// Concatenate here?
return 0;
}
How do I join the strings together?
如何将字符串连接在一起？
回答by
Assuming this is question 6.8, it doesn't say you have to use accumulate - it says use "a library algorithm". However, you can use accumulate:
假设这是问题 6.8，它并没有说您必须使用累积 - 它说使用“库算法”。但是，您可以使用累积：
#include <numeric>
int main () {
std::string str = "Hello World!";
std::vector<std::string> vec(10,str);
std::string a = std::accumulate(vec.begin(), vec.end(), std::string(""));
std::cout << a << std::endl;
}
All that accumulate does is set 'sum' to the third parameter, and then for all of the values 'val' from first parameter to second parameter, do:
累积所做的只是将“sum”设置为第三个参数，然后对于从第一个参数到第二个参数的所有值“val”，请执行以下操作：
sum = sum + val
it then returns 'sum'. Despite the fact that accumulate is declared in <numeric>it will work for anything that implements operator+()
然后返回“总和”。尽管在其中声明了积累这一事实，<numeric>但它适用于任何实现operator+()
回答by Sanjaya R
How about std::copy?
std::copy 怎么样？
std::ostringstream os;
std::copy( vec_strings.begin(), vec_string.end(), ostream_iterator<string>( os ) );
cout << os.str() << endl;
回答by pixelgrease
The following snippet compiles in Visual C++ 2012 and uses a lambda function:
以下代码段在 Visual C++ 2012 中编译并使用 lambda 函数：
int main () {
string str = "Hello World!";
vector<string> vec (10,str);
stringstream ss;
for_each(vec.begin(), vec.end(), [&ss] (const string& s) { cat(ss, s); });
cout << ss.str() << endl;
}
The accumulateexample in the 1st answer is elegant, but as sellibitzepointed out, it reallocates with each concatenation and scales at O(N2). This for_eachsnippet scales at about O(N). I profiled both solutions with 100K strings; the accumulateexample took 23.6 secs, but this for_eachsnippet took 0.054 sec.
第accumulate一个答案中的示例很优雅，但正如sellibitze指出的那样，它重新分配每个连接并按 O(N2) 进行缩放。此for_each代码段的缩放比例约为 O(N)。我用 100K 字符串分析了两种解决方案；该accumulate示例耗时 23.6 秒，但此for_each代码段耗时 0.054 秒。
回答by Prasoon Saurav
I am not sure about your question.Where lies the problem? Its just a matter of a loop.
我不确定你的问题。问题出在哪里？它只是一个循环的问题。
#include<vector>
#include<string>
#include<iostream>
int main ()
{
std::string str = "Hello World!";
std::vector<string> vec (10,str);
for(size_t i=0;i!=vec.size();++i)
str=str+vec[i];
std::cout<<str;
}
EDIT :
编辑 ：
Use for_each()from <algorithm>
使用for_each()自<algorithm>
Try this:
尝试这个：
#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
string i;
void func(string &k)
{
i+=k;
}
int main () {
string str = "Hello World!";
vector<string> vec (10,str);
for_each(vec.begin(),vec.end(),func);
cout<<i;
return 0;
}
