c语言逆波兰计算器程序,C语言实现的简单的逆波兰计算器
前后調(diào)試了好幾天，類型轉(zhuǎn)換問題頗為麻煩，無賴使用兩種不同類型的棧解決，但程序目前還有兩個(gè)問題：
1.目前不能進(jìn)行float類型的計(jì)算；
2.不能進(jìn)行多位數(shù)的運(yùn)算。
實(shí)際上就是目前只能進(jìn)行個(gè)位數(shù)的帶括號(hào)的加減乘除運(yùn)算，目前只能寫到這一步，期待后續(xù)改進(jìn)。
#include
#include
#include
#define CAPACITY 50
typedef struct stack
{
int top;
char arr[CAPACITY];
}*pStack,stack;
pStack initStack1()
{
pStack st = (pStack)malloc(sizeof(stack));
st->top = 0;
return st;
}
int stackEmpty1(pStack st)
{
return(st->top <= 0);
}
int stackFull1(pStack st)
{
return(st->top >= CAPACITY);
}
int pushStack1(pStack st, char a)
{
if(stackFull1(st))
{
printf("棧滿！\n");
return 0;
}
else
{
st->arr[st->top++] = a;
return 1;
}
}
char popStack1(pStack st)
{
char a;
if(stackEmpty1(st))
{
printf("?？?\n");
return 0;
}
else
{
a = st->arr[--st->top];
return a;
}
}
char getTop1(pStack st)
{
if(!stackEmpty1(st))
return st->arr[st->top-1];
else
printf("Empty stack");
return 0;
}
typedef struct
{
int top;
int data[CAPACITY];
}*dataStack;
dataStack initStack2()
{
dataStack st = (dataStack)malloc(sizeof(dataStack));
st->top = 0;
return st;
}
int stackEmpty2(dataStack st)
{
return(st->top <= 0);
}
int stackFull2(dataStack st)
{
return(st->top >= CAPACITY);
}
int pushStack2(dataStack st, int data)
{
if(stackFull2(st))
{
printf("棧滿！\n");
return 0;
}
else
{
st->data[st->top++] = data;
return 1;
}
}
int popStack2(dataStack st)
{
int data;
if(stackEmpty2(st))
{
printf("?？?\n");
return 0;
}
else
{
data = st->data[--st->top];
return data;
}
}
int getTop2(dataStack st)
{
if(!stackEmpty2(st))
return st->data[st->top-1];
else
printf("Empty stack");
return 0;
}
int priority(char op)
{
switch(op)
{
case '*':
case '/':return 2;
case '+':
case '-':return 1;
default: return 0;
}
}
int myAtoi(char a)
{
return a-'0';
}
int main(void)
{
int p = 0;
int i = 0;
int j;
int result, tmp1, tmp2;
char ch, temp;
char infix[CAPACITY] = "\0";
char output[CAPACITY];
pStack st = initStack1();
dataStack st2 = initStack2();
printf("請(qǐng)輸入中綴表達(dá)式：");
scanf("%s", &infix);
printf("后綴表達(dá)式為：");
for(ch = infix[p];ch !='\0';ch = infix[++p])
{
switch(ch)
{
case '(':pushStack1(st,ch);break;
case ')':
{
while(getTop1(st) != '(')
{
output[i++] = getTop1(st);
printf("%c", popStack1(st));
}
popStack1(st);
break;
}
case '+':
case '-':
case '*':
case '/':
{
while(!stackEmpty1(st))
{
temp = getTop1(st);
if(priority(temp) >= priority(ch))
{
output[i++] = temp;
printf("%c", popStack1(st));
}
else
{
break;
}
}
pushStack1(st, ch);
break;
}
default:output[i++] = ch,printf("%c",ch);
}
}
while(!stackEmpty1(st))
{
temp = popStack1(st);
output[i++] = temp;
if(temp == '(')
{
printf("表達(dá)式有誤！");
exit(EXIT_FAILURE);
}
printf("%c", temp);
printf("\n");
}
output[i] = '\0';
free(st);
for(j=0;j
{
switch(output[j])
{
case'+':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp1 + tmp2;pushStack2(st2, result);break;
case'-':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp2 - tmp1;pushStack2(st2, result);break;
case'*':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp2 * tmp1;pushStack2(st2, result);break;
case'/':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp2 / tmp1;pushStack2(st2, result);break;
default:pushStack2(st2,myAtoi(output[j]));break;
}
}
printf("輸出結(jié)果為: %d\n", getTop2(st2));
return 0;
}兩種堆棧可以用泛型實(shí)現(xiàn)共用函數(shù)，目前還不清楚怎么實(shí)現(xiàn)。
總結(jié)
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