使用namespace std删除会导致程序获取垃圾结果 [英] Removing using namespace std causes the program to get crap results

问题描述

我有一个复杂的最小二乘拟合程序，我最近调试。我使用命名空间std来使用

I have a complicated least-squares fitting program, which I was recently debugging. I was debugging stuff by cout-ing them to the console, and to make it easier, I used

using namespace std;

和乐趣，经过几个星期的编码，我想删除这个，因为调试完成，是删除它会导致结果是错误的！我在g ++和icpc（intel编译器）中做了一个完整的抽象检查，其中我删除和恢复这个指令，当它被删除，结果错了...

and funnily, after weeks of coding, I wanted to remove this because debugging is done, and the surprise was that removing it causes the result to be wrong! I did a complete abstract check in g++ and icpc (intel compiler) where I remove and restore this directive, and when it's removed, the result it wrong...

知道这是一个非常广泛的问题，不容易打的答案，但你会做什么在这种情况下？

I know it's a very broad question and not easy to hit the answer, but what would you do in this case? How would you debug such a dependence?

推荐答案

这就是为什么在全局命名空间中使用指令（或者任何其他不适当的范围）是一个坏主意。这不只是一个风格的问题;它可以以微妙的方式改变代码的含义。有关这些问题的详细讨论，请参见此问题。

This is why using-directives in the global namespace (or any other inappropriately wide scope) are a bad idea. It's not just a matter of style; it can change the meaning of the code in subtle ways. See this question for a thorough discussion of the issues.

几乎肯定，您的代码使用来自标准库的无限定名称，这些名称也在全局命名空间中声明。因此，使用命名空间std; 删除更改了那些不合格名称的含义，而不是仅仅使代码无法编译。

Almost certainly, your code uses names from the standard library without qualification, and those names are also declared in the global namespace. Thus, removing using namespace std; changes the meaning of those unqualified names, rather than just making the code fail to compile.

这些可能是您的代码的名称;或者他们是C库函数。许多（特别是< cmath> 中的）被重载用于各种类型，其中C库本身只声明一个单一的函数。它是未指定是否，一些或所有这些被转储到全局命名空间以及命名空间std 。

These might be names from your code; or perhaps they are C library functions. Many (particularly those in <cmath>) are overloaded for various types, where the C library itself only declares a single function. It's unspecified whether none, some, or all of these are dumped into the global namespace as well as namespace std.

例如，您可能有一个函数调用

For example, you might have a function call

float angle = whatever;
float sine = sin(angle);

使用 using namespace std; 选择已经转储到全局命名空间中的重载的 float std :: sin（float）。没有它，根据实现选择使用全局命名空间，它可能仍然调用;或者它可以从C库调用 double sin（float）;或者它可能无法编译。

With using namespace std;, this will select the overloaded float std::sin(float) that you've dumped into the global namespace. Without it, depending on what the implementation chooses to do with the global namespace, it might still call that; or it might call double sin(float) from the C library; or it might fail to compile.

在GCC的情况下，似乎C库函数，但不是 C ++重载，转储到全局命名空间中，正如我在回答此问题时发现的。

In the case of GCC, it seems that the C library function, but not the C++ overloads, are dumped into the global namespace, as I discovered while answering this question.

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