数据结构栈的相关操作还有例题源码

it2025-10-10 3

一、例题

【问题描述】表达式计算是实现程序设计语言的基本问题之一也是栈的应用的一个典型的例子设计一个程序，使用算法优先打法对算术表达式求值。【基本要求】（1）实现栈的顺序存储结构的基本操作包括初始化栈、清空栈、入栈、出栈、取栈顶元素等。（2）以字符序列的形式从终端输入语法正确的、不含变量的整数表达式利用教科书p35页表3.1给出的算符优先关系，实现对算术四则混合运算表达式的求值。【测试数据】 4*（7-2）：99-34；2038/24 【实现提示】（1）设置运算符栈和算数栈。（2）在读入表达式的字符的同时，完成运算符和运算数（整数）的识别处理以及相应的运算。（3）在识别出运算的同时，要将其字符序列形式转换成整数形式。*/

二、代码

下面展示一些内联代码片。

这里是基操 #include "stdio.h" #include "malloc.h" #define TRUE 1 #define FALSE 0 #define OK 1 #define ERROR 0 #define INFEASIBLE -1//不可行的 #define OVERFLOW -2//溢出 typedef int Status;//状态 #define STACK_INIT_SIZE 100 #define STACKINCREMENT 10 typedef struct{ SElemType *base; SElemType *top; int stacksize; }SqStack; //构造一个空栈 Status InitStack(SqStack *S){ S->base = (SElemType*)malloc(STACK_INIT_SIZE*sizeof(SElemType)); if(!S->base) exit(OVERFLOW); S->top=S->base; S->stacksize=STACK_INIT_SIZE; return OK; } //判断是否为空栈 Status StackEmpty(SqStack S){ if(S.top == S.base) return TRUE; else return FALSE; } //用e返回S的顶元素 Status GetTop(SqStack S, SElemType *e){ if(S.top == S.base) return ERROR; *e = *(S.top-1); return OK; } //插入e为新的顶元素 Status Push(SqStack *S, SElemType e){ if((S->top - S->base) >= S->stacksize){ S->base = ( SElemType*)realloc(S->base, (S->stacksize+STACKINCREMENT)*sizeof(SElemType) ); if(!S->base) exit(OVERFLOW); S->top = S->base +S->stacksize; S->stacksize += STACKINCREMENT; } *(S->top)=e; S->top++; return OK; } //删除S的顶元素，并用e返回其值 Status Pop(SqStack *S, SElemType *e){ if(S->top == S->base) return ERROR; S->top--; *e = *(S->top); return OK; } //从栈底到栈顶依次对S的每个元素调用函数Visit（），一旦失败操作无效 Status ListTraverse(SqStack S,Status (*visit)(SElemType)){ SElemType *p; p=S.base; for(p=S.base;p<S.top;p++) (*visit)(*p); return OK; } //输出元素e Status output(SElemType e){ printf("%d ",e); return OK; } 这里是源码1 #include<iostream> #include<string> #include<stdlib.h> #include"stack.h" #include<Windows.h> using namespace std; /* 是让命名空间中的名称曝光的最简单的方法。命名空间像是在众多名称的可见范围之间竖起一道道围墙。可以避免和应用程序发生命名冲突的问题所谓命名冲突是指在应用程序内两个不同的实体具有相同的名称导致程序无法区分两者。命名冲突发生时，程序必须等到该命名冲突获得解析之后，才得以继续执行*/ char Precede(char a, char b) /*这是一个定义了算术优先级的表单分别对应加减乘除和括号情况*/ { int i, j; char pre[][10] = { /*运算符之间的优先级制作成一张表格*/ { '>','>','<','<','<','>','<','<','<','>' }, { '>','>','<','<','<','>','<','<','<','>' }, { '>','>','>','>','<','>','<','<','<','>' }, { '>','>','>','>','<','>','<','<','<','>' }, { '<','<','<','<','<','>','<','<','<','0' }, { '>','>','>','>','0','=','>','>','>','>' } }; switch (a) { case '+': i = 0; break; case '-': i = 1; break; case '*': i = 2; break; case '/': i = 3; break; case '(': i = 4; break; case ')': i = 5; break; } switch (b) { case '+': j = 0; break; case '-': j = 1; break; case '*': j = 2; break; case '/': j = 3; break; case '(': j = 4; break; case ')': j = 5; break; } return pre[i][j]; } int tranNum(stackL *postexp, string exp, int i) //将中缀表达式中的多位数，按照从低位到高位顺序压入postexp栈中，方便后续计算 { int len = 0,j; char expRead = exp[i]; while (expRead <= '9'&&expRead >= '0')//获得数符长度 { //PushL(postexp, expRead); i++; expRead = exp[i]; len++; } for (j = i-1; j >= i - len;j--) { PushL(postexp, exp[j]); } i--; //使i恰好下次循环指向非数字 PushL(postexp, '#'); return i; } void tranPost(stackL *op, stackL*postexp, string exp)//将中缀表达式转化为后缀表达式 { int i; char expRead, tempChar,preOp; InitStackL(op); InitStackL(postexp); PushL(op, '#'); PushL(postexp, '#'); for (i = 0; i < (int)exp.length(); i++) { expRead = exp[i]; if (expRead == '#') //表达式尾端 break; else if (expRead == ' ') //忽略空格 continue; else if (expRead <= '9'&&expRead >= '0') //处理操作数，传入i，postexp，exp，返回i { i = tranNum(postexp, exp, i); } else if (expRead != '(' && expRead != ')' && expRead == exp[i + 1])//处理单目运算符（++、--） { if (expRead == '+') PushL(op, 'a'); else if (expRead == '-') PushL(op, 's'); else { cout << "error\n"; break; } i++; //使遍历跳过自增自减符号到下一字符 } else //处理普通操作符 { tempChar = GetTopL(op); preOp = Precede(tempChar, expRead); if (expRead == ')') { tempChar = PopL(op); while (tempChar != '(') { PushL(postexp, tempChar); //PushL(postexp, '#'); tempChar = PopL(op); } } else if (preOp == '>') { PushL(postexp, PopL(op)); PushL(op, expRead); //PushL(postexp, '#'); } else if (preOp == '<') PushL(op, expRead); } } while (op->next->data != '#') { tempChar = PopL(op); PushL(postexp, tempChar); } } int ary(int expn) { int i; int sum = 1; for (i = 0; i < expn; i++) { sum *= 10; } return sum; } int getValue(stackL *assistStack)//求得属于同一数值的连续数符的值 { int value = 0; int ex = 0; char charValue; charValue = PopL(assistStack); while (charValue != '#') { value += (charValue - 48)*ary(ex); ex++; charValue = PopL(assistStack); } return value; } void pushBack(stackL *postexp, int valueBack)//将计算出来的int值化作字符，并从低位到高位压入主栈中 { int i = 0; char valueChar; if (valueBack == 0) PushL(postexp, '0'); while (valueBack != 0) { valueChar= (valueBack % 10)+48; PushL(postexp, valueChar); valueBack = valueBack / 10; } PushL(postexp, '#'); } int power(int value, int expn) { int result = 1; for (int i = 0; i < expn; i++) { result *= value; } return result; } int partCalu(stackL *postexp,stackL *assistStack )//计算小段式的值,此时两个操作数分别在两个栈中，操作数在辅助栈里 { int value_1, value_2, value_3; char opera; PushL(assistStack, PopL(postexp)); while (GetTopL(postexp) <= '9'&&GetTopL(postexp) >= '0')//使两个操作数和运算符都在辅助栈中，方便操作数的还原 { PushL(assistStack, PopL(postexp)); } value_1 = getValue(assistStack); value_2 = getValue(assistStack); opera = PopL(assistStack); switch (opera) { case '+': value_3 = value_1 + value_2; break; case '-': value_3 = value_1 - value_2; break; case '*': value_3 = value_1 * value_2; break; case '/': if (value_2 == 0) { cout << "error:除数为0" << endl; return 0; } else value_3 = value_1 / value_2; break; case'^': value_3 = power(value_1, value_2); default: break; } if (assistStack->next == NULL && postexp->next == NULL) { cout << value_2; return 0; } else if (assistStack->next == NULL && postexp->next->next == NULL)//结束条件 { cout << value_3; return 0; } pushBack(postexp, value_3);//将部分求值结果重新压入主栈 if (GetTopL(assistStack) <= '9'&&GetTopL(assistStack) >= '0')//再次满足部分求值条件 partCalu(postexp, assistStack); } int caluValue(stackL *postexp) { int control = 1; int value = 0; stackL *assistStack; char postRead,opera; assistStack = (stackL*)malloc(sizeof(stackL)); InitStackL(assistStack); postRead = GetTopL(postexp); while (postexp->next != NULL) { /*主栈栈顶为#时进行是否部分求值检测，如果辅助栈顶为数符的话，则进行部分求值，求值后将值继续压入主栈，压入时注意低位先压入，这样就可以在连续求值时进行递归*/ if (postRead == '#'&&assistStack->next!=NULL) { if (GetTopL(assistStack) <= '9'&&GetTopL(assistStack) >= '0') { control =partCalu(postexp, assistStack); if (control == 0) return 0; } else if(GetTopL(assistStack) > '9'|| GetTopL(assistStack) < '0') PushL(assistStack, PopL(postexp)); } else if (postRead == 'a'||postRead == 's') { opera=PopL(postexp); if (GetTopL(postexp) != '#') { cout << "单目运算符运用错误，应放在数字后" << endl; exit(0); } else { PushL(assistStack,PopL(postexp)); while (GetTopL(postexp) <= '9'&&GetTopL(postexp) >= '0')//将数符弹出主栈并压入辅助栈中 { PushL(assistStack, PopL(postexp)); } if (opera == 'a') pushBack(postexp, getValue(assistStack) + 1); else if (opera == 's') pushBack(postexp, getValue(assistStack) - 1); } } else PushL(assistStack, PopL(postexp)); postRead = GetTopL(postexp); } } int main() { while (1) { string express; cout << "please enter the expression statement,and end with '#'" << endl; cin >> express; stackL *op, *postexp; op = (stackL*)malloc(sizeof(stackL)); // 操作符栈 postexp = (stackL*)malloc(sizeof(stackL)); //后缀表达式栈 tranPost(op, postexp, express); //将输入的中缀表达式转化为后缀表达式 cout << express << " = "; caluValue(postexp); //计算后缀表达式的值 cout << "\n\n"; free(op); free(postexp); } }

