操作系统-PV习题

semaphore mutex1 = mutex2 = mutex3 = mutex0 = 1; //约束每个信箱同一时间只能有一个进程操作
semaphore empty0 = 0 , empty1 = 3 , empty2 = empty3 =2; //信箱剩余空间
semaphore full0 = 3 , full1 = full2 = full3 =0; //信箱剩余信件数
int in0 = in1 = in2 = in3 = out0 = out1 = out2 = out3 =0; //信箱中存、取信的位置，初始均为0
cobegin
process P0( ) {
    while(true) {
        P(full0);
        P(mutex0);
        {从M0[out0]取一条消息};
        out0 = ( out0 + 1 ) % 3;
        V(mutex0);
        V(empty0);
        {加工消息};
        P(empty1);
        P(mutex1);
        {消息存M1[in1]};
        in1 = ( in1 + 1 ) % 3;
        V(mutex1);
        V(full1);
    }
}
/*P1-P3略*/
coend    

semaphore waits = 0;
Semaphore mutex = 1;
int numbersum = 0;
cobegin
process readeri(int number) {    
    P(mutex);
    while(numbersum+number>=M) {
        V(mutex);
        P(waits);
    }
    numbersum = numbersum + number；
    V(mutex)；
    Read file；
    P(mutex)；
    numbersum = numbersum - number；
    V(waits)； 
    V(mutex);
}
coend

semaphore s1 = 1; //可捡白子
semaphore s2 = 0; //可捡黑子
process p1() {
    while(true) {
        P(s1);
        捡白子；
        V(s2);
    }
}
process p2() {
    while(true) {
        P(s2);
        捡白子；
        V(s1);
    }
}

Semaphore allow = -1;
Semaphore mutex = 1;
int count = count_black = count_black = 0;
process A(){
    while(true){
        P(mutex);
        get(black)；
        V(mutex);
        if(取到了){count_black++;}
        else break;
    }
    V(allow);
} 
process B(){
    while(true){
        P(mutex);
        get(write)；
        V(mutex);
        if(取到了){count_write++;}
        else break;
    }
    V(allow);
}
process C(){
    P(allow);
    count = count_black + count_write;
    printf();
}

semaphore mutex , empty[n2] , full[n2];
int i;
mutex = 1;
for(i=1; i<n2; i++) {
    empty[i]=m;
    full[i]=0;
}
cobegin
send() {           
    for(int i=0;i<n2;i++) {
        P(empty[i]);
        P(mutex);
        {将消息放入缓冲区;}
        V(mutex);
    }
    for(int i=0;i<n2;i++){
        V(full[i]);
    }
}
receive() {       
    for(int i=0;i<n2;i++) {
        P(full[i]);
        P(mutex);
        {从缓冲区中取出信息;}
        V(mutex);
    }
    for(int i=0;i<n2;i++) {
        V(empty[i]);
    }
}
coend

int buf[9];
int count = getptr = putptr = 0;
semaphore s1 = s2 = 1;
Semaphore sput = 1;
Semaphore sget = 0;
cobegin
process producer_i() {
    while(true) {
        P(sput);
        P(s1);
        buf[putptr] = 1;
        putptr = ( putptr + 1 ) % 9;
        buf[putptr] = 2;
        putptr = ( putptr + 1 ) % 9;
        buf[putptr] = 3;
        putptr = ( putptr + 1 ) % 9;
       V(sget);
        V(sget);
        V(sget);
        V(s1);        
    }
}
process consumer_j() {
    int y;
    while(true) {
       P(sget);
        P(s2);
        y = buf[getptr];
        getptr = ( getptr + 1 ) % 9;
        count++;
        if(count==3){
            count = 0;
           V(sput);
        }
        V(s2);
        消费整数y;
    }
}
coend

semaphore mutex1 = mutex2 = 1;
semaphore s1 = N/5 , s2 = 4N/5 , s3 = s4 = 0;
int count = 0;
int in1 = in2 = out1 = out2 = 0;
cobegin
process worker1() {
    while(true) {
        加工一个车架；
        P(s1);
        P(mutex1);
        车架放入box[in1];
        in1 = (in1 +1) % (N/5);
        V(mutex1);
        V(s3);
}
process worker2() {
    while(true) {
        加工一个车轮；
        P(s2);
        P(mutex2);
        车架放入box[in2];
        in2 = (in2 +1) % (4N/5);
        count = count + 1;
        if(count==4) {
            count = 0;
            V(s4);
        }
        V(mutex2);
}
process worker3() {
    while(true) {
        P(s3);
        P(mutex1);
        从box[out1]拿车架；
        out1 = (out1 + 1) % (N/5);
        V(mutex1);
        V(s1);
        P(s4);
        P(mutex2);
        从box[out2]开始取四个车轮；
        out2 = (out2 + 4) % (4N/5);
        V(s2);
        V(mutex2);
        装配车子；
    }
}
ceend

semaphore mutex=1;         
semaphore sx = M-1;
semaphore sy = N-1;
cobegin
process storeX( ) {
    while(true) {
        P(sx);
        P(mutex);
       {将X产品入库}；
       V(mutex);
        V(sy);
    }
}
process storeY( ) { 
    while(true) {
       P(sy);
       P(mutex);
       {将Y产品入库}；
       V(mutex);
       V(sx);
    }
}
coend

semaphore empty1 , empty2 , full1 , full2 , mutex , sa , sb;
int in1 , in2 , out1 , out2;
empty1 = empty2 = m; //剩余可放空间
full1 = full2 = 0; //剩余产品数
sa = sb = n;
mutex = 1;
in1 = in2 = out1 = out2 = 0;
cobegin
process put_A() {
    while(true) {
        P(empty1);
        P(sa);
        P(mutex);
        把A零件放入buffA[in1]；
        in1 = (in1 + 1) % m；
        V(mutex);
        V(sb);
        V(full1);
    }
} 
process put_B() {
    while(true) {
        P(empty2);
        P(sb);
        P(mutex);
        把B零件放入buffB[in2]；
        in2 = (in2 + 1) % m；
        V(mutex);
        V(sa);
        V(full2);
    }
} 
process get() {
    while(true) {
        P(full1);
        P(full2);
        P(mutex);
        从buffA[out1]取A零件;
        从buffB[out2]取B零件;
        out1 = (out1 + 1) % m；
        out2 = (out2 + 1) % m；
        V(mutex);
        V(empty1);
        V(empty2);
        把A和B装配成产品;       
    }
}
coend

Semaphore empty = 500;
Semaphore mutex = 1;
cobegin
process_i(){
    P(empty);
    P(mutex);
    进门；
    V(mutex);
    参观；
    P(mutex);
    出门；
    V(mutex);
   V(empty);
}

typedef struct{
    float a;
    float b;
}cnum;
cnum x , y ,z;
cnum add(cnum p ,cnum q){
    cnum s;
    s.a = p.a + q.a;
    s.b = s.b + q.b;
    return s;
}

Semaphore mutex = 1;
Sempahore full = 1;
Semaphore empty = 100;
process P(){
    P(empty);
    P(mutex);
    read20();
    V(mutex);
    V(full);
}
Process Q1(){
    P(full); get20();
    P(full); get20();
    comp40t30();
    V(empty); print15;
    V(empty); print15;
}