用c语言实现简易计算器的四则运算要求用上数据结构的栈方法，应该是用数组和指针实现的吧，请高手们赐教！！谢
用c语言实现简易计算器的四则运算
要求用上数据结构的栈方法，应该是用数组和指针实现的吧，请高手们赐教！！谢啦
[解决办法]
- C/C++ code
# include <stdio.h># include <stdlib.h># define STACK_INIT_SIZE 20# define INCREASE_STACK_SIZE 10# define ERROR 0# define OK 1typedef int SElemType;struct SqStack{ SElemType *base; SElemType *top; int size;};int initStack(SqStack &s){ s.base = (SElemType*)malloc(STACK_INIT_SIZE * sizeof(SElemType)); if(! s.base) return ERROR; s.top = s.base; s.size = STACK_INIT_SIZE; return OK;}int push(SqStack &s, SElemType e){ if(s.top - s.base >= s.size) { s.base = (SElemType*)realloc(s.base, (s.size + INCREASE_STACK_SIZE) * sizeof(SElemType)); if(! s.base) return ERROR; s.top = s.base + s.size; s.size += INCREASE_STACK_SIZE; } * s.top ++ = e; return OK;}int pop(SqStack &s, SElemType &e){ if(s.top == s.base) return ERROR; e = * -- s.top; return OK;}int getTop(SqStack s, SElemType &e){ if(s.top == s.base) return ERROR; e = * (s.top - 1); return OK;}bool isOperator(SElemType e){ switch(e) { case '+': case '-': case '*': case '/': case '(': case ')': case '=': return true; default : return false; }}SElemType operate(SElemType a, SElemType op, SElemType b){ SElemType c; switch(op) { case '+': c = a + b; break; case '-': c = a - b; break; case '*': c = a * b; break; case '/': c = a / b; break; } return c;}SElemType precede(SElemType a, SElemType b){ SElemType f = '='; switch(b) { case '+': case '-': if(a == '(' || a == '=') f = '<'; else f = '>'; break; case '*': case '/': if(a == '*' || a == '/' || a == ')') f = '>'; else f = '<'; break; case '(': if(a == ')') { printf("表达式有错误\n"); exit(0); } else f = '<'; break; case ')': if(a == '(') f = '='; else if(a == '=') { printf("表达式有错误\n"); exit(0); } else f = '>'; break; case '=': if(a == '(') { printf("表达式有错误\n"); exit(0); } else if(a == '=') f = '='; else f = '>'; break; } return f;}SElemType expressionValue(){ char c; SElemType s1, s2, op, x; SqStack opera, num; initStack(opera); initStack(num); push(opera, '='); c = getchar(); getTop(opera, x); while((c != '=') || x != '=') { if(isOperator(c)) { switch(precede(x, c)) { case '<': push(opera, c); c = getchar(); break; case '=': pop(opera, x); c = getchar(); break; case '>': pop(num, s2); pop(num, s1); pop(opera, op); push(num, operate(s1, op, s2)); break; } } else if(c >= '0' && c <= '9') { s1 = c - '0'; c = getchar(); while(c <= '9' && c >= '0') { s1 = s1 * 10 + c - '0'; c = getchar(); } push(num, s1); } else { printf("表达式有错误\n"); exit(0); } getTop(opera, x); } pop(num, x); return x;}int main(){ printf("%d\n", expressionValue()); return 0;}
------解决方案--------------------
- C/C++ code
//表达式求值#include <stdio.h>#include <malloc.h>#include <string.h>/**功能:根据运算符计算*参数:a, b参与运算的数, ch运算符*返回值:计算结果,操作符错误则返回0*/int cal(int a, char ch, int b){ switch(ch) { case '+': return a+b; break; case '-': return a-b; break; case '*': return a*b; break; case '/': return a/b; break; } return 0;}/**功能:计算表达式的值(用数组模拟栈)*参数:表达式字符串*返回值:计算结果*/int evaluateExpression(char *str){ int i = 0, result, numSub = 0, operSub = 0; int tmp, len = strlen(str); int *operand = (int*)malloc(sizeof(int)*len); char *operat = (char*)malloc(sizeof(char)*len); while(str[i] != '\0') { switch(str[i]) { case '+': while(operSub > 0 && operat[operSub-1] != '(') { printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]); operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]); printf("%d\n", operand[numSub-2]); --numSub; --operSub; } operat[operSub++] = '+'; break; case '-': while(operSub > 0 && operat[operSub-1] != '(') { printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]); operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]); printf("%d\n", operand[numSub-2]); --numSub; --operSub; } operat[operSub++] = '-'; break; case '*': if(str[i+1] >= '0' && str[i+1] <= '9') { tmp = 0; while(str[i+1] >= '0' && str[i+1] <= '9') { tmp = tmp * 10 + str[i+1] - '0'; ++i; } --i; printf("%d * %d = ", operand[numSub-1], tmp); operand[numSub-1] = cal(operand[numSub-1], '*', tmp); printf("%d\n", operand[numSub-1]); ++i; } else operat[operSub++] = '*'; break; case '/': if(str[i+1] >= '0' && str[i+1] <= '9') { tmp = 0; while(str[i+1] >= '0' && str[i+1] <= '9') { tmp = tmp * 10 + str[i+1] - '0'; ++i; } --i; printf("%d / %d = ", operand[numSub-1], tmp); operand[numSub-1] = cal(operand[numSub-1], '/', tmp); printf("%d\n", operand[numSub-1]); ++i; } else operat[operSub++] = '/'; break; case '(': operat[operSub++] = '('; break; case ')': while(operat[operSub-1] != '(') { printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]); operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]); printf("%d\n", operand[numSub-2]); --numSub; --operSub; } --operSub; break; default: tmp = 0; while(str[i] >= '0' && str[i] <= '9') { tmp = tmp * 10 + str[i] - '0'; ++i; } --i; operand[numSub++] = tmp; break; } ++i; } while(numSub > 1 && operSub >= 1) { printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]); operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]); printf("%d\n", operand[numSub-2]); --numSub; --operSub; } result = operand[numSub-1]; free(operand); free(operat); return result;}int main(){ char *str = "225/15-20+(4-3)*2"; int result; printf("计算过程:\n"); result = evaluateExpression(str); printf("计算结果:result = %d\n", result); return 0;}计算过程:225 / 15 = 1515 - 20 = -54 - 3 = 11 * 2 = 2-5 + 2 = -3计算结果:result = -3