为什么是stream（“log.txt”，ios :: app | ios :: trunc）;总是失败？ [英] Why does ofstream("log.txt", ios::app|ios::trunc); always fail?

问题描述

以下代码是使用VC ++ 2012年11月CTP在Windows 7 x64上编译的。

  #include< fstream& 
 
 using namespace std; 
 
 int main（）
 {
 ofstream fout（log.txt，ios :: app | ios :: trunc）; 
 if（！fout）
 {
 cout< 发生错误！ << endl; //总是去这里！为什么？ 
} 
} 
  

cppreference.com网站不说 ios :: app 不能与 ios :: trunc 组合。

exact 语义 ios :: app 和 ios :: trunc ？

c> c>

c>

在表132中的C ++ 11：

  + ---------------- ------------------- + ------------------- + 
 | ios_base标志组合| stdio等效| 
 |二进制输出trunc应用程序| | 
 + ----------------------------------- + --------- ---------- + 
 | + | w| 
 | + + a| 
 | + | a| 
 | + + w| 
 | + | r| 
 | + + | r +| 
 | + + + w +| 
 | + + + a +| 
 | + + a +| 
 + ----------------------------------- + --------- ---------- + 
 | + + wb| 
 | + + + ab| 
 | + + ab| 
 | + + + wb| 
 | + + rb| 
 | + + + r + b| 
 | + + + + w + b| 
 | + + + + a + b| 
 | + + + a + b| 
 + ----------------------------------- + --------- ---------- + 
  

正如您所看到的，

。

[C ++ 11：27.9.1.4/ 2]： [..] 如果模式不是表中显示的某些标志组合， 。

这些是语义。

^† [C ++ 11：27.9.1.7/2] & [C ++ 11：27.9.1.11/2] 显示模式从流对象传递到缓冲对象。 b $ b

The following code was compiled with VC++ Nov 2012 CTP on Windows 7 x64.

#include <fstream>

using namespace std;

int main()
{
    ofstream fout("log.txt", ios::app|ios::trunc);
    if (!fout)
    {
        cout << "An error occurred!" << endl; // Always go here! Why?
    }
}

The cppreference.com website doesn't say that ios::app cannot be combined with ios::trunc.

What are the exact semantics of ios::app and ios::trunc?

解决方案

The filebuf constructor to which these flags are passed^† has behaviours based on those flags defined in Table 132 in C++11:

+-----------------------------------+-------------------+
|     ios_base flag combination     |  stdio equivalent |
| binary  in    out    trunc    app |                   |
+-----------------------------------+-------------------+
|               +                   |  "w"              |
|               +               +   |  "a"              |
|                               +   |  "a"              |
|               +       +           |  "w"              |
|        +                          |  "r"              |
|        +      +                   |  "r+"             |
|        +      +       +           |  "w+"             |
|        +      +               +   |  "a+"             |
|        +                      +   |  "a+"             |
+-----------------------------------+-------------------+
|   +           +                   |  "wb"             |
|   +           +               +   |  "ab"             |
|   +                           +   |  "ab"             |
|   +           +       +           |  "wb"             |
|   +    +                          |  "rb"             |
|   +    +      +                   |  "r+b"            |
|   +    +      +       +           |  "w+b"            |
|   +    +      +               +   |  "a+b"            |
|   +    +                      +   |  "a+b"            |
+-----------------------------------+-------------------+

As you can see, your flag combination is not found in that table.

[C++11: 27.9.1.4/2]: [..] If mode is not some combination of flags shown in the table then the open fails.

Those are the semantics.

^{† [C++11: 27.9.1.7/2] & [C++11: 27.9.1.11/2] show us that the mode is passed from the stream object to the buffer object.}

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