如何读取整个流到std :: string? [英] How to read entire stream into a std::string?
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问题描述
我想读取一个整个流(多行)到一个字符串。
我使用这个代码,它的工作原理,但它冒犯我的风格感...当然有一个更容易的方法?也许使用stringstreams?
void Obj :: loadFromStream(std :: istream& stream)
{
std :: string s;
int p = stream.tellg(); //记住我们在哪里
stream.seekg(0,std :: ios_base :: end); // go to the end
int sz = stream.tellg() - p; // work out the size
stream.seekg(p); // restore the position
s.resize(sz); // resize the string
stream.read(& s [0],sz); //最后,读入数据。
编辑:
实际上,一个 const引用到字符串也可以,这可以使事情更容易...
const std :: string& s(...这里出现一个奇迹...)
解决方案如何
std :: istreambuf_iterator< char> eos;
std :: string s(std :: istreambuf_iterator< char>(stream),eos);
(如果不是MVP,可以是单行)
后2011年编辑,这种方法现在拼写
std :: string s(std :: istreambuf_iterator< ; char>(stream),{});
I'm trying to read an entire stream (multiple lines) into a string.
I'm using this code, and it works, but it's offending my sense of style... Surely there's an easier way? Maybe using stringstreams?
void Obj::loadFromStream(std::istream & stream) { std::string s; int p = stream.tellg(); // remember where we are stream.seekg(0, std::ios_base::end); // go to the end int sz = stream.tellg() - p; // work out the size stream.seekg(p); // restore the position s.resize(sz); // resize the string stream.read(&s[0], sz); // and finally, read in the data.
EDIT:
Actually, a const reference to a string would do as well, and that may make things easier...
const std::string &s(... a miracle occurs here...)
解决方案How about
std::istreambuf_iterator<char> eos; std::string s(std::istreambuf_iterator<char>(stream), eos);
(could be a one-liner if not for MVP)
post-2011 edit, this approach is now spelled
std::string s(std::istreambuf_iterator<char>(stream), {});
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