For input string: "-4.00"

如果您正在使用Java编程语言，可以使用Double.parseDouble("-4.00")将字符串"-4.00"转换为double类型的数字-4.0。例如：

String input = "-4.00";
double num = Double.parseDouble(input);
System.out.println(num); // 输出: -4.0

如果您使用其他编程语言，请查看该语言中字符串转换为数字的相关文档或函数。

相关问题

For input string: "4.00"

This input string represents a number with a decimal point, specifically the number four with two decimal places. In programming, this input string would typically need to be converted to a numerical data type, such as a float or a double, in order to perform mathematical operations with it. The specific method for doing so may vary depending on the programming language being used.

用C语言写基于栈的中缀算术表达式求值，例如输入2+2= 或20*（4.5-3）= 输出4.00 30.00

你好！感谢你的提问。关于基于栈的中缀算术表达式求值，可以使用以下步骤：

1. 创建两个栈：一个用于存储操作数、一个用于存储操作符；
2. 从左到右扫描中缀表达式，遇到数字直接入操作数栈；
3. 遇到操作符时，如果操作符栈为空或者栈顶操作符是左括号，则直接入操作符栈；否则，将栈顶操作符弹出，并将操作数栈中的两个操作数弹出，进行计算，并将结果入操作数栈，直到操作符栈为空或者栈顶操作符是左括号，然后将当前操作符入操作符栈；
4. 遇到左括号直接入操作符栈；
5. 遇到右括号时，将操作符栈中的操作符弹出，直到遇到左括号，将其丢弃；
6. 扫描完成后，如果操作符栈中还有操作符，则取出操作数栈中的两个操作数进行计算，直到操作符栈为空。

下面是C语言代码实现：

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>

#define MAXSIZE 1024

typedef struct {
    double data[MAXSIZE];
    int top;
} OpndStack;

typedef struct {
    char data[MAXSIZE];
    int top;
} OprtStack;

void InitStack(OpndStack *s) {
    s->top = -1;
}

void Push(OpndStack *s, double x) {
    if (s->top == MAXSIZE - 1) {
        printf("Stack overflow!\n");
        exit(1);
    } else {
        s->top++;
        s->data[s->top] = x;
    }
}

double Pop(OpndStack *s) {
    if (s->top == -1) {
        printf("Stack underflow!\n");
        exit(1);
    } else {
        double x = s->data[s->top];
        s->top--;
        return x;
    }
}

double GetTop(OpndStack *s) {
    if (s->top == -1) {
        printf("Stack is empty!\n");
        exit(1);
    } else {
        return s->data[s->top];
    }
}

void InitStack2(OprtStack *s) {
    s->top = -1;
}

void Push2(OprtStack *s, char x) {
    if (s->top == MAXSIZE - 1) {
        printf("Stack overflow!\n");
        exit(1);
    } else {
        s->top++;
        s->data[s->top] = x;
    }
}

char Pop2(OprtStack *s) {
    if (s->top == -1) {
        printf("Stack underflow!\n");
        exit(1);
    } else {
        char x = s->data[s->top];
        s->top--;
        return x;
    }
}

char GetTop2(OprtStack *s) {
    if (s->top == -1) {
        printf("Stack is empty!\n");
        exit(1);
    } else {
        return s->data[s->top];
    }
}

int IsEmpty(OpndStack *s) {
    return s->top == -1 ? 1 : 0;
}

int IsEmpty2(OprtStack *s) {
    return s->top == -1 ? 1 : 0;
}

int Priority(char oprt) {
    switch (oprt) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        case '(':
        case ')':
            return 0;
        default:
            return -1;
    }
}

void Calculate(OpndStack *s, char oprt) {
    double opnd1, opnd2, result;
    opnd2 = Pop(s);
    opnd1 = Pop(s);
    switch (oprt) {
        case '+':
            result = opnd1 + opnd2;
            break;
        case '-':
            result = opnd1 - opnd2;
            break;
        case '*':
            result = opnd1 * opnd2;
            break;
        case '/':
            result = opnd1 / opnd2;
            break;
        default:
            printf("Invalid operator!\n");
            exit(1);
    }
    Push(s, result);
}

double EvaluateExpression(char *s) {
    OpndStack opndStack;
    OprtStack oprtStack;
    double num = 0.0, result;
    int i = 0, flag = 0;
    InitStack(&opndStack);
    InitStack2(&oprtStack);
    while (s[i]) {
        if (isdigit(s[i])) {
            flag = 1;
            num = num * 10 + s[i] - '0';
            i++;
        } else {
            if (flag) {
                Push(&opndStack, num);
                num = 0.0;
                flag = 0;
            }
            if (s[i] == ')') {
                while (GetTop2(&oprtStack) != '(') {
                    Calculate(&opndStack, Pop2(&oprtStack));
                }
                Pop2(&oprtStack);
            } else {
                while (!IsEmpty2(&oprtStack) && Priority(s[i]) <= Priority(GetTop2(&oprtStack))) {
                    Calculate(&opndStack, Pop2(&oprtStack));
                }
                Push2(&oprtStack, s[i]);
                i++;
            }
        }
    }
    if (flag) {
        Push(&opndStack, num);
    }
    while (!IsEmpty2(&oprtStack)) {
        Calculate(&opndStack, Pop2(&oprtStack));
    }
    result = Pop(&opndStack);
    return result;
}

int main() {
    char s[MAXSIZE];
    double result;
    printf("Please input an infix arithmetic expression:\n");
    fgets(s, MAXSIZE, stdin);
    s[strlen(s) - 1] = '\0';
    result = EvaluateExpression(s);
    printf("%.2f", result);
    return 0;
}

你可以输入2 2= 或20*（4.5-3）= 测试一下。