为什么这个带有'1234'的代码在C ++中编译？

Why does this compile:

为什么编译:

char ch = '1234'; //no error

But not anything more than 4 chars :

但不超过4个字符:

char ch = '12345'; //error: Too many chars in constant

(Yes I know ' ' is used for one char and " " is for strings; I was just experimenting)

(是的,我知道''用于一个字符串,“”用于字符串;我只是在尝试)

Does this have anything to do with the fact that chars are represented using ASCII numbers?

这是否与使用ASCII数字表示字符的事实有关?

2 个解决方案

#1

It's a multicharacter literal, and has a type of int.

它是一个多字符文字,并且有一种int类型。

C++11 §2.13.2 Character literals

A character literal is one or more characters enclosed in single quotes, as in ’x’, optionally preceded by the letter L, as in L’x’. A character literal that does not begin with L is an ordinary character literal, also referred to as a narrow-character literal. An ordinary character literal that contains a single c-char has type char, with value equal to the numerical value of the encoding of the c-char in the execution character set. An ordinary character literal that contains more than one c-char is a multicharacter literal. A multicharacter literal has type int and implementation-defined value.

字符文字是用单引号括起来的一个或多个字符,如'x',可选地以字母L开头,如'x'中所示。不以L开头的字符文字是普通字符文字,也称为窄字符文字。包含单个c-char的普通字符文字具有char类型,其值等于执行字符集中c-char的编码的数值。包含多个c-char的普通字符文字是多字符文字。多字符文字具有int类型和实现定义的值。

#2

C++ has something called "multicharacter literals". '1234' is an example of one. They have type int, and it is implementation-defined what value they have and how many characters they can contain.

C ++有一种称为“多字符文字”的东西。 '1234'是一个例子。它们具有int类型,并且它是实现定义的,它们具有什么值以及它们可以包含多少个字符。

It's nothing directly to do with the fact that characters are represented as integers, but chances are good that in your implementation the value of '1234' is defined to be either:

这与字符表示为整数这一事实没有直接关系,但在您的实现中,'1234'的值被定义为:

'1' + 256 * '2' + 256 * 256 * '3' + 256 * 256 * 256 * '4'

or:

'4' + 256 * '3' + 256 * 256 * '2' + 256 * 256 * 256 * '1'

#1