用户输入字符串转换为double

Question

以下代码有效，但计算错误。请告诉我应该如何计算它，以便输出与输入相同。请不要更改或添加新代码，除非它只是计算部分。

目前的结果只是例如：

键入一个将转换为double的字符串：12.23
字符串12.23 - ＆gt;数字0.012

#include <stdio.h>

void main()
{
    char input[100];
    printf("Type a String which will be converted to a double: ");
    scanf("%s", input);

    double number = 0.0;
    double divider = 1.0;
    double sign = 1.0;

    enum { Start, End, Error, A, B, C, D } state = Start;

    int i = 0;
    while (state != End && state != Error)
    {
        char ch = input[i];
        i++;
        switch (state)
        {
        case Start:
            if (ch == '+' || ch == '-')
            {
                state = A;
                if (ch == '-')
                {
                    sign = -1.0;
                }
            }
            else if (ch >= '0' && ch <= '9')
            {
                state = B;
                number = number * 10.0 + ch - '0';
            }
            else
            {
                state = Error;
            }
            break;
        case A:
            if (ch >= '0' && ch <= '9')
            {
                state = B;
                number = number * 10.0 + ch - '0';
            }
            else
            {
                state = Error;
            }
            break;
        case B:
            if (ch >= '0' && ch <= '9')
            {
                state = B;
                number = number * 10.0 + ch - '0';
            }
            else if (ch == '.')
            {
                state = C;
                divider = divider * 10.0;
            }
            else if (ch == '\0')
            {
                state = End;
            }
            else
            {
                state = Error;
            }
            break;
        case C:
            if (ch >= '0' && ch <= '9')
            {
                state = D;
                divider = divider * 10.0;
            }
            else
            {
                state = Error;
            }
            break;
        case D:
            if (ch >= '0' && ch <= '9')
            {
                state = D;
                divider = divider * 10.0;
            }
            else if (ch == '\0')
            {
                state = End;
            }
            else
            {
                state = Error;
            }
            break;
        }
    }

    if (state == End)
    {
        printf("string %s -> number %g \n", input, sign * number / divider);
    }
    else
    {
        printf("Enter a valid number! \n");
    }
}

Answer 1

纠正你奇怪的代码

状态$active_group = 'default'; $active_record = TRUE; $db['default']['failover'] = array( array( 'hostname' => 'localhost', 'username' => 'root', 'password' => '', 'database' => 'com', 'dbdriver' => 'mysqli', 'dbprefix' => '', 'pconnect' => TRUE, 'db_debug' => TRUE, 'cache_on' => FALSE, 'cachedir' => '', 'char_set' => 'utf8', 'dbcollat' => 'utf8_general_ci', 'swap_pre' => '', 'encrypt' => FALSE, 'compress' => FALSE, 'stricton' => FALSE ), array( 'hostname' => 'localhost2', 'username' => 'root', 'password' => '', 'database' => 'ABC', 'dbdriver' => 'mysqli', 'dbprefix' => '', 'pconnect' => TRUE, 'db_debug' => TRUE, 'cache_on' => FALSE, 'cachedir' => '', 'char_set' => 'utf8', 'dbcollat' => 'utf8_general_ci', 'swap_pre' => '', 'encrypt' => FALSE, 'compress' => FALSE, 'stricton' => FALSE ) ); 和C应为

D

状态case C: if (ch >= '0' && ch <= '9') { state = D; divider = divider * 10.0; number = number * 10.0 + ch - '0'; } else { state = Error; } break; case D: if (ch >= '0' && ch <= '9') { state = D; divider = divider * 10.0; number = number * 10.0 + ch - '0'; } else if (ch == '\0') { state = End; } else { state = Error; } break; 应该避免增加分隔符：

B