我制作了一个程序，让我可以读取键盘上的所有独立功能键（至少我想测试一下）。我对其进行了设计，以便我可以将任何单个键输入称为单个值。它处理Return, F1- F12, delete, backspace, 箭头等
我只是想测试输入的修改。我已经确保 shift 有效，但现在我决定测试Ctrl和Alt.
问题1 为什么不Alt修改任何输入的键码？
问题 2 为什么我不能捕获某些Ctrl+ 组合？例如。Ctrl+ s; Ctrl+ 1- 9;
Ctrl+2有效，但我认为这可能是因为我的键盘设置为英国。
这是我正在使用的代码。
请注意，我不一定要问如何捕获这些组合键（除非是一两个简单的修改）。我只想知道为什么我不能。
#include <iostream>
#include <conio.h>
#include <cwchar>
union wide_char
{
short Result;
char C[2];
};
int main()
{
wchar_t R;
int N;
wide_char user_input;
//Loops forever, this is only a proof of concept program proving this is possible to incorporate into a larger program
while(true)
{
user_input.C[0] = 0;
user_input.C[1] = 0;
//Loop twice, or until code causes the loop to exit
//Two times are neccessary for function keys unfortunately
for(int i = 0; i < 2; ++i)
{
//While there isn't a key pressed, loop doing nothing
while(!kbhit()){}
//Grab the next key from the buffer
//Since the loop is done, there must be at least one
user_input.C[i] = getch();
switch(user_input.C[i])
{
case 0:
case -32:
//The key pressed is a Function key because it matches one of these two cases
//This means getch() must be called twice
//Break switch, run the for loop again ++i
break;
default:
//The character obtained from getch() is from a regular key
//Or this is the second char from getch() because the for loop is on run #2
//Either way we need a wide char (16 bits / 2 Bytes)
if(user_input.C[1] != 0)
//Function keys {Arrows, F1-12, Esc}
//We now combine the bits of both chars obtained
//They must be combined Second+First else the F1-12 will be duplicate
//This is because on F1-12 getch() returns 0 thus won't affect the combination
R = user_input.Result;
else
//Regular key press
R = user_input.C[0];
//Display our unique results from each key press
N = R;
std::cout << R << " R = N " << N << std::endl;
if( R == 'a' )
std::cout << "a = " << N << std::endl;
//Manually break for loop
i = 3;
break;
}
}
//We need to reset the array in this situation
//Else regular key presses will be affected by the last function key press
}
}