C ++简单计算器 [英] C++ simple calculator

问题描述

您好我试图制作一个C ++计算器来自动解算计算。我有一个字符串/ char *，格式为例如1 + 1，我希望它能解决计算并显示它。

Hi im trying to make a C++ calculator to auto solve a calculation. I have a string/char* that is in the format of e.g 1+1 and i would like it to solve the calculation and display it.

int main()
{
	char input[] = "3+2";
	float num1, num2;
	char operation;

	printf("%c \n", input[0]);
	printf("%c \n", input[1]);
	printf("%c \n", input[2]);

	num1 = input[0];
	operation = input[1];
	num2 = input[2];
	//printf("Enter operator either + or - or * or divide : ");
	//scanf("%c", &operation);
	//printf("Enter two operands: ");
	//scanf("%f%f", &num1, &num2);
	if (operation == '+')
			printf("%.1f + %.1f = %.1f\n", num1, num2, num1 + num2);
		if (operation == '-')
			printf("=%f\n", num1 - num2);
		if (operation == '*')
			printf("=%f\n", num1*num2);
		if (operation == '/')
			printf("=%f\n", num1 / num2);
	system("pause");
	return 0;
}

输出

Output

3
+
2
51.0 + 50.0 = 101.0
Press any key to continue . . .

它将输入数字作为ASCII等价物然后添加它们我需要在计算发生之前转换数字，例如char如果是这样的话我怎么做而不转换整个字符？



我尝试过：



将单独的字符转换为浮点数：



char * input =3 + 2;



float num1 = atof（input [0]）;

float num2 = atof（input [2]）;

Its taking the input numbers as the ASCII equivalent and then adding them do i need to convert the numbers before the calculation takes place e.g char to float if so how do i do that without converting the entire char?

What I have tried:

Converting separate chars to floats:

char* input = "3+2";

float num1 = atof(input[0]);
float num2 = atof(input[2]);

推荐答案

看看：提升精神：平原计算器示例 [ ^ ]



这是提升精神 [ ^ ] - 这是一个相当不错的图书馆可以帮助你解决这类问题。



你要做的事情叫做解析 [ ^ ]



最好的问候

Espen Harlinn

Have a look at: boost spirit: Plain calculator example[^]

It's a simple demonstration of boost spirit[^] - which is a rather nice library that helps you with this kind of stuff.

What you are trying to do is called parsing[^]

Best regards
Espen Harlinn

您的解析目标看起来非常适中。你的代码都是ANSI C所以这可能是一个C问题。



如果你想接受的表达式是数字操作号，其中数字是任何小数值， op是通常的数学运算符之一，如+， - ，*，/ ...看看strtod函数。就像atof，只有更好。



strtod - C ++参考 [ ^ ]



此函数将解析一个数字 - 给定一个字符指针，并返回指向*后该*数字的另一个指针。

Your parsing goals look very modest. Your code is all ANSI C so maybe this is really a C question.

If the expressions you want to accept are number op number where number is any decimal value and "op" is one of the usual math operators like +, -, *, / ... take a look at the strtod function. It's like atof, only better.

strtod - C++ Reference[^]

This function will parse a number - given a character pointer, and return another pointer that points to *after* that number.

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

int main()
{
    char text[] = "123.0+456.0";
    char *cursor = text;
    double term1 = strtod(cursor, &cursor);
    char op = *cursor++;
    if (!op)
        return 2;
    double term2 = strtod(cursor, &cursor);
    double result = 0.f;
    switch (op)
    {
        case '+':
            result = term1 + term2;
            break;
        case '-':
            result = term1 - term2;
            break;
        case '*':
            result = term1 * term2;
            break;
        case '/':
            result = term1 / term2;
            break;
        default:
            return 1;
    }
    printf("%f%c%f=%f\n", term1, op, term2, result);
    return 0;
}

此代码旨在表达一个想法 - 不是一个完整的解决方案。它不考虑除零，空格或更复杂的表达式。我希望它能带领你走向更好的方向。

This code is meant to express an idea - not be a complete solution. It does not consider divide by zero, white space, or more complex expressions. I hope it leads you in a better direction.

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