使用字符串操作截断整数？

Question

我有一个类的数据成员需要四舍五入为 2 位整数，而与输入位数无关。

例如：

roundUpto2digit(12356463) == 12 
roundUpto2digit(12547984) == 13 // The 5 rounds the 12 up to 13.

目前我的代码如下：

int roundUpto2digit(int cents){
    // convert cents to string
    string truncatedValue = to_string(cents);
    // take first two elements corresponding to the Most Sign. Bits
    // convert char to int, by -'0', multiply the first by 10 and sum the second 
    int totalsum =  int(truncatedValue[0]-'0')*10 + int(truncatedValue[1]-'0');
    // if the third element greater the five, increment the sum by one
    if (truncatedValue[2]>=5) totalsum++; 
    return totalsum;
} 

任何使它不那么难看的建议将不胜感激。

Answer 1

您可以使用定点整数算术，这可能更快并且看起来更好。你想要 10^2 的数字，并且你也有 10 的任意比例幂，所以要四舍五入你只需要应用公式：

ROUNDED_VAL = (INITIAL_VAL + (10^(ORIG_SCALE - 2) / 2)) / 10^(ORIG_SCALE - 2)

所以你的代码可能看起来像这样：

int roundUpto2digit(int cents){
    int scale = 10;
    while(cents / scale > 0) {
        // Find out original scale. This can be done maybe faster with a divide and conquer algorithm 
        scale *= 10;
    }
    int applied_scale = scale / 100;
    if (applied_scale == 0) {
        // In case there is only one digit, scale it up to two
        return 10 * cents;
    }
    return ((cents + (applied_scale / 2)) / applied_scale);
} 

编辑：我写的10 * cents 行是我根据我的解释提出的问题的任意推断。如果这不是所需的行为，当然可以更改。

Answer 2

#include <math.h>

int roundUpto2digit(int value)
{
    value /= pow(10,(int)log10(value)-2);
    /*
    (int)log10(value) returns base ten logarithm (number of digits of your number)
    pow(10, N) returns number 1 followed by N zeros
    example:
    pow(10,(int)log10(123)-2) returns 1
    pow(10,(int)log10(1234)-2) returns 10
    pow(10,(int)log10(1234567)-2) returns 10000
    thus
    value / pow(10,(int)log10(value)-2) returns first 3 digits
    */
    return value%10>=5? (value+10)/10 : value/10;
    /*
    if(value%10>=5)// the last digit is >= 5
    {
    // increase previous number
    value = value + 10;
    }
    // return number without the last digit
    return value/10;
    */
}