下面展示一些内联代码片。

上面运行出错是少包了啦小笨蛋，文件名字就写成少包的名字 #include<stdlib.h> #include<iostream> using namespace std; typedef char Elemtype; typedef struct LinkStack { Elemtype data; struct LinkStack *next; }stackL; //初始化 void InitStackL(stackL *lst) { lst->next = NULL; } //判断是否为空，为空则返回1 int StackLEmpty(stackL *lst) { return (lst->next == NULL); } //压入 void PushL(stackL *lst, Elemtype x) { stackL *p; p = (stackL*)malloc(sizeof(stackL)); p->next = lst->next; p->data = x; lst->next = p; } //弹出 Elemtype PopL(stackL* lst) { stackL *toFree; if (lst->next == NULL); //cout << "the stack is empty\n"; else { Elemtype x; x = (lst->next)->data; toFree = lst->next; lst->next = (lst->next)->next; free(toFree); return x; } return 0; } //获取栈顶元素 Elemtype GetTopL(stackL* lst) { if (lst->next == NULL); //cout << "the stack is empty\n"; else { Elemtype x; x = lst->next->data; return x; } return 0; }

下面展示一些内联代码片。

源码二 typedef int SElemType; /*放入堆栈的元素的类型*/ #include <ctype.h> #include "stack_s.c" /*判断输入的某个字符是否是运算符 *c表示输入的字符 *op数组中存放系统能识别的运算符 */ Status in(char c,char op[]){ char *p; p=op; while(*p != '\0'){ if(c == *p) return TRUE; p++; } return FALSE; } /*比较两个运算符的优先级 *a，b中存放待比较的运算符 *'>'表示a>b *'0'表示不可能出现的比较 */ char Precede(char a, char b){ int i,j; char pre[][7]={ /*运算符之间的优先级制作成一张表格*/ {'>','>','<','<','<','>','>'}, {'>','>','<','<','<','>','>'}, {'>','>','>','>','<','>','>'}, {'>','>','>','>','<','>','>'}, {'<','<','<','<','<','=','0'}, {'>','>','>','>','0','>','>'}, {'<','<','<','<','<','0','='}}; switch(a){ case '+': i=0; break; case '-': i=1; break; case '*': i=2; break; case '/': i=3; break; case '(': i=4; break; case ')': i=5; break; case '#': i=6; break; } switch(b){ case '+': j=0; break; case '-': j=1; break; case '*': j=2; break; case '/': j=3; break; case '(': j=4; break; case ')': j=5; break; case '#': j=6; break; } return pre[i][j]; } /*进行实际的运算 *a，b中分别以整数的形式存放两个待运算的操作数 *theta中存放代表操作符的字符 *结果以整数的形式返回 */ int Operate(int a, char theta, int b){ int i,j,result; i=a; j=b; switch(theta) { case '+': result = i + j; break; case '-': result = i - j; break; case '*': result = i * j; break; case '/': result = i / j; break; } return result; } /*从输入缓冲区中获得下一个整数或运算符，并通过n带回到主调函数 *返回值为1表示获得的是运算符 *返回值为0表示获得的是整形操作数 */ int getNext(int *n){ char c; *n=0; while((c=getchar())==' '); /*跳过一个或多个空格*/ if(!isdigit(c)){ /*通过函数判断如果字符不是数字，那么只能是运算符*/ *n=c; return 1; } do { /*能执行到该条语句，说明字符是数字，此处用循环获得连续的数字*/ *n=*n*10+(c-'0'); /*把连续的数字字符转换成相对应的整数*/ c=getchar(); } while(isdigit(c)); /*如果下一个字符是数字，进入下一轮循环*/ ungetc(c,stdin); /*新读入的字符不是数字，可能是运算符,为了不影响下次读入，把该字符放回到输入缓冲区*/ return 0; } int EvaluateExpression(){ int n; int flag; int c; char x,theta; int a,b; char OP[]="+-*/()#"; SqStack OPTR; SqStack OPND; InitStack(&OPTR); Push(&OPTR,'#'); InitStack(&OPND); flag=getNext(&c); GetTop(OPTR, &x); while(c!