linux uinput:简单示例?

我有一些问题让 双方 使用的代码边uinput工作。

基于uinput入门:用户级别输入子系统[死链接;存档 ]我总结了以下 作家 (减去错误处理):

int main(int ac, char **av)

{

int fd = open("/dev/uinput", O_WRONLY | O_NONBLOCK);

int ret = ioctl(fd, UI_SET_EVBIT, EV_ABS);

ret = ioctl(fd, UI_SET_ABSBIT, ABS_X);

struct uinput_user_dev uidev = {0};

snprintf(uidev.name, UINPUT_MAX_NAME_SIZE, "uinput-rotary");

uidev.absmin[ABS_X] = 0;

uidev.absmax[ABS_X] = 255;

ret = write(fd, &uidev, sizeof(uidev));

ret = ioctl(fd, UI_DEV_CREATE);

struct input_event ev = {0};

ev.type = EV_ABS;

ev.code = ABS_X;

ev.value = 42;

ret = write(fd, &ev, sizeof(ev));

getchar();

ret = ioctl(fd, UI_DEV_DESTROY);

return EXIT_SUCCESS;

}

这似乎可行,至少input_event似乎已编写了完整的结构。

然后,我写出了我最能想到的事件的天真的 读者

int main(int ac, char **av)

{

int fd = open(av[1], O_RDONLY);

char name[256] = "unknown";

ioctl(fd, EVIOCGNAME(sizeof(name)), name);

printf("reading from %s\n", name);

struct input_event ev = {0};

int ret = read(fd, &ev, sizeof(ev));

printf("Read an event! %i\n", ret);

printf("ev.time.tv_sec: %li\n", ev.time.tv_sec);

printf("ev.time.tv_usec: %li\n", ev.time.tv_usec);

printf("ev.type: %hi\n", ev.type);

printf("ev.code: %hi\n", ev.code);

printf("ev.value: %li\n", ev.value);

return EXIT_SUCCESS;

}

不幸的是,读者方面根本无法工作。每次只能读取8个字节,这几乎不是完整的input_event结构。

我犯了什么愚蠢的错误?

回答:

您还应该在实际事件之后编写一个同步事件。在您的编写方代码中:

struct input_event ev = {0};

ev.type = EV_ABS;

ev.code = ABS_X;

ev.value = 42;

usleep(1500);

memset(&ev, 0, sizeof(ev));

ev.type = EV_SYN;

ev.code = 0;

ev.value = 0;

ret = write(fd, &ev, sizeof(ev));

getchar();

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