13

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.

问题:

  1. stringstream.str(a_value); 有什么区别?和 stringstream << a_value; 并且,具体来说,为什么第一个不允许通过 << 进行连接,但第二个允许?

  2. 为什么 ss_1 会自动获得 s_a 和 s_b 之间的空白,但是我们是否需要在可以替换行 ***: 的行中显式添加空白 ss_2 << ss_2.str() << " " << s_b;

4

2 回答 2

5

您遇到的问题是因为std::stringstream默认构造ios_base::openmode mode = ios_base::in|ios_base::out为非附加模式。

您对此处的输出模式感兴趣(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

  1. mode & ios_base::ate == false:(即:非附加输出流):

    str将设置pptr() == pbase(),以便后续输出将覆盖从 s 复制的字符

  2. 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

(注意这个附加模式是c++11以后新增的)

更多细节可以在这里找到。

如果您想要附加行为,请创建您的stringstreamwith ios_base::ate

std::stringstream ss(std::ios_base::out | std::ios_base::ate)

这里的简单示例应用程序:

#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);
}

如上所述,这将以 2 种不同的方式输出:

hello world
how are you?
hello world
hello worldhow are you?
于 2012-11-26T03:51:36.340 回答
4

建议您阅读字符串流参考:http ://en.cppreference.com/w/cpp/io/basic_stringstream

std::stringstream::str()替换底层字符串的内容

operator<<将数据插入流中。

于 2012-11-26T03:37:10.687 回答