c++中如何添加多个字符串
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How to add many strings in c++
提问by cpx
As I know that C++ only allows to add 2 strings together, i.e:
s = s1 + s2
据我所知，C++ 只允许将 2 个字符串相加，即：
s = s1 + s2
But how can I add many strings together? Like:
但是如何将多个字符串相加呢？喜欢：
s = s1 + s2 + s3 + s4 + ... + sn
回答by cpx
If you're trying to append string objects of std::string class, this should work.
如果您尝试附加 std::string 类的字符串对象，这应该可行。
string s1 = "string1";
string s2 = "string2";
string s3 = "string3";
string s = s1 + s2 + s3;
OR
或者
string s = string("s1") + string("s2") + string("s3") ...
回答by Billy ONeal
First of all, you can do the +sn thing just fine. Though it's going to take exponentialquadradic(see comments) time assuming you're using std::basic_string<t>strings on C++03.
首先，你可以很好地做 +sn 的事情。尽管假设您在 C++03 上使用字符串，它将花费指数二次（见评论）时间std::basic_string<t>。
You can use the std::basic_string<t>::appendin concert with std::basic_string<t>::reserveto concatenate your string in O(n) time.
您可以使用std::basic_string<t>::appendin concert withstd::basic_string<t>::reserve在 O(n) 时间内连接您的字符串。
EDIT: For example
编辑：例如
string a;
//either
a.append(s1).append(s2).append(s3);
//or
a.append("I'm a string!").append("I am another string!");
回答by R Samuel Klatchko
s = s1 + s2 + s3 + .. + sn;
will work although it could create a lot of temporaries (a good optimizing compiler should help) because it will effectively be interpreted as:
尽管它可能会创建很多临时文件（一个好的优化编译器应该会有所帮助），但它会起作用，因为它会被有效地解释为：
string tmp1 = s1 + s2;
string tmp2 = tmp1 + s3;
string tmp3 = tmp2 + s4;
...
s = tmpn + sn;
An alternate way that is guaranteed not to create temporaries is:
保证不创建临时文件的另一种方法是：
s = s1;
s += s2;
s += s3;
...
s += sn;
回答by Nikolai Fetissov
回答by QuentinUK
Use a template to add strings, char* and char's to form a string
使用模板添加字符串，char* 和char's 组成一个字符串
strlen:-
斯特伦：-
#include <iostream>
#include <cstring>
// it_pair to wrap a pair of iterators for a for(:) loop
template<typename IT>
class it_pair
{
IT b;
IT e;
public:
auto begin() const
{
return b;
}
auto end() const
{
return e;
}
};
// string length
template<typename S> auto strlen(const S& s) -> decltype(s.size())
{
return s.size();
}
auto strlen(char c) -> size_t
{
return 1u;
}
auto strlen(const std::initializer_list<char>& il) -> size_t
{
return il.size();
}
template<typename IT>
auto strlen(const it_pair<IT>& p)
{
auto len = size_t{};
for(const auto& s:p)
len += strlen(s);
return len;
}
template<typename S, typename ...SS> auto strlen(S s, SS&... ss) -> size_t
{
return strlen(s) + strlen(ss...);
}
appending strings
附加字符串
// terminate recursion
template<typename TA, typename TB>
void append(TA& a, TB& b)
{
a.append(b);
}
// special case for a character
template<>
void append<std::string, const char>(std::string& a, const char& b)
{
a.append(1, b);
}
// special case for a collection of strings
template<typename TA, typename TB>
void append(TA& a, const it_pair<TB>& p)
{
for(const auto& x: p)
a.append(x);
}
// recursion append
template<typename TA, typename TB, typename ...TT>
void append(TA& a, TB& b, TT&... tt)
{
append(a, b);
append(a, tt...);
}
template<typename ...TT>
std::string string_add(const TT& ... tt)
{
std::string s;
s.reserve(strlen(tt...));
append(s, tt...);
return s;
}
template<typename IT>
auto make_it_pair(IT b, IT e)
{
return it_pair<IT>{b, e};
}
template<typename T>
auto make_it_pair(const T& t)
{
using namespace std;
return make_it_pair(cbegin(t), cend(t));
}
main example
主要例子
int main()
{
const char * s[] = {"vw", "xyz"};
std::vector<std::string> v{"l", "mn", "opqr"};
std::string a("a");
std::string b("bc");
std::string c("def");
std::cout << string_add(a, b+c, "ghij", make_it_pair(v), 'k', make_it_pair(s));
}
回答by Michael Hall
If you want to be able to do this
如果你想能够做到这一点
- Efficiently, i.e without incurring quadratic time
- Without overwriting the original variable
- Without creating temporary variables
- 高效，即不产生二次时间
- 不覆盖原始变量
- 不创建临时变量
Then this will do the job
然后这将完成工作
auto s = std::string(s1).append(s2).append(s3).append(sn);
And if you like things nicely formatted
如果你喜欢格式很好的东西
auto s = std::string(s1).append(s2)
.append(s3)
.append(sn)