='#' || x != '#') { if(flag == 0) { Push(&OPND,c); flag = getNext(&c); } else { GetTop(OPTR, &x); switch(Precede(x, c)) { case '<'://栈顶元素优先级低 Push(&OPTR,c); flag = getNext(&c); break; case '='://脱括号并接受下一字符 Pop(&OPTR,&x); flag = getNext(&c); break; case '>'：// 退栈并将运算结果入栈 Pop(&OPTR, &theta); Pop(&OPND,&b); Pop(&OPND,&a); Push(&OPND, Operate(a, theta, b)); break; } } GetTop(OPTR, &x); } GetTop(OPND, &c); return c; } void main(){ int c; printf("Please input one expression:"); c=EvaluateExpression(); printf("Result=%d\n",c); getch(); } #include<iostream> #include<string> #include<stdlib.h> #include"stack.h" #include<Windows.h> using namespace std; char Precede(char a, char b) { int i, j; char pre[][10] = { /*运算符之间的优先级制作成一张表格*/ { '>','>','<','<','<','>','<','<','<','>' }, { '>','>','<','<','<','>','<','<','<','>' }, { '>','>','>','>','<','>','<','<','<','>' }, { '>','>','>','>','<','>','<','<','<','>' }, { '<','<','<','<','<','>','<','<','<','0' }, { '>','>','>','>','0','=','>','>','>','>' }, { '>','>','>','>','<','>','>','<','<','>' }, { '>','>','>','>','<','>','>','>','>','>' }, { '>','>','>','>','<','>','>','>','>','>' }, { '<','<','<','<','<','0','<','<','<','=' } }; switch (a) { case '+': i = 0; break; case '-': i = 1; break; case '*': i = 2; break; case '/': i = 3; break; case '(': i = 4; break; case ')': i = 5; break; case '^': i = 6; break; case 'a': i = 7; break;//用a代替++符号方便处理 case 's': i = 8; break;//用s代替--符号方便处理 case '#': i = 9; break; } switch (b) { case '+': j = 0; break; case '-': j = 1; break; case '*': j = 2; break; case '/': j = 3; break; case '(': j = 4; break; case ')': j = 5; break; case '^': j = 6; break; case '++': j = 7; break; case '--': j = 8; break; case '#': j = 9; break; } return pre[i][j]; } int tranNum(stackL *postexp, string exp, int i)//将中缀表达式中的多位数，按照从低位到高位顺序压入postexp栈中，方便后续计算 { int len = 0,j; char expRead = exp[i]; while (expRead <= '9'&&expRead >= '0')//获得数符长度 { //PushL(postexp, expRead); i++; expRead = exp[i]; len++; } for (j = i-1; j >= i - len;j--) { PushL(postexp, exp[j]); } i--; //使i恰好下次循环指向非数字 PushL(postexp, '#'); return i; } void tranPost(stackL *op, stackL*postexp, string exp)//将中缀表达式转化为后缀表达式 { int i; char expRead, tempChar,preOp; InitStackL(op); InitStackL(postexp); PushL(op, '#'); PushL(postexp, '#'); for (i = 0; i < (int)exp.length(); i++) { expRead = exp[i]; if (expRead == '#') //表达式尾端 break; else if (expRead == ' ') //忽略空格 continue; else if (expRead <= '9'&&expRead >= '0') //处理操作数，传入i，postexp，exp，返回i { i = tranNum(postexp, exp, i); } else if (expRead != '(' && expRead != ')' && expRead == exp[i + 1])//处理单目运算符（++、--） { if (expRead == '+') PushL(op, 'a'); else if (expRead == '-') PushL(op, 's'); else { cout << "error\n"; break; } i++; //使遍历跳过自增自减符号到下一字符 } else //处理普通操作符 { tempChar = GetTopL(op); preOp = Precede(tempChar, expRead); if (expRead == ')') { tempChar = PopL(op); while (tempChar != '(') { PushL(postexp, tempChar); //PushL(postexp, '#'); tempChar = PopL(op); } } else if (preOp == '>') { PushL(postexp, PopL(op)); PushL(op, expRead); //PushL(postexp, '#'); } else if (preOp == '<') PushL(op, expRead); } } while (op->next->data != '#') { tempChar = PopL(op); PushL(postexp, tempChar); } } int ary(int expn) { int i; int sum = 1; for (i = 0; i < expn; i++) { sum *= 10; } return sum; } int getValue(stackL *assistStack)//求得属于同一数值的连续数符的值 { int value = 0; int ex = 0; char charValue; charValue = PopL(assistStack); while (charValue != '#') { value += (charValue - 48)*ary(ex); ex++; charValue = PopL(assistStack); } return value; } void pushBack(stackL *postexp, int valueBack)//将计算出来的int值化作字符，并从低位到高位压入主栈中 { int i = 0; char valueChar; if (valueBack == 0) PushL(postexp, '0'); while (valueBack != 0) { valueChar= (valueBack % 10)+48; PushL(postexp, valueChar); valueBack = valueBack / 10; } PushL(postexp, '#'); } int power(int value, int expn) { int result = 1; for (int i = 0; i < expn; i++) { result *= value; } return result; } int partCalu(stackL *postexp,stackL *assistStack )//计算小段式的值,此时两个操作数分别在两个栈中，操作数在辅助栈里 { int value_1, value_2, value_3; char opera; PushL(assistStack, PopL(postexp)); while (GetTopL(postexp) <= '9'&&GetTopL(postexp) >= '0')//使两个操作数和运算符都在辅助栈中，方便操作数的还原 { PushL(assistStack, PopL(postexp)); } value_1 = getValue(assistStack); value_2 = getValue(assistStack); opera = PopL(assistStack); switch (opera) { case '+': value_3 = value_1 + value_2; break; case '-': value_3 = value_1 - value_2; break; case '*': value_3 = value_1 * value_2; break; case '/': if (value_2 == 0) { cout << "error:除数为0" << endl; return 0; } else value_3 = value_1 / value_2; break; case'^': value_3 = power(value_1, value_2); default: break; } if (assistStack->next == NULL && postexp->next == NULL) { cout << value_2; return 0; } else if (assistStack->next == NULL && postexp->next->next == NULL)//结束条件 { cout << value_3; return 0; } pushBack(postexp, value_3);//将部分求值结果重新压入主栈 if (GetTopL(assistStack) <= '9'&&GetTopL(assistStack) >= '0')//再次满足部分求值条件 partCalu(postexp, assistStack); } int caluValue(stackL *postexp) { int control = 1; int value = 0; stackL *assistStack; char postRead,opera; assistStack = (stackL*)malloc(sizeof(stackL)); InitStackL(assistStack); postRead = GetTopL(postexp); while (postexp->next != NULL) { /*主栈栈顶为#时进行是否部分求值检测，如果辅助栈顶为数符的话，则进行部分求值，求值后将值继续压入主栈，压入时注意低位先压入，这样就可以在连续求值时进行递归*/ if (postRead == '#'&&assistStack->next!=NULL) { if (GetTopL(assistStack) <= '9'&&GetTopL(assistStack) >= '0') { control =partCalu(postexp, assistStack); if (control == 0) return 0; } else if(GetTopL(assistStack) > '9'|| GetTopL(assistStack) < '0') PushL(assistStack, PopL(postexp)); } else if (postRead == 'a'||postRead == 's') { opera=PopL(postexp); if (GetTopL(postexp) != '#') { cout << "单目运算符运用错误，应放在数字后" << endl; exit(0); } else { PushL(assistStack,PopL(postexp)); while (GetTopL(postexp) <= '9'&&GetTopL(postexp) >= '0')//将数符弹出主栈并压入辅助栈中 { PushL(assistStack, PopL(postexp)); } if (opera == 'a') pushBack(postexp, getValue(assistStack) + 1); else if (opera == 's') pushBack(postexp, getValue(assistStack) - 1); } } else PushL(assistStack, PopL(postexp)); postRead = GetTopL(postexp); } } int main() { while (1) { string express; cout << "please enter the expression statement,and end with '#'" << endl; cin >> express; stackL *op, *postexp; op = (stackL*)malloc(sizeof(stackL)); // 操作符栈 postexp = (stackL*)malloc(sizeof(stackL)); //后缀表达式栈 tranPost(op, postexp, express); //将输入的中缀表达式转化为后缀表达式 cout << express << " = "; caluValue(postexp); //计算后缀表达式的值 cout << "\n\n"; free(op); free(postexp); } }

下面展示一些内联代码片。

这个是我认为初学者最好理解的，但对栈的使用相对较少，所以放到最后面 #include <stdio.h> #include <string.h> #include <stack> #include <stdlib.h> #include <ctype.h> using namespace std; stack<double>num; stack<char>ch; //将字符对应的行数以数字形式返回 int tonum(char c) { switch(c) { case '+':return 0;break; case '-':return 1;break; case '*':return 2;break; case '/':return 3;break; case '(':return 4;break; case ')':return 5;break; case '=':return 6;break; } } //计算 double cal(double a,double b,char c) { if(c=='+') return a+b; if(c=='-') return a-b; if(c=='*') return a*b; if(c=='/') return a/b; } //比较两个运算符的大小 char compare(char a,char b) { //运算符优先表 char oper[7][8]={ ">><<<>>", ">><<<>>", ">>>><>>", ">>>><>>", "<<<<<= ", ">>>> >>", "<<<<< ="}; int x=tonum(a); int y=tonum(b); //返回比较结果 return oper[x][y]; } int main() { char str[105];//拿到所有的输入 while(scanf("%s",str)!=EOF)//保证输入不为空 { //清空结果栈 while(!num.empty())//num栈里面不是空的 num.pop(); //清空运算符栈 while(!ch.empty())//ch栈里面不是空的 ch.pop(); ch.push('=');//ch栈放入=号 char temp[30];//这个数组是存放输入的东西里面的数字小数点的 int len=strlen(str);//这里是拿到整个输入的长度 int k=0; str[len]='=';str[len+1]='\0'; for(int i=0;i<=len;)//遍历整个输入找到数字类型的东西 { //如果是0-9或者小数点，计算继续for循环求其值 if(isdigit(str[i])||str[i]=='.') //这个循环是判断整个输入找到数字类型的东西放入temp数组 { temp[k++]=str[i]; i++; continue; } if(k!=0) { //将字符串temp通过atof函数转换成double类型的浮点数 num.push(atof(temp)); //初始化字符串temp memset(temp,0,sizeof(temp)); //初始下标k k=0; } // printf("%c",compare(ch.top(),str[i])); switch(compare(ch.top(),str[i])) { //如果栈顶运算符小于当前运算符，将当前字符加入运算符栈 case '<':ch.push(str[i]);i++;break; //如果栈顶运算符等于当前运算符，消去栈顶运算符 case '=':ch.pop();i++;break; //如果栈顶运算符大于当前运算符，计算当前值，并使其结果进结果栈 case '>':{ double a=num.top();num.pop(); double b=num.top();num.pop(); // printf("%lf %c %lf\n",b,ch.top(),a); num.push(cal(b,a,ch.top())); ch.pop(); break; } } } //保留两位小数输出结果 printf("%.2lf\n",num.top()); num.pop(); } return 0; }

三、解说

认真看看！！！代码里面有注释！！！！

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