[英]How to evaluate a postfix expression using character stack using ASCII conversions
[英]How to evaluate arithmetic expression using stack c?
到現在為止,我剛剛完成了將表達式轉換為后綴表達式的操作,我嘗試進行評估,但是出了點問題並使我困惑了很長時間,而且我只知道如何解決它。
這是我的代碼轉向后綴表達式:
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
#define STACK_INIT_SIZE 20
#define STACKINCREMENT 10
#define MAXBUFFER 10
#define OK 1
#define ERROR 0
typedef char ElemType;
typedef int Status;
typedef struct {
ElemType *base;
ElemType *top;
int stackSize;
}sqStack;
Status InitStack(sqStack *s) {
s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));
if ( !s->base ) {
exit(0);
}
s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
return OK;
}
Status Push(sqStack *s, ElemType e) {
//if the stack is full
if ( s->top - s->base >= s->stackSize ) {
s->base = (ElemType *)realloc(s->base, (s->stackSize + STACKINCREMENT) * sizeof(ElemType));
if ( !s->base ) {
exit(0);
}
s->top = s->base + s->stackSize;
s->stackSize += STACKINCREMENT;
}
//store data
*(s->top) = e;
s->top++;
return OK;
}
Status Pop(sqStack *s, ElemType *e) {
if ( s->top == s->base ) {
return ERROR;
}
*e = *--(s->top);
return OK;
}
int StackLen(sqStack s) {
return (s.top - s.base);
}
int main() {
sqStack s;
char c;
char e;
InitStack(&s);
printf("Please input your calculate expression(# to quit):\n");
scanf("%c", &c);
while ( c != '#' ) {
while ( c >= '0' && c <= '9' ) {
printf("%c", c);
scanf("%c", &c);
if ( c < '0' || c > '9' ) {
printf(" ");
}
}
if ( ')' == c ) {
Pop(&s, &e);
while ( '(' != e ) {
printf("%c ", e);
Pop(&s, &e);
}
} else if ( '+' == c || '-' == c ) {
if ( !StackLen(s) ) {
Push(&s, c);
} else {
do {
Pop(&s, &e);
if ( '(' == e ) {
Push(&s, e);
} else {
printf("%c", e);
}
}while ( StackLen(s) && '(' != e );
Push(&s, c);
}
} else if ( '*' == c || '/' == c || '(' == c ) {
Push(&s, c);
} else if ( '#' == c ) {
break;
} else {
printf("\nInput format error!\n");
return -1;
}
scanf("%c", &c);
}
while ( StackLen(s) ) {
Pop(&s, &e);
printf("%c ", e);
}
return 0;
}
當我放3 *(7-2)#時,它返回3 7 2-
一切順利,但是我不知道下一步如何評估它,我只是將其轉換為后綴表達式,我想使用堆棧對其進行評估。
在轉換為波蘭語表示法時,您根本沒有考慮到運算符的優先級。 很難提供完整的代碼,但這是我假設我們沿着strBuild
構造一個字符串的strBuild
。
if (token is integer)
{
strBuild.Append(token);
strBuild.Append(" ");
}
else if (token is Operator)
{
while (isOperator(Stack.Peek()))
{
if (GetExprPrecedence(token) <= GetExprPrecedence(Stack.Peek()))
{
strBuild.Append(Stack.Pop());
strBuild.Append(" ");
}
else
break;
}
Stack.Push(token);
}
else if (token == '(')
{
Stack.Push(token);
}
else if (token == ')')
{
while (Stack.Peek() != '(')
{
if (Stack.Count > 0)
{
strBuild.Append(Stack.Pop());
strBuild.Append(" ");
}
else
{
Show("Syntax Error while Parsing");
break;
}
}
Stack.Pop();
}
while (Stack.Count > 0 && isOperator(Stack.Peek()))
{
if (Stack.Peek() == '(' || Stack.Peek() == ')')
{
MessageBox.Show("All Tokens Read - Syntax Error");
break;
}
Stack.Pop();
}
return strBuild;
strBuild
應該是RPN字符串。 現在進行評估。
for (int i = 0; i < StrPostFix.Length; i++)
{
token = StrPostFix[i];
if (isOperator(token))
{
switch (token)
{
case "/":
op2 = Stack.Pop();
op1 = Stack.Pop();
val = op1 / op2;
break;
case "*":
op2 = Stack.Pop();
op1 = Stack.Pop();
val = op1 * op2;
break;
case "+":
op2 = Stack.Pop();
op1 = Stack.Pop();
val = op1 + op2;
break;
case "-":
op2 = Stack.Pop();
op1 = Stack.Pop();
val = op1 - op2;
break;
}
Stack.Push(val);
}
else
Stack.Push(token);
}
return Stack.Pop();
return Stack.Pop();
彈出評估值並返回。 現在,對於您主要的疑問,由於您沒有處理,我沒有回答,這是您可以處理優先級問題的方式:
enum OperatorPrecedence { Add, Minus, Mult, Div, Brackets };
int GetExprPrecedence(string op)
{
int p = 0;
switch (op)
{
case "(":
p = (int)OperatorPrecedence .Brackets;
break;
case "/":
p = (int)OperatorPrecedence .Div;
break;
case "*":
p = (int)OperatorPrecedence .Mult;
break;
case "-":
p = (int)OperatorPrecedence .Minus;
break;
case "+":
p = (int)OperatorPrecedence .Add;
break;
}
return p;
}
請注意,偽代碼類似於C-Sharp,因為這是我的工作。 我已盡力使它看起來像算法,而且也不模糊,以便您可以與代碼關聯。 根據我的算法得出的結果:
注意,我使用Box Brackets [
代替Round (
對於我的表情。最終答案是15。
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