#include "stdafx.h"
#include <stdio.h>
#include <windows.h>
#include <math.h>
#include <conio.h>
typedef struct {
short int pitch;
short int duration;
} NOTE;
NOTE notes[] = {{14, 500}, {16, 500}, {12, 500}, {0, 500}, {7, 1000}};
void setfreq(int hz)
{
hz = 1193180 / hz; // clocked at 1.19MHz
_outp(0x43, 0xb6); // timer 2, square wave
_outp(0x42, hz);
_outp(0x42, hz >> 8);
}
void playnote(NOTE note)
{
_outp(0x61, _inp(0x61) | 0x03); // start speaker going
setfreq((int)(400 * pow(2, note.pitch / 12.0)));
Sleep(note.duration);
_outp(0x61, _inp(0x61) & ~0x03); // stop that racket!
}
int main(int argc, char* argv[])
{
int i;
HANDLE h;
h = CreateFile("\\\\.\\giveio", GENERIC_READ, 0, NULL,
OPEN_EXISTING, FILE_ATTRIBUTE_NORMAL, NULL);
if(h == INVALID_HANDLE_VALUE)
{
printf("Couldn't access giveio device\n");
return -1;
}
for(i=0; i < sizeof(notes)/sizeof(int); ++i)
playnote(notes[i]);
CloseHandle(h);
return 0;
}
有几点需要注意：
1，\\\\.\\giveio是打开的设备，由于是用\\转义表示\，所以上述其实是表示\\ . \
2，这种底层的编程很少。以上_outp访问的是底层寄存器，功能如注
操作系统的一个经典问题是"生产者-消费者"问题, 这涉及同步信号量和互斥信号量的应用, 在这里,我用线程的同步和互斥来实现.
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <pthread.h>
#include <semaphore.h>
#define N 2 // 消费者� ......
