经典进程同步问题（一）——生产者、消费者问题

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今天学习了如题的问题，在此记录。

这里写图片描述

下面是每个部分问题的算法实现：

PS:P()、V()操作分别是wait()和signal()原语操作

V1.0

//注意此处的信号量的初始值Semaphore full = 0;//下边的两个进程执行顺序没有先后Produce:    ...;    put product to buffer;    V(full);    ...;Consumer:    ...;    P(full);    take product from buffer;    ...;

V2.0

Semaphore empty = 0,full = 0;Producter:    while(true){    ...;    P(empty);    put product into buffer;    V(full);    ...;    }Consumer:    while(true){    ...;    P(full);    get product from buffer;    V(empty);    ...;    }

V3.0

Semaphore empty = 0,full = 0;Producter:    while(true){    ...;    P(empty);    put product into buffer;    V(full);    ...;    }Consumer:    while(true){    ...;    P(full);    get product from buffer;    V(empty);    ...;    }

V4.0

//信号量Semaphore empty = 0,full = 0,mutex = 1;//公共变量item buffer[n];int i = 0,j = 0;Producter:    while(true){    ...;    P(empty);//限制一次进入一个进程    P(mutex);//限制一次只能进入一个缓冲区    buffer[i] = product;    i = (i + 1) mod n;    V(mutex);    V(full);    ...;    }Consumer:    while(true){    ...;    P(full);    P(mutex);    take buffer[j];    j = (j + 1) mod n;    V(mutex);    V(empty);    ...;    }

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