问题描述
这是我写的一个小功能，灵感来自线程
紧急帮助！需要Caesar Cipher PLEASE
$ cat /home/keisar/bin/c/ymse/rot13.h
char rot13（char字符）
{
int change;
changed = character - ''a''+''n'';
返回已更改;
}
我发现此代码有两个奇怪之处：
1）我不知道不必指定b应该被替换为n，
c by o等等。怎么样？
2）该函数返回一个char（char rot13），但是改变了
是一个整数。怎么可能？
This is a little function I wrote, inspired by the thread
"Urgent HELP! required for Caesar Cipher PLEASE"
$ cat /home/keisar/bin/c/ymse/rot13.h
char rot13(char character)
{
int changed;
changed = character - ''a'' + ''n'';
return changed;
}
I find two things strange about this code:
1) I don''t have to specify that b should be replaced by n,
c by o and so on. How come?
2) The function returns a char(char rot13), but changed
is an integer. How is that possible?
推荐答案
因为字符是整数类型。与Pascal不同，C不需要
字符之间的转换函数及其数值，
但是将它们视为可互换的。
-
/ - Joona Palaste（pa*****@cc.helsinki.fi）-------------芬兰----- --- \
\ - ---------------------规则！ -------- /
给我看一个好的mouser，我会告诉你一只口臭的猫。
- 加菲猫
Because chars are integer types. Unlike Pascal, C does not require
conversion functions between characters and their numeric values,
but treats them as interchangable by themselves.
--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"Show me a good mouser and I''ll show you a cat with bad breath."
- Garfield
这篇关于为什么这样做？ （rot13功能）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！