我想编写一些“hello world”程序裸机申请于英特尔伽利略木板。当然，使用 UEFI 打印文本（到 UART-1）效果很好，但我想“手动”访问 UART，而不需要 UEFI 的任何帮助。
在 QEMU 中我的代码运行良好：
.h file
#define COM1_PORT (0x03F8)
#define UART_PORT (COM1_PORT)
enum uart_port_offs_t
{ // DLAB RW
THR = 0, // 0 W Transmitter Holding Buffer
RBR = 0, // 0 R Receiver Buffer
DLL = 0, // 1 RW Divisor Latch Low Byte
IER = 1, // 0 RW Interrupt Enable Register
DLH = 1, // 1 RW Divisor Latch High Byte
IIR = 2, // - R Interrupt Identification Register
FCR = 2, // - RW FIFO Control Register
LCR = 3, // - RW Line Control Register
MCR = 4, // - RW Modem Control Register
LSR = 5, // - R Line Status Register
MSR = 6, // - R Modem Status Register
SR = 7, // - RW Scratch Register
};
.c file
void uart_init(void)
{
outb(UART_PORT + IER, 0x00); // Disable all interrupts
outb(UART_PORT + LCR, LCR_DLAB);
outb(UART_PORT + DLL, BAUD_LL); // Set divisor (lo byte)
outb(UART_PORT + DLH, BAUD_HL); // (hi byte)
outb(UART_PORT + LCR, LCR_WORD_BITS_8 | LCR_PAR_NONE | LCR_STOP_BITS_1);
outb(UART_PORT + FCR, FCR_ENABLE | FCR_CLR_RECV | FCR_CLR_SEND | FCR_TRIGGER_16);
outb(UART_PORT + MCR, MCR_DSR | MCR_RTS | MCR_AUX2);
}
ssize_t uart_write(const char *buf, size_t len)
{
size_t written = 0;
while (written < len) {
while (!is_output_empty()) {
asm volatile ("pause");
}
outb(UART_PORT + THR, buf[written]);
++written;
}
return written;
}
main
SystemTable->ConOut->OutputString(SystemTable->ConOut, L"Exiting EFI boot services ...\r\n");
SystemTable->BootServices->ExitBootServices(ImageHandle, map_key);
uart_init();
while (1) {
const char s[] = "UART\r\n";
uart_write(s, sizeof (s) - 1);
}
规格对我没有太大帮助。我猜想 Intel Galileo 板上的 UART 不使用/模拟普通/传统 COM 端口 3F8h、2F8h、3E8h 或 2E8h。
谁能告诉我我做错了什么，或者甚至发布一个最小的裸机 hello world 示例？
