Question

我的程序基本上采用一个表达式，最初是一个中缀表达式，将其转换为后缀，然后使用该后缀表达式来计算最终答案。我的程序不允许使用除数组之外的任何其他数据结构，所以我相信堆栈具有预定的大小。计算最终答案是有效的，前提是输入已经是后缀表达式：

Enter an expression: 5392-*+
Postfix expression is 5 3 9 2 - * +
The final answer is 26.

但是如果我输入中缀，我会得到什么：

Enter an expression: 5+3*(9-2)
Postfix expression is 5 3 9 2 - * +
The final answer is 3.

所以我意识到我需要我的“评估”函数来读取后缀表达式而不是中缀表达式，之前我的“infix_to_postfix”函数只是逐个打印出来的字符而不是将这些字符放入自己的数组中。 / p>

int main() {
int i, j, leftPar = 0, rightPar = 0, operand = 0, operator = 0;
char exp[20], operators[] = {'*', '/', '+', '-', '^'};
printf("Enter an expression: ");
scanf("%s", exp);
for (i = 0; i < sizeof(exp); i++) {
    if (isalnum(exp[i]) == 1)
        operand = operand + 1;
    for (j = 0; j < 5; j++) {
        if (exp[i] == operators[j])
            operator = operator + 1;
    }
    if (exp[i] == '(')
        leftPar = leftPar + 1;
    if (exp[i] == ')')
        rightPar = rightPar + 1;
}
if (operand > 7 && operand > 6){
    printf("Too many operands and operators. The program will now     close.\n");
    return 0;
}
if (leftPar != rightPar) {
    printf("Imbalanced parentheses. The program will now close.\n");
    return 0;
}
printf("Postfix expression is ");
infix_to_postfix(exp);
printf("\n");
printf("The final answer is %d.\n", evaluate(exp));
}

char infix_to_postfix(char init[]) {
int x, y;
while (init[x] != '\0') {
    if (isalnum(init[x]))
        printf("%c ", init[x]); // Operand printed out immediately.
    else if (init[x] == '(')
        push(init[x]); // '(' character pushed.
    else if (init[x] == ')') {
        while ((y = pop()) != '(') // Popping elements from stack until reaching '('
            printf("%c ", y);
    }
    else {
        while (priority(init[x]) <= priority(stack[top])) // If expression operator has higher precedence than stack operator, expression operator is pushed onto stack. Else stack operator is popped and printed out.
            printf("%c ", pop());
        push(init[x]);
    }
    x++;
}
while(top != -1)
    printf("%c ", pop()); // Remaining operators printed out.
return 0;
}
int evaluate(char init[]) {
int x, n1, n2, n3 = 0, num;
while(init[x] != '\0')
{
    if(isdigit(init[x]))
    {
        num = init[x] - 48;
        push(num);
    }
    else
    {
        n1 = pop();
        n2 = pop();
        switch(init[x])
        {
            case '+':
            {
                n3 = n1 + n2;
                break;
            }
            case '-':
            {
                n3 = n2 - n1;
                break;
            }
            case '*':
            {
                n3 = n1 * n2;
                break;
            }
            case '/':
            {
                n3 = n2 / n1;
                break;
            }
        }
        push(n3);
    }
    x++;
}
return pop();
}

忽略第29行的“evaluate（exp）”，如何更改我的“infix_to_postfix”函数，甚至我的函数声明，以便它返回一个字符数组（或字符串，我不知道） “评估”功能可以作为参数吗？

将字符数组传递给函数并返回字符数组

0 个答案: