istream运算符>>之间的差异。 （double& val）在libc ++和libstdc ++之间 [英] Discrepancy between istream's operator>> (double& val) between libc++ and libstdc++

问题描述

由于我最近升级到Mac OS X 10.9，默认标准C ++库已从libstdc ++更改为libc ++。从那时起，我观察到在下面的代码示例中记录的stringstream运算符>>（double）的意外行为。

总而言之，当double值后跟字母时，libc ++似乎在从stringstreams中提取双精度值有问题。

我已经检查标准（2003年），但是我找不到任何具体信息，如果提取在这种情况下应该工作。

感谢任何输入，无论这是libc ++还是libstdc ++中的错误。

  #include< sstream& 
 #include< iostream> 
 
 using namespace std; 
 
 void extract_double（const string& s）
 {
 stringstream ss; 
 double d; 
 
 ss<< s; 
 ss>> d; 
 if（！ss.fail（））
 cout<< '< ss.str（）<< '转换为<< d<< endl; 
 else 
 cout<< '< ss.str（）<< '无法转换为double<< endl; 
} 
 
 int main（）
 {
 extract_double（ -  4.9）; 
 extract_double（ -  4.9 X）; 
 extract_double（ -  4.9_）; 
 extract_double（ -  4.9d）; 
 extract_double（ -  4.9X）; 
} 
  

使用 c ++编译代码--stdlib = libc ++ streamtest.cxx 给出

 ' -  4.9'转换为-4.9 
' X'转换为-4.9 
'-4.9_'转换为-4.9 
'-4.9d'无法转换为double 
'-4.9X'无法转换为double 
  

使用 c ++编译代码--stdlib = libstdc ++ streamtest.cxx 给予

 ' -  4.9'转换为-4.9 
'-4.9 X'转换为-4.9 
'-4.9_'转换为-4.9 
'-4.9d'转换为-4.9 
'-4.9X'转换为-4.9 
  

编译器版本为

  $ c ++ --version 
 Apple LLVM版本5.0（clang-500.2.79）（基于LLVM 3.3svn）
目标：x86_64-apple-darwin13.0.0 
线程模型：posix 
  
 
 
解决方案
看起来libstdc ++是正确的，libc ++是错误的，根据22.4.2.1.2 
 
 
 
在阶段2，
 
 
 [字符 -  nm]不被丢弃，则进行检查以确定是否允许c作为阶段1返回的转换说明符的输入字段的下一个字符[％g在这种情况下为nm。 
 
 
由于％g 转换说明符允许 d 或 X 个字符，字符不会累积。它也不会被丢弃（只能删除组分隔符）。因此，阶段2必须在此时结束。
 
 
 
然后在阶段3，累积的字符将被转换。
 
 
 
像libc ++在阶段2错误地累积 d 和 X ，然后尝试转换它们，并且失败。
 
With my recent upgrade to Mac OS X 10.9 the default standard C++ library changed from libstdc++ to libc++. Since then I observe unexpected behaviour of the stringstream operator>>(double) documented in the code example below. 

In summary the libc++ seems to have problems with extracting double values from stringstreams when the double value is followed by a letter.

I already checked the standard (2003) but I can't find any specific information if extraction should work in this case or not.

So I would be grateful for any input whether this is a bug in libc++ or libstdc++.
#include <sstream>
#include <iostream>

using namespace std;

void extract_double(const string & s)
{
  stringstream ss;
  double d;

  ss << s;
  ss >> d;
  if(!ss.fail())
    cout << "'" << ss.str() << "' converted to " << d << endl;
  else
    cout << "'" << ss.str() << "' failed to convert to double" << endl;
}

int main()
{
  extract_double("-4.9");
  extract_double("-4.9 X");
  extract_double("-4.9_");
  extract_double("-4.9d");
  extract_double("-4.9X");
}
Compiling the code with c++ --stdlib=libc++ streamtest.cxx gives
'-4.9' converted to -4.9
'-4.9 X' converted to -4.9
'-4.9_' converted to -4.9
'-4.9d' failed to convert to double
'-4.9X' failed to convert to double
Compiling the code with c++ --stdlib=libstdc++ streamtest.cxx gives 
'-4.9' converted to -4.9
'-4.9 X' converted to -4.9
'-4.9_' converted to -4.9
'-4.9d' converted to -4.9
'-4.9X' converted to -4.9
Compiler version is 
$ c++ --version
Apple LLVM version 5.0 (clang-500.2.79) (based on LLVM 3.3svn)
Target: x86_64-apple-darwin13.0.0
Thread model: posix

解决方案
It looks like that libstdc++ is right and libc++ is wrong, according to the 22.4.2.1.2 of the (2011) standard.

At stage 2, 

  
If it [the character - n.m.] is not discarded, then a check is made to determine if c is allowed as the next character of an input field of the conversion specifier returned by Stage 1 ["%g" in this case - n.m.] . If so, it is accumulated.
Since %g conversion specifier does not admit d or X characters, the character is not accumulated. It is not discarded either (only group separator characters can be discarded). Therefore Stage 2 must end at this point.

Then at stage 3 accumulated characters are converted.

It looks like libc++ erroneously accumulates d and X at stage 2, then attempts to convert them, and this fails.

                        
这篇关于istream运算符&gt;&gt;之间的差异。 （double&amp; val）在libc ++和libstdc ++之间的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！