隐式转换为basic_istream / ifstream / ofstream的布尔值在Visual Studio 2013中不起作用 [英] Implicit cast to bool of basic_istream/ifstream/ofstream doesn't work in Visual Studio 2013

问题描述

下面的代码在VS 2012中编译，但在VS 2013中编译

The code below compiles in VS 2012 but not in VS 2013

std::ofstream stm;
if(stm != NULL)
{
}

在VS 2013中，您会遇到以下编译错误：

In VS 2013 you get this compilation error:

binary'！='没有找到采用左侧操作数为'std'的运算符:: ofstream'（或没有可接受的转换）

binary '!=' no operator found which takes a left-hand operand of type 'std::ofstream' (or there is no acceptable conversion)

我看了看标题，并在< xiobase> ，我发现了以下内容：

I looked at the headers and in <xiobase> and I found the following:

VS2012

ios_base::operator void *() const;

VS2013

operator void *（）const 已被删除，并添加了带有显式的布尔运算符：

operator void *() const has been removed and the operator bool with explicit was added instead:

ios_base::explicit operator bool() const;

现在我的问题：

1. 我在互联网上找不到有关此更改的任何信息。您知道是否在任何地方都有关于此更改的官方文章吗？


2. 我有旧代码，其中频繁使用if（stm！= NULL）。出于不相关的原因，最好不要更改代码。有没有一种方法可以使其在VS 2013中进行编译而不进行更改？我找不到任何可以恢复运算符 void * 或从运算符bool（）删除 explicit 的条件编译指令。

1. I couldn't find any information about this change in the internet. Do you know if there is an official article about this change anywhere?
2. I have legacy code where if(stm != NULL) is used a lot. For unrelated reasons it's preferable not to change the code. Is there a way to make it compile in VS 2013 without changing it? I couldn't find any conditional compilation directives that could restore operator void* or remove explicit from operator bool().

PS：gcc 4.9.0仍然具有 operator void *（）const 。所以它不会有这个问题。

PS: gcc 4.9.0 still has operator void*() const. So it will not have this problem.

更新：

遗留代码编译我按照建议实现了以下重载：

To make my legacy code compile I implemented the following overloads as it was suggested:

#include <xiosbase>

bool operator==(const std::basic_ios<char, char_traits<char>> &stm, int null_val)
{
    return static_cast<bool>(stm) == null_val;
}

bool operator==(int null_val, const std::basic_ios<char, char_traits<char>> &stm)
{
    return operator==(stm, null_val);
}

bool operator!=(int null_val, const std::basic_ios<char, char_traits<char>> &stm)
{
    return !operator==(stm, null_val);
}

bool operator!=(const std::basic_ios<char, char_traits<char>> &stm, int null_val)
{
    return !operator==(stm, null_val);
}

在我的情况下， char 值类型足够，第二个参数是int，因为无论如何都不支持非NULL。

In my case the char value type was enough and the second parameter is int because something that is not NULL is not supported anyway.

推荐答案

有很多旧代码，您可能可以添加一个自定义 operator！= （和 operator == ）函数，接受正确的参数：

If you have a lot of legacy code, you could probably add a custom operator!= (and operator==) function which takes the correct arguments:

bool operator!=(std::basic_ios const& ios, const void* ptr);
bool operator!=(const void* ptr, std::basic_ios const& ios);

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