机器人工程的工作与考研之困惑→汇总篇←
【摘要】
有困惑，说明还在思考，麻木才是最恐怖的自我放弃。
如果在思想上不能做自己的主人，那么在身体上就只能做他人的奴仆。
还挺拗口的O(∩_∩)O
☞ 机器人工程的工作与考研之困惑“卷”
☞ 机器人工程的工作与考研之困惑“歧视”
☞ 机器人工程的工作与考研之困惑“取舍”
☞ ...
有困惑，说明还在思考，麻木才是最恐怖的自我放弃。
如果在思想上不能做自己的主人，那么在身体上就只能做他人的奴仆。
还挺拗口的O(∩_∩)O
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#include<stdio.h>
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#define n 4
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int compltedPhilo = 0,i;
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struct fork{
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int taken;
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}ForkAvil[n];
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struct philosp{
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int left;
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int right;
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}Philostatus[n];
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void goForDinner(int philID){ //same like threads concept here cases implemented
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if(Philostatus[philID].left==10 && Philostatus[philID].right==10)
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printf("Philosopher %d completed his dinner\n",philID+1);
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//if already completed dinner
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else if(Philostatus[philID].left==1 && Philostatus[philID].right==1){
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//if just taken two forks
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printf("Philosopher %d completed his dinner\n",philID+1);
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Philostatus[philID].left = Philostatus[philID].right = 10; //remembering that he completed dinner by assigning value 10
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int otherFork = philID-1;
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if(otherFork== -1)
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otherFork=(n-1);
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ForkAvil[philID].taken = ForkAvil[otherFork].taken = 0; //releasing forks
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printf("Philosopher %d released fork %d and fork %d\n",philID+1,philID+1,otherFork+1);
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compltedPhilo++;
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}
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else if(Philostatus[philID].left==1 && Philostatus[philID].right==0){ //left already taken, trying for right fork
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if(philID==(n-1)){
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if(ForkAvil[philID].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
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ForkAvil[philID].taken = Philostatus[philID].right = 1;
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printf("Fork %d taken by philosopher %d\n",philID+1,philID+1);
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}else{
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printf("Philosopher %d is waiting for fork %d\n",philID+1,philID+1);
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}
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}else{ //except last philosopher case
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int dupphilID = philID;
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philID-=1;
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if(philID== -1)
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philID=(n-1);
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if(ForkAvil[philID].taken == 0){
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ForkAvil[philID].taken = Philostatus[dupphilID].right = 1;
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printf("Fork %d taken by Philosopher %d\n",philID+1,dupphilID+1);
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}else{
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printf("Philosopher %d is waiting for Fork %d\n",dupphilID+1,philID+1);
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}
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}
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}
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else if(Philostatus[philID].left==0){ //nothing taken yet
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if(philID==(n-1)){
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if(ForkAvil[philID-1].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
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ForkAvil[philID-1].taken = Philostatus[philID].left = 1;
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printf("Fork %d taken by philosopher %d\n",philID,philID+1);
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}else{
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printf("Philosopher %d is waiting for fork %d\n",philID+1,philID);
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}
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}else{ //except last philosopher case
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if(ForkAvil[philID].taken == 0){
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ForkAvil[philID].taken = Philostatus[philID].left = 1;
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printf("Fork %d taken by Philosopher %d\n",philID+1,philID+1);
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}else{
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printf("Philosopher %d is waiting for Fork %d\n",philID+1,philID+1);
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}
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}
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}else{}
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}
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int main(){
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for(i=0;i<n;i++)
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ForkAvil[i].taken=Philostatus[i].left=Philostatus[i].right=0;
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while(compltedPhilo<n){
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/* Observe here carefully, while loop will run until all philosophers complete dinner
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Actually problem of deadlock occur only thy try to take at same time
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This for loop will say that they are trying at same time. And remaining status will print by go for dinner function
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*/
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for(i=0;i<n;i++)
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goForDinner(i);
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printf("\nTill now num of philosophers completed dinner are %d\n\n",compltedPhilo);
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}
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return 0;
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}
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#include<iostream>
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#define n 4
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using namespace std;
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int compltedPhilo = 0,i;
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struct fork{
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int taken;
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}ForkAvil[n];
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struct philosp{
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int left;
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int right;
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}Philostatus[n];
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void goForDinner(int philID){ //same like threads concept here cases implemented
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if(Philostatus[philID].left==10 && Philostatus[philID].right==10)
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cout<<"Philosopher "<<philID+1<<" completed his dinner\n";
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//if already completed dinner
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else if(Philostatus[philID].left==1 && Philostatus[philID].right==1){
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//if just taken two forks
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cout<<"Philosopher "<<philID+1<<" completed his dinner\n";
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Philostatus[philID].left = Philostatus[philID].right = 10; //remembering that he completed dinner by assigning value 10
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int otherFork = philID-1;
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if(otherFork== -1)
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otherFork=(n-1);
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ForkAvil[philID].taken = ForkAvil[otherFork].taken = 0; //releasing forks
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cout<<"Philosopher "<<philID+1<<" released fork "<<philID+1<<" and fork "<<otherFork+1<<"\n";
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compltedPhilo++;
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}
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else if(Philostatus[philID].left==1 && Philostatus[philID].right==0){ //left already taken, trying for right fork
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if(philID==(n-1)){
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if(ForkAvil[philID].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
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ForkAvil[philID].taken = Philostatus[philID].right = 1;
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cout<<"Fork "<<philID+1<<" taken by philosopher "<<philID+1<<"\n";
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}else{
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cout<<"Philosopher "<<philID+1<<" is waiting for fork "<<philID+1<<"\n";
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}
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}else{ //except last philosopher case
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int dupphilID = philID;
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philID-=1;
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if(philID== -1)
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philID=(n-1);
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if(ForkAvil[philID].taken == 0){
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ForkAvil[philID].taken = Philostatus[dupphilID].right = 1;
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cout<<"Fork "<<philID+1<<" taken by Philosopher "<<dupphilID+1<<"\n";
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}else{
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cout<<"Philosopher "<<dupphilID+1<<" is waiting for Fork "<<philID+1<<"\n";
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}
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}
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}
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else if(Philostatus[philID].left==0){ //nothing taken yet
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if(philID==(n-1)){
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if(ForkAvil[philID-1].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
