c - 实现以下函数，将 float 的字符串表示形式转换为 double :

Question

Implement the following function which converts a string representation of a floating-point number into a double:

double strToDouble(char[] str)

Your implementation should be able to properly handle strings such as:

• “123456”
• “-123.456”
• “123.456e13”
• “123.456e-13”

Any improperly formatted input strings should result in the value 0.0 being returned.

Hints and Suggestions:

• Use the function strToInt above as a starting point.
• Use if statements after the first loop to check for things like decimal points, e’s (for scientific notation exponents), and negative signs.
• Use a separate loop to process things after a possible decimal point.
• Use yet another loop to process possible exponent values after an e.

调用函数 strToDouble 后我是如何回答的

1)正确的输出应该是这样的:

123456
-123.456000
1234560000000000.000000
0.000000

2)为什么我们需要放[i + 2]和[i+1]？我是随机放的，但我没有对此进行解释。

-

double strToDouble(const char str[]) {
    int i = 0, k = 10;
    double result = 0, doubleValue, expValue = 0, resultDecimal = 0;
    double exp = 1;

    if (str[0] == '-')
    {
        i++;
    }

    while (str[i] >= '0' && str[i] <= '9')
    {
        doubleValue = str[i] - '0';
        result = result * 10 + doubleValue;
        i++;
    }

    if (str[i] == '.')
    {
        while (str[i + 1] >= '0' && str[i + 1] <= '9')
        {
            doubleValue = (str[i + 1] - '0');
            resultDecimal = resultDecimal + doubleValue / k;
            k *= 10;
            i++;
        }
    }


    if (str[i + 1] == 'e')
    {
        if (str[i + 2] == '-')
        {
            while (str[i + 3] >= '0' && str[i + 3] <= '9')
            {
                doubleValue = str[i + 3] - '0';
                expValue = expValue * 10 + doubleValue;
                i++;
            }
            for (k = 0; k < expValue; k++)
            {
                exp /= 10;
            }
        }
        else
        {
            while (str[i + 2] >= '0' && str[i + 2] <= '9')
            {
                doubleValue = str[i + 2] - '0';
                expValue = expValue * 10 + doubleValue;
                i++;
            }
            for (k = 0; k < expValue; k++)
            {
                exp *= 10;
            }
        }
    }

    result = (result + resultDecimal) * exp;

Answer 1

Why we need to put [i + 2] and [i+1]?I put randomly, but i do not have explanation to that.

这些随机偏移会破坏您的代码。它适用于给定的示例，但如果您不指定句点而是指定指数(例如“1e3”)，它就会中断，因为即使没有指定，您也会继承解析句点后尾数的代码的偏移量期间:

if (str[i] == '.') {
    // parse mantissa, using i + 1 as index
}

if (str[i + 1] == 'e')     // i + 1 may be one beyond the character you
                           // should be looking at.

只有当您确实找到一个字符时，才应该在句点之后前进当前字符:

if (str[i] == '.') {
    i++;            // step over period
    // parse mantissa, using i as index
}

if (str[i] == 'e') {
    i++;           // step over 'e'
    // parse exponent, using i as index
}

同样的情况也适用于跨过符号 + (你似乎根本不处理它)和 - (你检测到，但不考虑它)在结果中)。

换句话说:您当前正在检查的字符应始终位于索引 i 处。你可以向前看，但只能在上下文中。当您向前看时，您还应该确保您看到的字符不在可能的空终止符之后。