如何将ascii字符转换为C中的int

How can I convert an ascii character into an int in C?

如何将ascii字符转换为C中的int？

8 个解决方案

#1

What about:

关于什么：

int a_as_int = (int)'a';

#2

Are you searching for this:

你在找这个：

int c = some_ascii_character;

Or just converting without assignment:

或者只是转换而不分配：

(int)some_aschii_character;

#3

I agree to Ashot and Cwan, but maybe you like to convert an ascii-cipher like '7' into an int like 7?

我同意Ashot和Cwan，但也许你想将像'7'这样的ascii-cipher转换成像7这样的int？

Then I recoomend:

然后我重新开始：

char seven = '7';
int i = seven - '0';

~~or, maybe you get a warning,~~

或者，也许你得到一个警告，

~~int i = (int) (seven - '0');~~

corrected after comments, thanks.

评论后纠正，谢谢。

#4

Use the ASCII to integer function atoi():

使用ASCII到整数函数atoi（）：

#include <stdlib.h>

int num = atoi("23");

#5

You mean the ASCII ordinal value? Try type casting like this one:

你的意思是ASCII序数值？尝试像这样的类型转换：

int x = 'a';

#6

As everyone else told you, you can convert it directly... UNLESS you meant something like "how can I convert an ASCII Extended character to its UTF-16 or UTF-32 value". This is a TOTALLY different question (one at least as good). And one quite difficult, if I remember correctly, if you are using only "pure" C. Then you could start here: https://*.com/questions/114611/what-is-the-best-unicode-library-for-c/114643#114643

正如其他人告诉你的那样，你可以直接转换它......除非你的意思是“如何将ASCII扩展字符转换为UTF-16或UTF-32值”。这是一个完全不同的问题（一个至少同样好）。如果我没记错的话，一个相当困难，如果你只使用“纯粹的”C.那么你可以从这里开始：https：//*.com/questions/114611/what-is-the-best-unicode-library-对于-C / 114643＃114643

(for ASCII Extended I mean one of the many "extensions" to the ASCII set. The 0-127 characters of the base ASCII set are directly convertible to Unicode, while the 128-255 are not.). For example ISO_8859-1 http://en.wikipedia.org/wiki/ISO_8859-1 is an 8 bit extensions to the 7 bit ASCII set, or the (quite famous) codepages 437 and 850.

（对于ASCII扩展，我指的是ASCII集的许多“扩展”之一。基本ASCII集的0-127个字符可直接转换为Unicode，而128-255不是。）。例如，ISO_8859-1 http://en.wikipedia.org/wiki/ISO_8859-1是7位ASCII集的8位扩展，或（非常着名的）代码页437和850。

#7

A char value in C is implicitly convertible to an int. e.g, char c; ... printf("%d", c) prints the decimal ASCII value of c, and int i = c; puts the ASCII integer value of c in i. You can also explicitly convert it with (int)c. If you mean something else, such as how to convert an ASCII digit to an int, that would be c - '0', which implicitly converts c to an int and then subtracts the ASCII value of '0', namely 48 (in C, character constants such as '0' are of type int, not char, for historical reasons).

C中的char值可隐式转换为int。例如，char c; ... printf（“％d”，c）打印c的十进制ASCII值，int i = c;将i的ASCII整数值放在i中。您也可以使用（int）c显式转换它。如果您指的是其他内容，例如如何将ASCII数字转换为int，那将是c - '0'，它隐式地将c转换为int，然后减去ASCII值'0'，即48（在C中），由于历史原因，诸如'0'的字符常量是int类型，而不是char。

#8

It is not possible with the C99 standard library, unless you manually write a map from character constants to the corresponding ASCII int value.

除非您手动将字符常量中的映射写入相应的ASCII int值，否则无法使用C99标准库。

Character constants in C like 'a' are not guaranteed to be ASCII.

像'a'这样的C中的字符常量不保证是ASCII。

C99 only makes some guarantees about those constants, e.g. that digits be contiguous.

C99只对这些常数做出了一些保证，例如：这些数字是连续的。

The word ASCII only appears on the C99 N1256 standard draft in footer notes, and footer note 173) says:

单词ASCII仅出现在页脚注释中的C99 N1256标准草案上，页脚注释173）表示：

In an implementation that uses the seven-bit US ASCII character set, the printing characters are those whose values lie from 0x20 (space) through 0x7E (tilde); the control characters are those whose values lie from 0 (NUL) through 0x1F (US), and the character 0x7F (DEL).

在使用7位US ASCII字符集的实现中，打印字符是其值为0x20（空格）到0x7E（代字号）的字符;控制字符是其值从0（NUL）到0x1F（US）的字符，以及字符0x7F（DEL）。

implying that ASCII is not the only possibility

暗示ASCII不是唯一的可能性

#1