如何在 C 中使用 scanf() 读取负双精度数？

发布于 2024-09-03 08:30:03 字数 693 浏览 2 评论 0原文

我正在学习 C 的基础知识并编写一个简单的一阶方程求解器。我希望输入正好是 ax+b=c 或 ax-b=c，其中 a、b、c 是 double 类型。我正在使用 scanf() 读取用户输入并检查其形式是否正确。但是，如果我输入负双精度数，例如 -4.6，作为等式中的“a”，scanf() 将无法正确读取 a、b、c。我在 scanf() 中使用 %lf。那么我如何读取负双精度呢？非常感谢。

我的代码：

#include <stdio.h>
int main(void)
{
    double a,b,c,x;
    printf("Enter the expression:\n");
    if (scanf("%lfx%lf=%lf", &a, &b, &c) == 3)
    {
    x = (c - b)/a;
    printf("x=%.2f\n", x);
    }
    else if (scanf("%lfx+%lf=%lf", &a, &b, &c) == 3)
    {
        x = (c - b)/a;
        printf("x=%.2f\n", x);
    }
    else
        printf("Invalid expression\n");
    return 0;

}

原文

I'm learning basics of C and writing a simple first order equation solver. I want the input to be exactly ax+b=c or ax-b=c, where a, b, c are double type. I'm employing scanf() to read in user input and to check if it's of the correct form. However, if I enter a negative double, -4.6 say, as the "a" in the equation, scanf() won't read the a,b,c correctly. I'm using %lf inside scanf(). How do I read a negative double, then? Many thanks.

My code:

#include <stdio.h>
int main(void)
{
    double a,b,c,x;
    printf("Enter the expression:\n");
    if (scanf("%lfx%lf=%lf", &a, &b, &c) == 3)
    {
    x = (c - b)/a;
    printf("x=%.2f\n", x);
    }
    else if (scanf("%lfx+%lf=%lf", &a, &b, &c) == 3)
    {
        x = (c - b)/a;
        printf("x=%.2f\n", x);
    }
    else
        printf("Invalid expression\n");
    return 0;

}

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╭ゆ眷念 2024-09-10 08:30:03

直接使用 scanf()（而不是 fgets() 和 sscanf()）的困难之一是找出它在哪里当它出错的时候。

您还可以通过注意 scanf() 返回成功转换的数量来帮助自己。您应该进行测试：

if (scanf("%lfx+%lf=%lf", &a, &b, &c) != 3)
    ...something went wrong...
else
   ...more code using successfully read data...

只要数据正确并且没有遇到 EOF，您所拥有的就应该可以正常工作。也就是说，调用是格式良好的；它发生的背景可能不那么清晰。

另外，由于您希望在两个术语之间允许使用“+”或“-”，因此您确实需要转向读取下一个标记（可能是负数、“x”、“+”或'-'，和另一个（可能是负数）数字）。然后你可以解释这些。如果用户输入“-2.3x-4.5=9.1”，则使用 scanf() 是一场灾难，那么当扫描在（第二个）“-”上失败时，您将无法返回并以负数格式重新读取整个数据。或者您需要将分隔符“+”或“-”读入字符值。

One of the difficulties with using scanf() directly (instead of, say, fgets() and sscanf()) is working out where it got to when it went wrong.

You could also help yourself by noting that scanf() returns the number of successful conversions. You should be testing:

if (scanf("%lfx+%lf=%lf", &a, &b, &c) != 3)
    ...something went wrong...
else
   ...more code using successfully read data...

What you've got should work OK as long as the data is correct and you don't encounter EOF. That is, the call is well formed; the context in which it occurs may not so clean.

Also, since you want to allow either '+' or '-' between the two terms, you really need to move towards a tokenizing mechanism that reads the next token (a possibly negative number, an 'x', a '+' or '-', and another (possibly negative) number). You can then interpret these. Using scanf() is a disaster if the user entered '-2.3x-4.5=9.1', then when the scan fails on on the (second) '-', you have no way to go back and reread the whole data with the negative format. Or you need to read the separator symbol '+' or '-' into a char value.

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