您的kbhit(显然,甚至是Windows 的原始kbhit)不会检测是否按下了某个键,而只会检测stdin 上是否有新内容可供阅读。这只会是每秒 25 次左右,具体取决于您的自动重复设置。在您的示例代码中使 stdout 无缓冲将使这一点更加明显 (000000W00000000W000000000W)
如何解决这个问题,以便在按住键盘键时,kbhit() 始终返回 0?
这是不可移植的。在Linux中,可以使用/dev/input下的设备文件来完成
请参阅下面的示例程序。它将注册所有键,甚至是一个单独的 SHIFT 请注意,这实际上是一个键盘记录器,因此您必须以 root 身份运行(或设置程序 setuid)。然后它将注册所有击键,即使它没有键盘焦点。哎呀!
注意:以下示例基于 Kevin Cox 的 keystate.c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <sysexits.h>
#include <glob.h>
#include <linux/input.h>
#include <sys/stat.h>
#include <fcntl.h>
/* Return -1 if no key is being pressed, or else the lowest keycode
(c.f. linux/input-event-codes.h) of all the keys that are being pressed */
int keycode_of_key_being_pressed() {
FILE *kbd;
glob_t kbddev; // Glob structure for keyboard devices
glob("/dev/input/by-path/*-kbd", 0, 0, &kbddev); // Glob select all keyboards
int keycode = -1; // keycode of key being pressed
for (int i = 0; i < kbddev.gl_pathc ; i++ ) { // Loop through all the keyboard devices ...
if (!(kbd = fopen(kbddev.gl_pathv[i], "r"))) { // ... and open them in turn (slow!)
perror("Run as root to read keyboard devices");
exit(1);
}
char key_map[KEY_MAX/8 + 1]; // Create a bit array the size of the number of keys
memset(key_map, 0, sizeof(key_map)); // Fill keymap[] with zero's
ioctl(fileno(kbd), EVIOCGKEY(sizeof(key_map)), key_map); // Read keyboard state into keymap[]
for (int k = 0; k < KEY_MAX/8 + 1 && keycode < 0; k++) { // scan bytes in key_map[] from left to right
for (int j = 0; j <8 ; j++) { // scan each byte from lsb to msb
if (key_map[k] & (1 << j)) { // if this bit is set: key was being pressed
keycode = 8*k + j ; // calculate corresponding keycode
break; // don't scan for any other keys
}
}
}
fclose(kbd);
if (keycode)
break; // don't scan for any other keyboards
}
return keycode;
}
void main()
{
setvbuf(stdout, NULL, _IONBF, 0); // Set stdout unbuffered
while (1) {
int key = keycode_of_key_being_pressed();
printf((key < 0 ? "no key\n" : "keycode: %d\n"), key);
if (key == KEY_X)
exit(0);
}
}