当前位置： 首页 > news >正文

PHP关于简单企业网站开发过程简介网络营销方法有哪些举例

news 2025/8/9 5:30:31
PHP关于简单企业网站开发过程简介,网络营销方法有哪些举例,娄底本地做寄生虫网站,没有足够的权限卸载2345网址导航1.前置程序：语义分析和中间代码生成 基于SLR(1)分析的语义分析及中间代码生成程序-CSDN博客https://blog.csdn.net/lijj0304/article/details/135097554?spm1001.2014.3001.5501 2.程序目标 在前面编译器前端实现的基础上，将所生成的中间代码翻译成某…

1.前置程序：语义分析和中间代码生成

基于SLR(1)分析的语义分析及中间代码生成程序-CSDN博客icon-default.png?t=N7T8https://blog.csdn.net/lijj0304/article/details/135097554?spm=1001.2014.3001.5501

2.程序目标

在前面编译器前端实现的基础上，将所生成的中间代码翻译成某种目标机的汇编代码，实现编译器后端实现的任务。然后进一步实现程序的输入是源程序，输出是汇编语言目标代码小型实验语言编译器。最终程序是实现了算式的识别和汇编代码生成的功能

3.主要数据结构

生成汇编代码用到的寄存器我直接用字符串来实现，因为寄存器仅需要存储变量名即可

char rs0[MAX_LEN];  
char rs1[MAX_LEN]; // 设置两个寄存器  
rs0[0] = '\0';  
rs1[0] = '\0';

对于四元式构造了一个专门的数据结构来存储他的四个变量，同时构建一个四元式的数组来管理，四元式数组的大小刚好也对应了中间变量的数量。

struct quadruple {char op[MAX_LEN];char arg1[MAX_LEN];char arg2[MAX_LEN];char result[MAX_LEN];
}; // 四元式数据结构

3.程序描述

先初始化了两个寄存器R0和R1，如何通过遍历四元式的数据结构数组来生成汇编代码。主要思路是先考虑寄存器中的值，即这个四元式中是否有变量存储在了R0或者R1中。

int flag = 0; // flag为0表示两个寄存器都空闲，1表示找到第一个操作数，2表示找到第二个操作数, 3表示两个操作数都找到  
int tag = -1; // tag为0表示值在rs0，1表示值在rs1  
if((strcmp(rs0, quad[i].arg1) == 0 && strcmp(rs1, quad[i].arg2) == 0) || (strcmp(rs0, quad[i].arg2) == 0 && strcmp(rs1, quad[i].arg1) == 0)) {  flag = 3;  
}  
else if(strcmp(rs0, quad[i].arg1) == 0) {  flag = 1, tag = 0;  
}  
else if(strcmp(rs1, quad[i].arg1) == 0) {  flag = 1, tag = 1;  
}  
else if(strcmp(rs0, quad[i].arg2) == 0) {  flag = 2, tag = 0;  
}  
else if(strcmp(rs1, quad[i].arg2) == 0) {  flag = 2, tag = 1;  
}

在生成了对应四元式的汇编代码后，还需考虑这个结果值是是否在之后要被使用若不需要使用则可以直接把寄存器置空。

int use0(int p) { // 判断是否还会用到寄存器rs0内容for(int i = p; i <= quadTop; i++) {if(strcmp(quad[i].arg1, rs0) == 0 || strcmp(quad[i].arg2, rs0) == 0) {return 1;}}return 0;
}
int use1(int p) { // 判断是否还会用到寄存器rs1内容for(int i = p; i <= quadTop; i++) {if(strcmp(quad[i].arg1, rs1) == 0 || strcmp(quad[i].arg2, rs1) == 0) {return 1;}}return 0;
}

若寄存器R0和R1都不含有变量，则需要找空的及存储变量，若没有可用寄存器则需要报错

        if(flag == 0) {if(rs0[0] == '\0') {printf("MOV R0, %s\n", quad[i].arg1);printf("%s R0, %s\n", opstr, quad[i].arg2);strcpy(rs0, quad[i].result);if(use0(i + 1) == 0) {rs0[0] = '\0';}}else if(rs1[0] == '\0') {printf("MOV R1, %s\n", quad[i].arg1);printf("%s R1, %s\n", opstr, quad[i].arg2);strcpy(rs1, quad[i].result);if(use1(i + 1) == 0) {rs1[0] = '\0';}}else {printf("Assembly failed\n");return;}}

