book: redo string literal explanation (changkun#237)

Timothy-Liuxf · web-flow · commit 41b88c861e75 · 2022-07-17T13:18:50.000+02:00
diff --git a/book/en-us/03-runtime.md b/book/en-us/03-runtime.md
@@ -263,22 +263,34 @@ Temporary variables returned by non-references, temporary variables generated
 by operation expressions, original literals, and Lambda expressions
 are all pure rvalue values.
 
-Note that a string literal became rvalue in a class, and remains an lvalue in other cases (e.g., in a function)：
+Note that a literal (except a string literal) is a prvalue. However, a string
+literal is an lvalue with type `const char` array. Consider the following examples:
 
 ```cpp
-class Foo {
-        const char*&& right = "this is a rvalue";
-public:
-        void bar() {
-            right = "still rvalue"; // the string literal is a rvalue
-        }
-};
+#include <type_traits>
 
 int main() {
-    const char* const &left = "this is an lvalue"; // the string literal is an lvalue
+    // Correct. The type of "01234" is const char [6], so it is an lvalue
+    const char (&left)[6] = "01234";
+
+    // Assert success. It is a const char [6] indeed. Note that decltype(expr)
+    // yields lvalue reference if expr is an lvalue and neither an unparenthesized
+    // id-expression nor an unparenthesized class member access expression.
+    static_assert(std::is_same<decltype("01234"), const char(&)[6]>::value, "");
+
+    // Error. "01234" is an lvalue, which cannot be referenced by an rvalue reference
+    // const char (&&right)[6] = "01234";
 }
 ```
 
+However, an array can be implicitly converted to a corresponding pointer.The result, if not an lvalue reference, is an rvalue (xvalue if the result is an rvalue reference, prvalue otherwise):
+
+```cpp
+const char*   p    = "01234"; // Correct. "01234" is implicitly converted to const char*
+const char*&& pr   = "01234"; // Correct. "01234" is implicitly converted to const char*, which is a prvalue.
+// const char*& pl = "01234"; // Error. There is no type const char* lvalue
+```
+
 **xvalue, expiring value** is the concept proposed by C++11 to introduce
 rvalue references (so in traditional C++, pure rvalue and rvalue are the same concepts),
 a value that is destroyed but can be moved.
diff --git a/book/zh-cn/03-runtime.md b/book/zh-cn/03-runtime.md
@@ -224,22 +224,32 @@ int main() {
 要么是求值结果相当于字面量或匿名临时对象，例如 `1+2`。非引用返回的临时变量、运算表达式产生的临时变量、
 原始字面量、Lambda 表达式都属于纯右值。
 
-需要注意的是，字符串字面量只有在类中才是右值，当其位于普通函数中是左值。例如：
+需要注意的是，字面量除了字符串字面量以外，均为纯右值。而字符串字面量是一个左值，类型为 `const char` 数组。例如：
 
 ```cpp
-class Foo {
-        const char*&& right = "this is a rvalue"; // 此处字符串字面量为右值
-public:
-        void bar() {
-            right = "still rvalue"; // 此处字符串字面量为右值
-        }
-};
+#include <type_traits>
 
 int main() {
-    const char* const &left = "this is an lvalue"; // 此处字符串字面量为左值
+    // 正确，"01234" 类型为 const char [6]，因此是左值
+    const char (&left)[6] = "01234";
+
+    // 断言正确，确实是 const char [6] 类型，注意 decltype(expr) 在 expr 是左值
+    // 且非无括号包裹的 id 表达式与类成员表达式时，会返回左值引用
+    static_assert(std::is_same<decltype("01234"), const char(&)[6]>::value, "");
+
+    // 错误，"01234" 是左值，不可被右值引用
+    // const char (&&right)[6] = "01234";
 }
 ```
 
+但是注意，数组可以被隐式转换成相对应的指针类型，而转换表达式的结果（如果不是左值引用）则一定是个右值（右值引用为将亡值，否则为纯右值）。例如：
+
+```cpp
+const char*   p   = "01234";  // 正确，"01234" 被隐式转换为 const char*
+const char*&& pr  = "01234";  // 正确，"01234" 被隐式转换为 const char*，该转换的结果是纯右值
+// const char*& pl = "01234"; // 错误，此处不存在 const char* 类型的左值
+```
+
 **将亡值(xvalue, expiring value)**，是 C++11 为了引入右值引用而提出的概念（因此在传统 C++ 中，
 纯右值和右值是同一个概念），也就是即将被销毁、却能够被移动的值。
 
