为什么包含字符串常量的初始化程序有效初始化一个char数组？

Question

事实证明char c[] = {"a"};在C++03和C++11中都是完全有效的。

我不希望它是，因为它是char而不是char const*的数组，我期望一个括号初始化程序要求它的每个“项目”都有一个兼容的类型。它有一个项目，这是一个char const*而不是一个char。

那么是什么让这个初始化有效？是否有这样的理由？

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^{同样，char c[] = {"aa"};编译，并打印c结果在输出 “aa”。}

^{我会期望char c[]{"a"}是在C++ 11，当然有效，但它是不一样的！同样，char c[] = {'a'}在两者中都是显而易见的，因为是char c[] = "a"。}

Answer 1

标量类型也可以使用花括号进行初始化（就像结构和数组一样）。

struct S { int x, char c }; 
S s = {5, 'a'}; 

int arr[] = {5, 6, 7}; 

/* (my guess) out of consistency */ 
int z = { 4 }; 

而且因为字符串文字可以被分配到字符数组的指针和

char arr[] = "literal"; 
char* ptr = "another"; 

似乎拟合允许char arr[] = { "literal" };了。

Answer 2

虽然它可能不一定是直观的，但它只是是允许的;两种标准都有明确的规定：

[2003: 8.5.2/1]:A char array (whether plain char , signed char , or unsigned char) can be initialized by a string-literal (optionally enclosed in braces); a wchar_t array can be initialized by a wide string-literal (optionally enclosed in braces); successive characters of the string-literal initialize the members of the array. [..]

[n3290: 8.5.2/1]:A char array (whether plain char , signed char , or unsigned char), char16_t array, char32_t array, or wchar_t array can be initialized by a narrow character literal, char16_t string literal, char32_t string literal, or wide string literal, respectively, or by an appropriately-typed string literal enclosed in braces. Successive characters of the value of the string literal initialize the elements of the array.

虽然我不能解释为什么委员会这样做。

Answer 3

我想这是C兼容性？实际上，T x = { value of T };也适用于其他类型的T。在C99标准中，

6.7.8/11: The initializer for a scalar shall be a single expression, optionally enclosed in braces.

6.7.8/14: An array of character type may be initialized by a character string literal, optionally enclosed in braces.

6.7.8/15: An array with element type compatible with wchar_t may be initialized by a wide string literal, optionally enclosed in braces.

我不知道为什么C有这个虽然。