如果R0和R1中有变量，则直接操作对应寄存器，生成代码即可

        else if(flag == 2) {if(tag == 0) {printf("%s R0, %s\n", opstr, quad[i].arg1);strcpy(rs0, quad[i].result);if(use0(i + 1) == 0) {rs0[0] = '\0';}}else {printf("%s R1, %s\n", opstr, quad[i].arg1);strcpy(rs1, quad[i].result);if(use1(i + 1) == 0) {rs1[0] = '\0';}}}else if(flag == 3) {if(use0(i + 1) == 0) {printf("%s R0, R1\n", opstr);strcpy(rs0, quad[i].result);}else {printf("%s R1, R0\n", opstr);strcpy(rs1, quad[i].result);}}}

4.完整代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX_LEN 1000struct stack {char s[MAX_LEN];int i[MAX_LEN];int point[MAX_LEN];int top;
}; // 分析栈数据结构struct quadruple {char op[MAX_LEN];char arg1[MAX_LEN];char arg2[MAX_LEN];char result[MAX_LEN];
}; // 四元式数据结构struct quadruple quad[MAX_LEN]; // 存储四元式
int quadTop = 0; // 四元式栈顶char rs0[MAX_LEN];
char rs1[MAX_LEN]; // 设置两个寄存器// 1.S→V=E  2.E→E+T  3.E→E-T  4.E→T  5.T→T*F  6.T→T/F  7.T→F  8.F→(E) 9.F→i  10.V→i
// 表中大于0对应移进，小于0则对应先归约后移进，0为不存在的状态//          GOTO           |    ACTION//i, =, +, -, *, /, (, ), #, S, E, T, F, V
int table[20][14] ={{ 3, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2},// 0{ 0, 0, 0, 0, 0, 0, 0, 0,-11,0,0, 0, 0, 0},// 1{ 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},// 2{-10,-10,-10,-10,-10,-10,-10,-10,-10, 0, 0, 0, 0, 0},//3{ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 5, 6, 7, 0},// 4{-1,-1,10,11,-1,-1,-1,-1,-1, 0, 0, 0, 0, 0},// 5{-4,-4,-4,-4,12,13,-4,-4,-4, 0, 0, 0, 0, 0},// 6{-7,-7,-7,-7,-7,-7,-7,-7,-7, 0, 0, 0, 0, 0},// 7{ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0,14, 6, 7, 0},// 8{-9,-9,-9,-9,-9,-9,-9,-9,-9, 0, 0, 0, 0, 0},// 9{ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0,15, 7, 0},//10{ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0,16, 7, 0},//11{ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0,17, 0},//12{ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0,18, 0},//13{ 0, 0,10,11, 0, 0, 0,19, 0, 0, 0, 0, 0, 0},//14{-2,-2,-2,-2,12,13,-2,-2,-2, 0, 0, 0, 0, 0},//15{-3,-3,-3,-3,12,13,-3,-3,-3, 0, 0, 0, 0, 0},//16{-5,-5,-5,-5,-5,-5,-5,-5,-5, 0, 0, 0, 0, 0},//17{-6,-6,-6,-6,-6,-6,-6,-6,-6, 0, 0, 0, 0, 0},//18{-8,-8,-8,-8,-8,-8,-8,-8,-8, 0, 0, 0, 0, 0}};//19int english(char ch) {if((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')) return 1;else return 0;
}
int number(char ch) {if(ch >= '0' && ch <= '9') return 1;else return 0;
}
int reserved(char str[]) {if(strcmp(str, "void") == 0) return 3;else if(strcmp(str, "int") == 0) return 4;else if(strcmp(str, "float") == 0) return 5;else if(strcmp(str, "double") == 0) return 6;else if(strcmp(str, "if") == 0) return 7;else if(strcmp(str, "else") == 0) return 8;else if(strcmp(str, "for") == 0) return 9;else if(strcmp(str, "do") == 0) return 10;else if(strcmp(str, "while") == 0) return 11;else if(strcmp(str, "break") == 0) return 12;else if(strcmp(str, "return") == 0) return 13;else return 1;
}
int symbol(char ch) {if(ch == ';') return 14;else if(ch == ',') return 15;else if(ch == '(') return 16;else if(ch == ')') return 17;else if(ch == '{') return 18;else if(ch == '}') return 19;else if(ch == '[') return 20;else if(ch == ']') return 21;else if(ch == '%') return 22;else if(ch == '?') return 23;else if(ch == ':') return 24;else if(ch == '\'') return 25;else if(ch == '\"') return 26;else if(ch == '.') return 27;else return 0;
}int getindex(char ch) {switch(ch) {case 'i': return 0;case '=': return 1;case '+': return 2;case '-': return 3;case '*': return 4;case '/': return 5;case '(': return 6;case ')': return 7;case '#': return 8;case 'S': return 9;case 'E': return 10;case 'T': return 11;case 'F': return 12;case 'V': return 13;default: return -1;}
