Producer Consumer Problem | Producer Consumer Problem C++
The producer consumer problem (bounded buffer problem) is a great case of a multi-process synchronization issue. The issue portrays two procedures, the producer and the consumer, which share a typical, settled size buffer utilized as a line. |
| Producer Consumer Problem |
Problem: To ensure that the producer won't attempt to include data into the buffer if it's full and that the consumer won't endeavor to expel data from an empty buffer.
Solution: The producer is to either rest or dispose of data if the buffer is full. Whenever the consumer expels a thing from the buffer, it informs the producer, who begins to fill the buffer once more. Similarly, the consumer can rest in the event that it observes the buffer to be empty. Whenever the producer places data into the buffer, it awakens the sleeping consumer. An insufficient solution could result in a halt where the two procedures are holding on to be stirred.
#include<iostream>
#include<cstdlib>
using namespace std;
int in=0,out=0;
int s[size] = {};
void consumer()
{
Solution: The producer is to either rest or dispose of data if the buffer is full. Whenever the consumer expels a thing from the buffer, it informs the producer, who begins to fill the buffer once more. Similarly, the consumer can rest in the event that it observes the buffer to be empty. Whenever the producer places data into the buffer, it awakens the sleeping consumer. An insufficient solution could result in a halt where the two procedures are holding on to be stirred.
Producer Consumer Problem C++ Program
#include<iostream>
#include<cstdlib>
#include<ctime>
#define size 50using namespace std;
int in=0,out=0;
int s[size] = {};
void consumer()
{
if(in<=out)
return;
else
{
int a=s[out];
int a=s[out];
cout<<a<<" is consumed"<<endl;
s[out]=0;
out++;
}}
}}
void producer(int n)
{
if(in>size)
return;
else
{
out=0;
s[in]=n;
cout<<n<<" is produced"<<endl;
in++;
}
}
int main()
{
int a;
srand(time(NULL));
int r[12]={1,1,2,1,1,2,2,2,1,1,2,2};
int values[12];
for(int i=0;i<12;i++)
{
values[i] = rand() % 10 + 1;
}
for(int i=0;i<12;i++)
{ a=r[i];
switch(a)
{
case 1:producer(values[i]);
break;
case 2:consumer();
break;
default:break;
} }}
OUTPUT :
9 is produced
9 is produced
9 is consumed
5 is produced
1 is produced
9 is consumed
5 is consumed
1 is consumed
4 is produced
10 is produced
4 is consumed
10 is consumed
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