《十进制表达式四则运算》源代码.doc

塔里木大学信息工程学院课程设计源代码：#include #define MAXSIZE 16typedef struct int dataMAXSIZE; int top; int base; seqstack; /* 顺序栈的定义*/*以下为函数声明*/void InitStack(seqstack *); int Empty(seqstack *);void Push(seqstack *, int );int Pop(seqstack *);int GetTop(seqstack *);int Operate(int ,char ,int );char Proceed(char ,char );int In(char );int EvalExpres(void);/* 定义两个栈分别存放运算符和操作数*/seqstack StackR,StackD; /*主函数*/int main() int v; char ch;while(1) printf(t本程序的功能为：用顺序栈实现整型算术表达式的求值n); v = EvalExpres(); printf(The result is:%d,v); /*以下为程序控制*/ printf(nInput q to quit and ENTER run again:); do scanf(%c,&ch); if(ch = q | ch = Q) exit(0); while(ch!=n); system(cls); return 0;void InitStack(seqstack *s) s-top = 0; s-base = 0; /* 初始化栈*/int Empty(seqstack *s) if(s-top = s-base) return 1; else return 0; /* 判断栈是否为空*/void Push(seqstack *s, int x) if(s-top = MAXSIZE) printf(OVER FLOW!n); exit(0); else s-datas-top = x; s-top+; /* 进栈 */int Pop(seqstack *s) int e; if(Empty(s) printf(Under flow!n); return 0; /* 下溢*/ else s-top-; e = s-datas-top; return e; /* 出栈*/int GetTop(seqstack *s) /*取栈顶元素*/ if(Empty(s) printf(Under flow!n); return 0; else return s-datas-top-1; int EvalExpres(void) /* 表达式求解函数*/ int a,b,i=0,s=0; char c80,r; InitStack(&StackR); Push(&StackR,#); InitStack(&StackD); printf( 请输入表达式并以#结束：); gets(c); while(ci!=# | GetTop(&StackR)!=#) if(!In(ci) /* 判断读入的字符不是运算符 是则进栈*/ if(ci = 0 & ci = 9) s += ci-0; while(!In(c+i) /*此处实现的功能为当输入的数字为多位数时*/ s*=10; s += ci-0; Push(&StackD,s+0); s = 0; else printf(你输入的表达式有误!n); return 0; else switch(Proceed(GetTop(&StackR),ci) /* 此函数用来比较读取的运算符和栈顶运算符的优先级*/ case : /* 栈顶的优先级低则出栈并将结果写入栈内*/ r = Pop(&StackR); a = Pop(&StackD)-0; b = Pop(&StackD)-0; Push(&StackD,Operate(a,r,b) ; break; return (GetTop(&StackD)-0); /* 返回运算结果*/int In(char c) /*问题2：解决In函数问题:判断C是否为运算符是返回1否则返回0*/ char ch7=+,-,*,/,#,(,); int i; for(i = 0; i ; break; case *: case /: case (: ch = ; break; case (: ch = ; else if(op=() /*栈顶元素为(的时候*/ switch(c) case +: case -: case *: case /: case (: ch = ; break; case (: printf(Error!n); exit(0); else if(op=#) /*栈顶元素为#的时候*/ switch(c) case +: case -: case *: case /: case (: ch = ; break; case ): printf(Error!n); exit(0); return ch; /* 问题3：解决Operate函数*/int Operate(int a,char r,int b) /*返回由aRb的值 */ int s; int d1 = a; int d2 = b; /*把字符ab变为对应数字*/ switch(r)