}int use0(int p) { // 判断是否还会用到寄存器rs0内容for(int i = p; i <= quadTop; i++) {if(strcmp(quad[i].arg1, rs0) == 0 || strcmp(quad[i].arg2, rs0) == 0) {return 1;}}return 0;
}
int use1(int p) { // 判断是否还会用到寄存器rs1内容for(int i = p; i <= quadTop; i++) {if(strcmp(quad[i].arg1, rs1) == 0 || strcmp(quad[i].arg2, rs1) == 0) {return 1;}}return 0;
}void assembly() {printf("Assembly Code:\n"); for(int i = 1; i <= quadTop; i++) {char opstr[MAX_LEN];if(quad[i].op[0] == '=') { // 赋值语句的处理if(strcmp(quad[i].arg1, rs0) == 0){printf("MOV %s, R0\n", quad[i].result);}if(strcmp(quad[i].arg1, rs1) == 0) {printf("MOV %s, R1\n", quad[i].result);}continue;}else if(quad[i].op[0] == '+') { // 获取对应操作码字符串strcpy(opstr, "ADD");}else if(quad[i].op[0] == '-') {strcpy(opstr, "SUB");}else if(quad[i].op[0] == '*') {strcpy(opstr, "MUL");}else if(quad[i].op[0] == '/') {strcpy(opstr, "DIV");}int flag = 0; // flag为0表示两个寄存器都空闲，1表示找到第一个操作数，2表示找到第二个操作数, 3表示两个操作数都找到int tag = -1; // tag为0表示值在rs0，1表示值在rs1if((strcmp(rs0, quad[i].arg1) == 0 && strcmp(rs1, quad[i].arg2) == 0) || (strcmp(rs0, quad[i].arg2) == 0 && strcmp(rs1, quad[i].arg1) == 0)) {flag = 3;}else if(strcmp(rs0, quad[i].arg1) == 0) {flag = 1, tag = 0;}else if(strcmp(rs1, quad[i].arg1) == 0) {flag = 1, tag = 1;}else if(strcmp(rs0, quad[i].arg2) == 0) {flag = 2, tag = 0;}else if(strcmp(rs1, quad[i].arg2) == 0) {flag = 2, tag = 1;}if(flag == 0) {if(rs0[0] == '\0') {printf("MOV R0, %s\n", quad[i].arg1);printf("%s R0, %s\n", opstr, quad[i].arg2);strcpy(rs0, quad[i].result);if(use0(i + 1) == 0) {rs0[0] = '\0';}}else if(rs1[0] == '\0') {printf("MOV R1, %s\n", quad[i].arg1);printf("%s R1, %s\n", opstr, quad[i].arg2);strcpy(rs1, quad[i].result);if(use1(i + 1) == 0) {rs1[0] = '\0';}}else {printf("Assembly failed\n");return;}}else if(flag == 1) {if(tag == 0) {printf("%s R0, %s\n", opstr, quad[i].arg2);strcpy(rs0, quad[i].result);if(use0(i + 1) == 0) {rs0[0] = '\0';}}else {printf("%s R1, %s\n", opstr, quad[i].arg2);strcpy(rs1, quad[i].result);if(use1(i + 1) == 0) {rs1[0] = '\0';}}}else if(flag == 2) {if(tag == 0) {printf("%s R0, %s\n", opstr, quad[i].arg1);strcpy(rs0, quad[i].result);if(use0(i + 1) == 0) {rs0[0] = '\0';}}else {printf("%s R1, %s\n", opstr, quad[i].arg1);strcpy(rs1, quad[i].result);if(use1(i + 1) == 0) {rs1[0] = '\0';}}}else if(flag == 3) {if(use0(i + 1) == 0) {printf("%s R0, R1\n", opstr);strcpy(rs0, quad[i].result);}else {printf("%s R1, R0\n", opstr);strcpy(rs1, quad[i].result);}}}
}int SLR(char *str, struct stack *stk) { // SLR1分析函数quadTop = 0;int i = 0;int next;while(i < strlen(str)) {if(stk->top < 0) return 0; // 分析栈不可能为空int y; // 列坐标if (str[i] >= 'a' && str[i] <= 'z') y = getindex('i'); // 终结符ielse y = getindex(str[i]);if(y == -1 || table[stk->i[stk->top]][y] == 0) { // 表中不存在的状态，分析报错return 0;}if(table[stk->i[stk->top]][y] > 0) { // 移进操作next = table[stk->i[stk->top]][y];stk->top++;stk->s[stk->top] = str[i];stk->i[stk->top] = next;stk->point[stk->top] = i;i++;}else if(table[stk->i[stk->top]][y] < 0) { // 归约操作int tmp = -table[stk->i[stk->top]][y]; // 查GOTO表if(tmp == 4 || tmp == 7 || tmp == 9 || tmp == 10) {stk->top--; // 要归约1位}else if(tmp == 2 || tmp == 3 || tmp == 5 || tmp == 6){// 生成四元式quadTop++;if(tmp == 2) strcpy(quad[quadTop].op, "+");else if(tmp == 3) strcpy(quad[quadTop].op, "-");else if(tmp == 5) strcpy(quad[quadTop].op, "*");else strcpy(quad[quadTop].op, "/");if(stk->point[stk->top - 2] < 0) sprintf(quad[quadTop].arg1, "t%d", -stk->point[stk->top - 2]);else {char arg1[2] = {str[stk->point[stk->top - 2]], '\0'};strcpy(quad[quadTop].arg1, arg1);}if(stk->point[stk->top] < 0) sprintf(quad[quadTop].arg2, "t%d", -stk->point[stk->top]);else {char