编译原理实验六—代码优化

实验目的：

1. 通过上机实习，加深对代码优化的理解，掌握基本块优化、循环优化的方法。 

2.  掌握利用 DAG 进行基本块优化的技术。 

坑....闷头写了两天总算模拟出了个能跑得起来的差不多的代码，里面还有很多地方可以优化，函数和结构都有很多....先懒着这样吧

实现代码（待改进版）

#include<bits/stdc++.h>
#define PB push_back
using namespace std;
char opera[10]={'=','+','-','*','/','_'};
int n;
int cnt;
map<string,int> mp;
//该变量或常量所对应的结点编号
map<int,int> numbermp;
//该数字对应的结点编号

/*
Solved:
1.多变量对应同一常量
2.根据变量的值计算另一个变量的值

*/

struct Tac{
string str;        //四元式的完整式子
int id;            //序号
int operation;    //操作
string op1;        //操作数1
int num1;        //若操作数1为数字
string op2;        //操作数2
int num2;        //若操作数2为数字
string res;        //结果
int ans;        //若结果为数字
}t[100];

struct node{
int id;            //结点序号
int operation;    //操作

int type;        //type=0:常量num,type=1:变量var
int num;        //结点若代表常量
vector<string> var;
//若结点代表变量
int lchild;        //左。右孩子序号
int rchild;

bool operator < (const node& y) const{
return id<y.id;
}

node(){
id = 0;
operation = 0;
type = 0;
num = 0;
var.clear();
lchild = 0;
rchild = 0;
}
};
vector<node> Dag;

//vector<node> nd;

void debug(){
for(int i = 1;i < n; i++){
cout<<t[i].str<<' '<<t[i].id<<' '<<t[i].op1<<' '<<opera[t[i].operation]<<' '<<t[i].op2<<' '<<t[i].res<<endl;
}
}
void superdebug(){
for(int i = 0 ; i < Dag.size();i++){
cout<<"id = "<<Dag[i].id<<" type = "<<Dag[i].type<<" num = "<<Dag[i].num;
cout<<" var_num = "<<Dag[i].var.size()<<" first_var = "<<Dag[i].var[0];
cout<<" lc = "<<Dag[i].lchild<<" rc = "<<Dag[i].rchild;
cout<<" op "<<opera[Dag[i].operation]<<' '<<mp["T0"]<<endl;
}

}

void process(Tac &stm){
string s = stm.str;
char op = s[1];
for(int i=0;i<5;i++)
if(op==opera[i]) stm.operation = i;
int ptr = 3;
while(s[ptr]==' ') ptr++;
while(s[ptr]!=',') stm.op1+=s[ptr++];
ptr++;
while(s[ptr]==' ') ptr++;
while(s[ptr]!=',') stm.op2+=s[ptr++];
ptr++;
while(s[ptr]==' ') ptr++;
while(s[ptr]!=')') stm.res+=s[ptr++];

if(stm.operation == 1 && stm.op2==" ") stm.operation = 5;
}

void read(){
freopen("exp6_2.in","r",stdin);
char s[100];
n = 1;
while(gets(s)){
for(int i=0;i<strlen(s);i++){
t
.str += s[i];
}
t
.id = n;
process(t
);
n++;
}
debug();
}

node & fiDag_node(int id){
for(int i=0;i<Dag.size();i++){
if(Dag[i].id == id) return Dag[i];
}
}

//变量（常量）s是否已经存在，flag = 1存在
node new_node_var(string s,int &flag){
if(mp[s]) {
flag = 1;
int id = mp[s];
return fiDag_node(id);
}
else{
node tmp;
//cout<<"var = "<<s<<' '<<"cnt = "<<cnt<<endl;
tmp.type = 1;
tmp.var.push_back(s);
mp[s] = cnt;
tmp.id = cnt;
flag = 0;
cnt++;
Dag.push_back(tmp);
return tmp;
}
}

//变量（常量）s是否已经存在，flag = 1存在
node new_node_num(string str,int &flag){
int s = 0;
for(int i=0;i<str.length();i++) s = s*10+str[i]-'0';
if(numbermp[s]) {
flag = 1;
int id = numbermp[s];
return fiDag_node(id);
}
else{
node tmp;
tmp.num = s;
tmp.type = 0;
//cout<<"num = "<<tmp.num<<' '<<"cnt = "<<cnt<<endl;
numbermp[s] = cnt;
tmp.id = cnt;
flag = 0;
cnt++;
Dag.push_back(tmp);
return tmp;
}
}

void pop_node(node &n1){
Dag.pop_back();
cnt--;
if(n1.type)
for(int i = 0;i<n1.var.size();i++) mp[n1.var[i]] = 0;
else
numbermp[n1.num] = 0;
}

void erase_node(string str){
queue<node> q;
for(int i = 0 ; i < Dag.size(); i++){
if(!Dag[i].var.size() || Dag[i].var[0] != str) q.push(Dag[i]);
}
Dag.clear();
while(!q.empty()){
node temp = q.front();
q.pop();
Dag.PB(temp);
}
}

void relocate_node(string str){
int id = mp[str];
if(!id) return ;

