c++ - 预递增字符串文字

Question

N3337 标准草案的 § 2.14.5/8 指出:

Ordinary string literals and UTF-8 string literals are also referred to as narrow string literals. A narrow string literal has type “array of n const char”, where n is the size of the string as defined below, and has static storage duration (3.7).

§ 4.2

An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to a prvalue of type “pointer to T”. The result is a pointer to the first element of the array.

我期望 ++"hello world!" 将 "hello world!" 衰减为 char* 但这并没有发生。必须使用额外的加号 +++"hello world!" 强制执行。有些情况下不应用数组到指针的转换，例如 sizeof 运算符。

§ 5.3.2 简单地说:

The operand of prefix ++ is modified by adding 1, or set to true if it is bool (this use is deprecated). The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type or a pointer to a completely-defined object type. The result is the updated operand; it is an lvalue, and it is a bit-field if the operand is a bit-field. If x is not of type bool, the expression ++x is equivalent to x+=1

那么为什么没有发生这种转化呢？

Answer 1

当需要将左值转换为纯右值时，将应用数组到指针(和其他衰减)转换，按照 5/8:

Whenever a glvalue expression appears as an operand of an operator that expects a prvalue for that operand, the lvalue-to-rvalue (4.1), array-to-pointer (4.2), or function-to-pointer (4.3) standard conversions are applied to convert the expression to a prvalue.

请注意，根据您自己对 5.3.2 的引用，++ 的操作数必须是可修改的左值。数组到指针的转换用于将 glvalue 转换为 prvalue，这不是 ++ 所需要的。因此，没有什么需要数组到指针的转换。应用它实际上是无效的，因为该转换的结果是纯右值，++ 不能对其执行操作。