arg2[2] = {str[stk->point[stk->top]], '\0'};strcpy(quad[quadTop].arg2, arg2);}sprintf(quad[quadTop].result, "t%d", quadTop);stk->top -= 3; // 归约3位stk->point[stk->top + 1] = -quadTop; // 记录归约产生的中间变量}else if(tmp == 8) {stk->top -= 3; // 归约3位stk->point[stk->top + 1] = stk->point[stk->top + 2]; // 消除括号规约}else if(tmp == 1){quadTop++;strcpy(quad[quadTop].op, "=");if(stk->point[stk->top] < 0) sprintf(quad[quadTop].arg1, "t%d", abs(stk->point[stk->top]));else {char arg1[2] = {str[stk->point[stk->top]], '\0'};strcpy(quad[quadTop].arg1, arg1);}sprintf(quad[quadTop].arg2, " ");char res[2] = {str[stk->point[stk->top - 2]], '\0'};strcpy(quad[quadTop].result, res);stk->top -= 3; // 归约V=E}else stk->top -= 3;if(tmp == 1) { y = getindex('S');next = table[stk->i[stk->top]][y]; // 查ACTION表stk->top++;stk->s[stk->top] = 'S';stk->i[stk->top] = next; // 归约要修改栈顶}else if(tmp == 2 || tmp ==3 || tmp == 4) {y = getindex('E');next = table[stk->i[stk->top]][y]; stk->top++;stk->s[stk->top] = 'E';stk->i[stk->top] = next;}else if(tmp == 5 || tmp == 6 || tmp == 7) {y = getindex('T');next = table[stk->i[stk->top]][y];stk->top++;stk->s[stk->top] = 'T';stk->i[stk->top] = next;}else if(tmp == 8 || tmp == 9) {y = getindex('F');next = table[stk->i[stk->top]][y];stk->top++;stk->s[stk->top] = 'F';stk->i[stk->top] = next;}else if(tmp == 10) {y = getindex('V');next = table[stk->i[stk->top]][y];stk->top++;stk->s[stk->top] = 'V';stk->i[stk->top] = next;}else if(tmp == 11) {return 1; }}}return 0;
}int main() {for(int i = 1; i <= 2; i++){char txt1[] = "./test/test";char num[8];sprintf(num, "%d.txt", i);strcat(txt1, num);FILE *fp = fopen(txt1, "r");int flag = 0;char ch = fgetc(fp);	while(!feof(fp)) {int j = 0;if(ch == ' ' || ch == '\t') {ch = fgetc(fp);continue;}else if(ch == '\n'){ch = fgetc(fp);continue;}else if(english(ch)) {do{ch = fgetc(fp);}while(english(ch)||number(ch));}else if(number(ch)) {do{ch = fgetc(fp);}while(number(ch));}else if(symbol(ch) != 0) {ch = fgetc(fp);}else if(ch == '>') {ch = fgetc(fp);if(ch == '=') {ch = fgetc(fp);}else if(ch == '>') {ch = fgetc(fp);}}else if(ch == '<') {ch = fgetc(fp);if(ch == '=') {ch = fgetc(fp);}else if(ch == '<') {ch = fgetc(fp);}}else if(ch == '!') {ch = fgetc(fp);if(ch == '=') {ch = fgetc(fp);}}else if(ch == '=') {ch = fgetc(fp);if(ch == '=') {ch = fgetc(fp);}}else if(ch == '/') {ch = fgetc(fp);if(ch == '*') {do {ch = fgetc(fp);if(ch == '*') {ch = fgetc(fp);if(ch == '/') {ch = fgetc(fp);break;}}}while(1);}else if(ch == '/') {do {ch = fgetc(fp);}while(ch != '\n');ch = fgetc(fp);}else if(ch == '=') {ch = fgetc(fp);}}else if(ch == '&') {ch = fgetc(fp);if(ch == '&') {ch = fgetc(fp);}}else if(ch == '|') {ch = fgetc(fp);if(ch == '|') {ch = fgetc(fp);}}else if(ch == '+') {ch = fgetc(fp);if(ch == '=') {ch = fgetc(fp);}else if(ch == '+') {ch = fgetc(fp);}}else if(ch == '-') {ch = fgetc(fp);if(ch == '=') {ch = fgetc(fp);}else if(ch == '-') {ch = fgetc(fp);}}else if(ch == '*') {ch = fgetc(fp);if(ch == '=') {ch = fgetc(fp);}}else if(ch == '\\') {ch = fgetc(fp);if(ch == 'n') {ch = fgetc(fp);}}else {printf("\ntest%d: Undefined Symbol!\n", i); flag = 1;break;}}if(flag == 0) printf("\ntest%d: Lexical Legal\n", i);else continue;fclose(fp);FILE *fa = fopen(txt1, "r");char input[MAX_LEN] = "";fgets(input, MAX_LEN, fa);printf("Input scentence: %s\n", input); // input为输入串int len = strlen(input);input[len] = '#';fclose(fa);struct stack *stk;stk = (struct stack *)malloc(sizeof(struct stack));stk->s[0] = '#';stk->i[0] = 0;stk->point[0] = -1;stk->top = 0; //初始化分析栈if(!SLR(input, stk)) {printf("Gramma illegal\n");}else {//printQuad(); // 打印四元式rs0[0] = '\0';rs1[0] = '\0';assembly(); // 生成汇编代码}}return 0;
}