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ForkAvil[philID-1].taken = Philostatus[philID].left = 1;
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cout<<"Fork "<<philID<<" taken by philosopher "<<philID+1<<"\n";
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}else{
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cout<<"Philosopher "<<philID+1<<" is waiting for fork "<<philID<<"\n";
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}
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}else{ //except last philosopher case
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if(ForkAvil[philID].taken == 0){
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ForkAvil[philID].taken = Philostatus[philID].left = 1;
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cout<<"Fork "<<philID+1<<" taken by Philosopher "<<philID+1<<"\n";
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}else{
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cout<<"Philosopher "<<philID+1<<" is waiting for Fork "<<philID+1<<"\n";
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}
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}
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}else{}
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}
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int main(){
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for(i=0;i<n;i++)
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ForkAvil[i].taken=Philostatus[i].left=Philostatus[i].right=0;
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while(compltedPhilo<n){
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/* Observe here carefully, while loop will run until all philosophers complete dinner
-
Actually problem of deadlock occur only thy try to take at same time
-
This for loop will say that they are trying at same time. And remaining status will print by go for dinner function
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*/
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for(i=0;i<n;i++)
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goForDinner(i);
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cout<<"\nTill now num of philosophers completed dinner are "<<compltedPhilo<<"\n\n";
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}
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return 0;
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}
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#include <stdio.h>
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int current[5][5], maximum_claim[5][5], available[5];
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int allocation[5] = {0, 0, 0, 0, 0};
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int maxres[5], running[5], safe = 0;
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int counter = 0, i, j, exec, resources, processes, k = 1;
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int main()
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{
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printf("\nEnter number of processes: ");
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scanf("%d", &processes);
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for (i = 0; i < processes; i++)
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{
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running[i] = 1;
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counter++;
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}
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printf("\nEnter number of resources: ");
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scanf("%d", &resources);
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printf("\nEnter Claim Vector:");
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for (i = 0; i < resources; i++)
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{
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scanf("%d", &maxres[i]);
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}
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printf("\nEnter Allocated Resource Table:\n");
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for (i = 0; i < processes; i++)
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{
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for(j = 0; j < resources; j++)
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{
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scanf("%d", ¤t[i][j]);
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}
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}
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printf("\nEnter Maximum Claim Table:\n");
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for (i = 0; i < processes; i++)
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{
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for(j = 0; j < resources; j++)
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{
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scanf("%d", &maximum_claim[i][j]);
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}
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}
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printf("\nThe Claim Vector is: ");
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for (i = 0; i < resources; i++)
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{
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printf("\t%d", maxres[i]);
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}
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printf("\nThe Allocated Resource Table:\n");
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for (i = 0; i < processes; i++)
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{
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for (j = 0; j < resources; j++)
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{
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printf("\t%d", current[i][j]);
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}
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printf("\n");
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}
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printf("\nThe Maximum Claim Table:\n");
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for (i = 0; i < processes; i++)
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{
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for (j = 0; j < resources; j++)
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{
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printf("\t%d", maximum_claim[i][j]);
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}
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printf("\n");
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}
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for (i = 0; i < processes; i++)
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{
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for (j = 0; j < resources; j++)
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{
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allocation[j] += current[i][j];
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}
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}
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printf("\nAllocated resources:");
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for (i = 0; i < resources; i++)
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{
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printf("\t%d", allocation[i]);
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}
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for (i = 0; i < resources; i++)
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{
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available[i] = maxres[i] - allocation[i];
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}
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printf("\nAvailable resources:");
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for (i = 0; i < resources; i++)
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{
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printf("\t%d", available[i]);
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}
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printf("\n");
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while (counter != 0)
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{
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safe = 0;
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for (i = 0; i < processes; i++)
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{
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if (running[i])
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{
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exec = 1;
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for (j = 0; j < resources; j++)
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{
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if (maximum_claim[i][j] - current[i][j] > available[j])
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{
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exec = 0;
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break;
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}
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}
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if (exec)
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{
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printf("\nProcess%d is executing\n", i + 1);
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running[i] = 0;
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counter--;
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safe = 1;
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for (j = 0; j < resources; j++)
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{
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available[j] += current[i][j];
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}
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break;
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}
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}
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}
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if (!safe)
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{
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printf("\nThe processes are in unsafe state.\n");
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break;
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}
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else
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{
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printf("\nThe process is in safe state");
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printf("\nAvailable vector:");
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for (i = 0; i < resources; i++)
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{
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printf("\t%d", available[i]);
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}
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printf("\n");
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}
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}
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return 0;
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}
文章来源: zhangrelay.blog.csdn.net，作者：zhangrelay，版权归原作者所有，如需转载，请联系作者。
原文链接：zhangrelay.blog.csdn.net/article/details/124890695
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