C++ 从字符串流中删除字符并附加一些数据
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remove char from stringstream and append some data
提问by lord.didger
In my code there is a loop that adds sth like that "number," to stringstream. When it ends, I need to extract ',' add '}' and add '{' if the loop is to repeated.
在我的代码中，有一个循环将诸如“数字”之类的东西添加到 stringstream。当它结束时，如果要重复循环，我需要提取 ',' 添加 '}' 并添加 '{'。
I thought i can use ignore() to remove ',' but it didn't work. Do you know how I can do what I describe?
我以为我可以使用 ignore() 来删除 ',' 但它没有用。你知道我怎样才能做到我所描述的吗？
example:
例子：
douCoh << '{';
for(unsigned int i=0;i<dataSize;i++)
if(v[i].test) douCoh << i+1 << ',';
douCoh.get(); douCoh << '}';
回答by bcmpinc
You can seek the stringstreamand go back 1 character, using stringstream::seekp. Note that it does not remove the last character, but only moves the write head. This is sufficient in this case, as we overwrite the last character with an }.
您可以使用 查找stringstream并返回 1 个字符stringstream::seekp。请注意，它不会删除最后一个字符，而只会移动写入头。这在这种情况下就足够了，因为我们用}.
douCoh << '{';
for(unsigned int i=0;i<dataSize;i++)
if(v[i].test) douCoh << i+1 << ',';
douCoh.seekp(-1,douCoh.cur); douCoh << '}';
回答by Sylvain Defresne
You can extract the string (with the str()member), remove the last char with std::string::eraseand then reset the new string as buffer to the std::ostringstream.
您可以提取字符串（使用str()成员），删除最后一个字符，std::string::erase然后将新字符串作为缓冲区重置为std::ostringstream.
However, a better solution would be to not insert the superfluous ','in the first place, by doing something like that :
但是，更好的解决方案是首先不要插入多余','的内容，通过执行以下操作：
std::ostringstream douCoh;
const char* separator = "";
douCoh << '{';
for (size_t i = 0; i < dataSize; ++ i)
{
if (v[i].test)
{
douCoh << separator << i + 1;
separator = ",";
}
}
douCoh << '}';
回答by jdehesa
I have had this very problem and I found out that you can simply do:
我遇到了这个问题，我发现你可以简单地做：
douCoh.seekp(-1, std::ios_base::end);
And the keep inserting data. As others stated, avoiding inserting the bad data in the first place is probably the ideal solution, but in my case was the result of a 3rd party library function, and also I wanted to avoid copying the data to strings.
并不断插入数据。正如其他人所说，首先避免插入错误数据可能是理想的解决方案，但在我的情况下是第 3 方库函数的结果，而且我想避免将数据复制到字符串。
回答by Drum Inc
stringstream douCoh;
for(unsigned int i=0;i<dataSize;i++)
if(v[i].test)
douCoh << ( douCoh.tellp()==0 ? '{' : ',' ) << i+1;
douCoh << '}';
回答by DarioP
I've found this way using pop_back()string's method since c++11. Probably not so good as smarter ones above, but useful in much more complicated cases and/or for lazy people.
我pop_back()从 c++11 开始就使用字符串的方法找到了这种方式。可能不如上面更聪明的那么好，但在更复杂的情况下和/或对懒惰的人有用。
douCoh << '{';
for(unsigned int i=0;i<dataSize;i++)
if(v[i].test) douCoh << i+1 << ',';
string foo(douCoh.str());
foo.pop_back();
douCoh.str(foo);
douCoh.seekp (0, douCoh.end);
douCoh << '}';
回答by Beno?t
Have fun with std::copy, iterators and traits. You either have to assume that your data is reverse iterable (end - 1) or that your output can be rewinded. I choose it was easier to rewind.
享受 std::copy、迭代器和特征带来的乐趣。您要么必须假设您的数据是可反向迭代的（end - 1），要么您的输出可以倒回。我选择倒带更容易。
#include <ostream>
#include <algorithm>
#include <iterator>
namespace My
{
template<typename Iterator>
void print(std::ostream &out, Iterator begin, Iterator end)
{
out << '{';
if (begin != end) {
Iterator last = end - 1;
if (begin != last) {
std::copy(begin, last, std::ostream_iterator< typename std::iterator_traits<Iterator>::value_type >(out, ", "));
}
out << *last;
}
out << '}';
}
}
#include <iostream>
int main(int argc, char** argv)
{
My::print(std::cout, &argv[0], &argv[argc]);
std::cout << '\n';
}
回答by Daniel Hornik
#include <sstream>
#include <vector>
#include <iterator>
#include <algorithm>
template<typename T>
std::string implode(std::vector<T> vec, std::string&& delim)
{
std::stringstream ss;
std::copy(vec.begin(), vec.end(), std::ostream_iterator<std::string>(ss, delim.c_str()));
if (!vec.empty()) {
ss.seekp(-1*delim.size(), std::ios_base::end);
ss<<'douCoh << '{';
for(unsigned int i=0;i<dataSize;i++){
if(v[i].test){
douCoh << i+1;
if(i != dataSize - 1) douCoh << ',';
}
}
/*douCoh.get();*/ douCoh << '}';
';
}
return ss.str();
}
int main()
{
std::cout<<implode(std::vector<std::string>{"1", "2", "3"}, ", ");
return 0;
}
回答by Bj?rn Pollex
You could use std::string::eraseto remove the last character directly from the underlying string.
您可以使用std::string::erase直接从底层字符串中删除最后一个字符。
回答by RedX
Why not just check the counter? And not insert the ','
为什么不直接查柜台呢？而不是插入'，'
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