mp[str] = 0;
int szflag = 0;//该结点只有它自己本身一个变量名
int nodeloc = -1;
int strloc = -1;
for(int i = 0 ; i < Dag.size(); i++){
int sz = Dag[i].var.size();
for(int j = 0; j < sz; j++){
if(Dag[i].var[j] == str){
if(sz == 1){
szflag = 1;
}
nodeloc = i;
strloc = j;
break;
}
}
if(nodeloc != -1) break;
}
if(nodeloc == -1) return ;
//若当前结点只含有一个变量名
if(szflag){
for(int i = 0 ; i < Dag.size(); i++){
//若当前是某个的子节点不能删除
if(Dag[i].lchild == id || Dag[i].rchild == id){
//    cout<<"是某个节点的子节点"<<' '<<id<<' '<<str<<endl;
return ;
}
}
//cout<<"非子节点"<<' '<<str<<endl;
erase_node(str);
}
//从当前结点的多个变量中删除变量str
else{
queue<string> qq;
for(int j = 0 ; j < strloc ; j++){
string stemp = Dag[nodeloc].var[j];
qq.push(stemp);
}
for(int j = strloc+1; j < Dag[nodeloc].var.size(); j++){
string stemp = Dag[nodeloc].var[j];
qq.push(stemp);
}
Dag[nodeloc].var.clear();
while(!qq.empty()){
string stemp = qq.front();
qq.pop();
Dag[nodeloc].var.PB(stemp);
}
}
}

int cal(int a,int b,int op){
char opr = opera[op];
if(opr == '+') return a+b;
if(opr == '-') return a-b;
if(opr == '*') return a*b;
if(opr == '/') return a/b;
}

node new_node(string str,int &flag){
int f = 0;
node tmp;
if(isdigit(str[0])) tmp = new_node_num(str,f);
else tmp = new_node_var(str,f);
flag = f;
return tmp;
}

void add_edge(Tac &stm){
if(!stm.operation){//赋值
int flag = 0;
node n1 = new_node(stm.op1,flag);
//考虑是否出现过，若出现过是否被引用
relocate_node(stm.res);
node &tmp = fiDag_node(n1.id);
tmp.var.PB(stm.res);
mp[stm.res] = tmp.id;
}
else{//运算
int flag1 = 0;
int flag2 = 0;
node n1 = new_node(stm.op1,flag1);
node n2 = new_node(stm.op2,flag2);

//两个操作数都是数字或代表常量的变量
if(!n1.type && !n2.type){
//cout<<"All op are numbers"<<endl;
int p = cal(n1.num,n2.num,stm.operation);
if(!flag2) pop_node(n2);
if(!flag1) pop_node(n1);
if(!numbermp[p]) {
int flag = 0;
node ret = new_node(stm.res,flag);
numbermp[p] = ret.id;
ret.num = p;
ret.type = 0;
Dag.pop_back();
Dag.PB(ret);
}
else{
int id = numbermp[p];
node &temp = fiDag_node(id);
temp.var.PB(stm.res);
mp[stm.res] = id;
}
}
//两个操作数不同时为数字
else{
//cout<<"Not all numbers Debugging"<<endl;
int flag = 0;
node ret = new_node(stm.res,flag);
ret.lchild = n1.id;
ret.rchild = n2.id;
ret.operation = stm.operation;
//若结果非已存在结点
if(!flag) {
Dag.pop_back();
int fouDag = -1;
//两个操作数是否已经进行过运算
for(int i = 0; i < Dag.size();i++){
if(Dag[i].lchild && Dag[i].rchild
&& Dag[i].lchild == ret.lchild && Dag[i].rchild == ret.rchild
&& Dag[i].operation == ret.operation){
fouDag = i;
break;
}
}
if(fouDag != -1){
Dag[fouDag].var.PB(stm.res);
mp[stm.res] = Dag[fouDag].id;
}
else{
Dag.push_back(ret);
}
}
else{
cout<<"TAT"<<endl;
//结果已经存在 ,删除已有的结点，新建结点并连接左右孩子
relocate_node(stm.res);
int flag = 0;
node freshnode = new_node(stm.res,flag);
freshnode.lchild = n1.id;
freshnode.rchild = n2.id;
freshnode.operation = stm.operation;
Dag.pop_back();
Dag.PB(freshnode);
}
}
}
//superdebug();
//cout<<endl<<endl;
}

void solve(){
for(int i = 1; i < n; i++){
add_edge(t[i]);
}
cout<<endl<<"---ready to debug---"<<endl;
superdebug();

}

void display(){
for(int i = 0 ; i < Dag.size(); i++){
if(!Dag[i].operation && !Dag[i].type){
int sz = Dag[i].var.size();
int num = Dag[i].num;
for(int j = 0; j < sz; j++){
cout<<"(=,"<<num<<", ,"<<Dag[i].var[j]<<")"<<endl;
}
}
if(Dag[i].operation){
int sz = Dag[i].var.size();
string first_var = Dag[i].var[0];
int lchild = Dag[i].lchild;
int rchild = Dag[i].rchild;
string lc = fiDag_node(lchild).var[0];
string rc = fiDag_node(rchild).var[0];
cout<<"("<<opera[Dag[i].operation]<<","<<lc<<","<<rc<<","<<first_var<<")"<<endl;
for(int j = 1; j < sz;j++){
string tmp_var = Dag[i].var[j];
cout<<"(=,"<<first_var<<", ,"<<tmp_var<<")"<<endl;
}
}
}
}

int main(){
cnt = 1;
read();
solve();
display();

return 0;
}
/*
(=,3, ,T0)
(*,2,T0,T1)
(+,R,r,T2)
(*,T1,T2,A)
(=,A, ,B)
(*,2,T0,T3)
(+,R,r,T4)
(*,T3,T4,T5)
(-,R,r,T6)
(*,T5,T6,B)

(*,A,B,T1)
(/,6,2,T2)
(-,T1,T2,T3)
(=,T3, ,X)
(=,5, ,C)
(*,A,B,T4)
(=,2,  ,C)
(+,18,C,T5)

(-,A,C,D)
(*,A,C,E)
(*,D,E,F)
(=,2, ,S)
(-,A,C,T)
(*,A,C,Q)
(*,2,S,G)
(*,T,Q,J)
(*,G,5,K)
(+,K,J,L)
(=,L, ,M)
*/