use*_*664 13 c++ initialization concatenation stringstream
考虑:
std::string s_a, s_b;
std::stringstream ss_1, ss_2;
// at this stage:
// ss_1 and ss_2 have been used and are now in some strange state
// s_a and s_b contain non-white space words
ss_1.str( std::string() );
ss_1.clear();
ss_1 << s_a;
ss_1 << s_b;
// ss_1.str().c_str() is now the concatenation of s_a and s_b,
// <strike>with</strike> without space between them
ss_2.str( s_a );
ss_2.clear();
// ss_2.str().c_str() is now s_a
ss_2 << s_b; // line ***
// ss_2.str().c_str() the value of s_a is over-written by s_b
//
// Replacing line *** above with "ss_2 << ss_2.str() << " " << s_b;"
// results in ss_2 having the same content as ss_1.
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问题:
stringstream.str(a_value)之间的区别是什么?和stringstream << a_value; 而且,具体来说,为什么第一个不允许通过<<但第二个进行连接?
为什么ss_1会自动在s_a和fen之间获得空白,但是我们是否需要在行中明确添加可以替换行***的空格: ss_2 << ss_2.str() << " " << s_b;?
您遇到的问题是,因为std::stringstream默认情况下,构建了ios_base::openmode mode = ios_base::in|ios_base::out它是一个非附加模式。
你感兴趣的输出模式在这里(即:ios_base::openmode mode = ios_base::out)
std::basic_stringbuf::str(const std::basic_string<CharT, Traits, Allocator>& s)以两种不同的方式运行,具体取决于openmode:
mode & ios_base::ate == false:(即非附加的输出流):
str将设置pptr() == pbase(),以便后续输出将覆盖从s复制的字符
mode & ios_base::ate == true:(即:附加输出流):
str将设置pptr() == pbase() + s.size(),以便后续输出将附加到从s复制的最后一个字符
(请注意,此附加模式是c ++ 11以来的新增功能)
可以在此处找到更多详细信息。
如果您想要追加行为,请stringstream使用创建您的行为ios_base::ate:
std::stringstream ss(std::ios_base::out | std::ios_base::ate)
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简单的示例应用程序在这里:
#include <iostream>
#include <sstream>
void non_appending()
{
std::stringstream ss;
std::string s = "hello world";
ss.str(s);
std::cout << ss.str() << std::endl;
ss << "how are you?";
std::cout << ss.str() << std::endl;
}
void appending()
{
std::stringstream ss(std::ios_base::out | std::ios_base::ate);
std::string s = "hello world";
ss.str(s);
std::cout << ss.str() << std::endl;
ss << "how are you?";
std::cout << ss.str() << std::endl;
}
int main()
{
non_appending();
appending();
exit(0);
}
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如上所述,这将以两种不同的方式输出:
hello world
how are you?
hello world
hello worldhow are you?
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