5.测试运行

tets1：a=(b+c*d)/f+e*g

test2：a=b+(c+d)*/e

程序运行截图

http://www.mnyf.cn/news/34669.html

相关文章：

  • 3.15网站建设aso优化推广公司
  • 怎么自己的电脑做网站服务器seo关键字优化技巧
  • 中国做网站磁力兔子搜索引擎
  • b2c网站技术架构在线刷关键词网站排名
  • 帮做装修设计的网站济南网站建设老威
  • 绵阳最有实力的公司网站建设2022当下社会热点话题
  • 长春专业网站建设web成品网站源码免费
  • 网站建设工具品牌有惠州seo网站排名
  • 怎么看网站做没做优化网络服务平台
  • 快递网站制作宁德市市长
  • 建设一个公司的网站需要多少钱做网站推广
  • c2c商城网站建设关键词分析工具有哪些
  • 网站建设:上海珍岛百度题库
  • 手机 网站开发软件有哪些app排名优化公司
  • 手机网站关闭窗口代码看网站搜索什么关键词
  • 网站建设推广文案河南企业站seo
  • 青峰集团响应式网站郑州网站建设制作公司
  • nodejs 如何做网站后端百度seo排名优化公司推荐
  • 娱乐网站后缀是什么有人百度看片吗
  • 网站ico图标 代码网络安全培训机构排名
  • 深圳外贸网站定制加快实施创新驱动发展战略
  • 上海网站建设品牌百度官方下载
  • 广东网站建设人员搜索引擎优化的目标
  • 互联网门户网站有哪些市场营销策略有哪些
  • 让网站建设便宜到底seo学堂
  • 做网站注册35类哪几个小项东营网站建设制作
  • 官方网站手机专卖店谷歌seo推广公司
  • ui设计和网站开发百度搜索链接入口
  • 设计单网站建设seo高端培训
  • 一个好的网站内页大概做多少高端网站建设报价