我想知道如何在程序中获得光标位置(x，y)，而无需在屏幕上写任何东西，也不能一直跟踪它.
I would like to know how to get the cursor position (x, y) in my program, without writing anything on the screen neither tracking it all the time.
我找到了一种使用此功能获取位置的方法(我不检查读，写等的返回情况来在此主题上编写较小的代码，但我在程序中做到了):
I found out a way to get its position with this function (I don't check the return of read, write, etc here to write a smaller code on this subject but I do it in my program) :
void get_cursor_position(int *col, int *rows) { int a = 0; int i = 0; char buf[4] write(1, "\033[6n", 4); // string asking for the cursor position read(1, buf, 4); while (buf[i]) { if (buf[i] >= 48 && buf[i] <= 57) { if (a == 0) *rows = atoi(&buf[i]) - 1; else *col = atoi(&(buf[i]) - 1; a++; } i++; } }此功能为我提供了准确的光标位置(*行= y，* col = x)，但它在屏幕上书写.
This function gives me the exact cursor position (*rows = y, *col = x), but it writes on the screen.
如何在不在屏幕上书写任何内容的情况下获得光标位置? (因为如果光标在打印的字符之一上，它将被覆盖).
How can I get the cursor position without writting anything on the screen ? (because if the cursor is on one of the printed characters, it will overwrite it).
这是一个学校项目，所以我只能使用termcap，不能使用ncurses函数，唯一允许的函数是tputs，tgoto，tgetstr，tgetnum，tgetflag.
This is a school project, so I only can use termcap, I can't use ncurses functions, the only allowed functions are tputs, tgoto, tgetstr, tgetnum, tgetflag.
推荐答案有几个问题:
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规范模式为缓冲(请参见下文)
read在用于标准输出的文件描述符上完成(可能会起作用—有时—但不要指望它)
the read is done on the file-descriptor for standard output (that may happen to work — sometimes — but don't count on it)
read读取的字符不足以获得典型的响应
the read does not read enough characters to get a typical response
响应将有两个十进制整数，以分号 ;
the response would have two decimal integers, separated by semicolon ;
响应将带有最后一个字符(如果read实际要求提供足够的字符，这将成为问题……)
the response would have a final character (which would become an issue if the read actually asked for enough characters...)
进一步阅读:
- 常规终端接口 单个UNIX ®规范，版本2
在规范模式输入处理中，终端输入以行为单位进行处理.行由换行符(NL)，文件结束符(EOF)或行结束符(EOL)分隔.有关EOF和EOL的更多信息，请参见特殊字符.这意味着read请求将不返回，直到键入了整行或收到信号为止.同样，无论在read()调用中请求多少个字节，最多都将返回一行.但是，不必一次读取整行.可以在read()中请求任意数量的字节，甚至一个字节，而不会丢失信息.
In canonical mode input processing, terminal input is processed in units of lines. A line is delimited by a newline character (NL), an end-of-file character (EOF), or an end-of-line (EOL) character. See Special Characters for more information on EOF and EOL. This means that a read request will not return until an entire line has been typed or a signal has been received. Also, no matter how many bytes are requested in the read() call, at most one line will be returned. It is not, however, necessary to read a whole line at once; any number of bytes, even one, may be requested in a read() without losing information.
- XTerm控制序列
- XTerm Control Sequences CSI Ps n Device Status Report (DSR). Ps = 5 -> Status Report. Result ("OK") is CSI 0 n Ps = 6 -> Report Cursor Position (CPR) [row;column]. Result is CSI r ; c R
也就是说，您的程序应该准备读取 Escape [ ，后跟两个十进制整数(其长度没有固定限制)，以及另外两个字符;和R.
That is, your program should be prepared to read Escape[ followed by two decimal integers (with no fixed limit on their length), and two other characters ; and R.
顺便说一句， termcap 本身对您的解决方案无济于事.尽管ncurses在终端数据库中定义了一些相关功能:
By the way, termcap by itself will do little for your solution. While ncurses has some relevant capabilities defined in the terminal database:
# u9 terminal enquire string (equiv. to ANSI/ECMA-48 DA) # u8 terminal answerback description # u7 cursor position request (equiv. to VT100/ANSI/ECMA-48 DSR 6) # u6 cursor position report (equiv. to ANSI/ECMA-48 CPR)很少有程序使用它们，在任何情况下，您都会发现很难在termcap应用程序中使用光标位置报告.
few programs use those, and in any case you would find it difficult to use the cursor position report in a termcap application.
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