转换匿名数组初始化器列表 [英] Casting an anonymous array initializer list

问题描述

我可以成功地对char字符串数组的初始化列表进行C转换，但似乎无法使其与C ++转换（static_cast）一起使用：

I can successfully do a C cast of an initializer list for an array of char strings, but can't seem to get it to work with a C++ cast (static_cast):

int main()
{
   char x[] = "test 123";

   // This works fine:

   char **foo = (char *[]) { "a", x, "abc" };
   std::cout << "[0]: " << foo[0] << "    [1]: " << foo[1]
             << "    [2]: " << foo[2] << std::endl;

   // This will not compile ("expected primary-expression before '{' token"):

   //char **bar = static_cast<char *[]>( { "a", x, "abc" } );
   //std::cout << "[0]: " << bar[0] << "    [1]: " << bar[1]
   //          << "    [2]: " << bar[2] << std::endl;
}

是否可以在此处使用C ++强制转换？如果是这样，正确的语法是什么？如果不是，为什么不这样做，C转换是否让我摆脱了我不应该做的事情？

Is it possible to use a C++ cast here? If so, what's the correct syntax? If not, why not, and is the C cast letting me get away with something I shouldn't be doing?

最终，我问这个的原因是我正在调用将char数组指针作为参数的函数，并且我想使用匿名数组作为调用参数。

Ultimately, the reason I'm asking this is that I am calling a function that has a char array pointer as a parameter, and I would like to use an anonymous array as the calling argument.

我正在使用GCC 4.4.6。

I'm using GCC 4.4.6.

推荐答案

我可以成功地对一个char字符串数组的初始化列表进行C转换

I can successfully do a C cast of an initializer list for an array of char strings

不，可以没错您根本没有使用初始化列表或C强制转换。您使用的是复合文字。这是C语言所不具备的C语言功能。某些编译器确实支持C ++作为语言扩展。

No, you can't. You didn't use an initializer list nor a C cast at all. What you used was a compound literal. It is a C language feature that doesn't exist in C++. Some compilers do support them in C++ as a language extension.

我强烈建议您使用编译器选项，该选项至少在使用非标准功能时会发出警告，以避免混淆

I highly recommend you to use a compiler option that at least warns when you use non-standard features to avoid confusion like this.

，但似乎无法使其与C ++ cast一起使用

but can't seem to get it to work with a C++ cast

您不能强制转换初始化列表表达式。您将通常必须先初始化一个命名数组，然后再初始化该指针-尽管您几乎不需要真正使用单独的指针变量，因为无论如何该数组在大多数情况下都隐式地衰减为指针。

You cannot cast an initializer list expression. You will have to initialize a named array normally, then the pointer - although you hardly ever really need a separate pointer variable since the array implicitly decays to a pointer in most contexts anyway.

const char* arr[] = { "a", x, "abc" };
const char** foo = arr;

---

---

我的原因我问这是我正在调用一个以char数组指针作为参数的函数，并且我想使用匿名数组作为调用参数。

the reason I'm asking this is that I am calling a function that has a char array pointer as a parameter, and I would like to use an anonymous array as the calling argument.

如果可以修改该函数，则可以使用多种方法来允许调用而无需命名数组。您可以接受 std :: initializer_list ，也可以接受可以从初始化列表构造的类型，例如 std :: array 。

If you can modify the function, then there are ways to allow the call without a named array. You could accept an std::initializer_list, or a type that can be constructed from an initializer list such as an instance of std::array.

PS。从字符串文字到 char * 的隐式转换在C ++中也是不允许的-但某些编译器允许作为语言扩展。在此处使用 const char * 。

PS. Implicit conversion from string literal to char* is not allowed in C++ either - but allowed by some compilers as a language extension. Use const char* here.

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