g++ 将返回的字符串文字视为 const char 指针而不是

g++ treats returned string literal as const char pointer not const char array(g++ 将返回的字符串文字视为 const char 指针而不是 const char 数组)

问题描述

当从一个应该使用 g++(版本 4.7.3)执行隐式转换的函数返回字符串文字时，我看到了一些奇怪的行为.谁能解释为什么下面的代码:

I'm seeing some odd behaviour when returning a string literal from a function that should perform an implicit conversion with g++ (version 4.7.3). Can anyone explain why the following code:

#include <stdio.h>

class Test
{
public:
  template <unsigned int N>
  Test(const char (&foo)[N])
  {
    printf("Template const char array constructor
");
  }

  Test(char* foo)
  {
    printf("char* constructor
");
  }
};

Test fn()
{
  return "foo";
}

int main()
{
  Test t("bar");
  Test u = fn();

  return 0;
}

产生结果:

Template const char array constructor
char* constructor

在 g++ 上?令人惊讶的是，在从 fn() 生成返回值时，优先选择 char* 构造函数而不是 const char 数组构造函数.诚然，有一个警告，不推荐使用从字符串常量到 'char*' 的转换"

on g++? The surprising thing being that the char* constructor is chosen in preference to the const char array constructor when generating the return value from fn(). Admittedly there is a warning, "deprecated conversion from string constant to 'char*'"

更令人惊讶的是，如果您删除 char* 构造函数，那么代码将无法使用 g++ 编译.

Even more surprisingly if you remove the char* constructor then the code doesn't compile with g++.

它在 clang 中按预期工作(两次都使用模板构造函数)，这让我认为这是一个编译器错误，但也许它只是 C++ 规范的一个奇怪的角落 - 有人可以确认吗?

It works as expected with clang (Template constructor used both times), which makes me think this is a compiler bug, but maybe it's just a weird corner of the C++ spec - could anyone confirm?

推荐答案

看来这是一个影响多个版本的 gcc 的错误，已被反复报告，最近一次是大约一个月前针对最新版本，4.8.2.请参阅 http://gcc.gnu.org/bugzilla/show_bug.cgi?id=24666

It appears that this is a bug affecting several versions of gcc which has been reported over and over again, most recently about a month ago against the most recent version, 4.8.2. See http://gcc.gnu.org/bugzilla/show_bug.cgi?id=24666

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