Question

所以我无法弄清楚为什么会这样。该程序要求用户输入一个整数，并根据数字是偶数还是奇数分别将数字加倍或三倍。问题是，对于只有偶数整数循环，它为每个循环迭代减去第一个数字1。奇数整数的循环与偶数整数完全相同，除了数字三倍而不是加倍，它工作得很好。有什么想法吗？

前：

输入：12
预期产量：1122
实际产量：1121

输入：13
预期产量：111333
实际产量：111333

//This program asks the user to enter an integer and depending on whether the integer is even or odd, doubles or triples
//each digit in the integer respectively. It then reads out the result and asks the user if they would like to enter
//another integer and run the program again.
int main()
{
    string restart;
    int integer, remainder;

    while (restart != "n" && restart !="N")
    {
        cout << "Enter an integer: ";
        cin >> integer;
        cout << endl;

        //Creates variable "temp" and "mycount" and divides "integer" by 10 until remainder is 0 and counts number of
        //steps to do this to count the number of significant digits of the integer
        int temp = integer, mycount = 0;
        while (temp != 0)
        {
            temp = temp / 10;
            mycount++;
        }

        //Computes if the integer is even or odd by determining the remainder: 0 remainder = even, Non-0 remainder = odd
        //Creates two integer variables "exponent" and "sum" and sets them to 0
        //Assigns variable "temp" to the integer value
        remainder = integer % 2;
        int exponent = 0;
        int sum = 0;
        temp = integer;

        //If integer is even, doubles each significant digit
        if (remainder == 0)
        {
            //Begins for loop which runs for the number of times equal to the number of significant digits stored in "mycount"
            for (int i = mycount; i > 0; i--)
            {
                //Stores current significant digit in "digit"
                //Removes current significant digit by dividing by 10
                //Multiplies current significant digit by 11 and stores it in "timesTwo"
                //Multiplies current significant digit by 10^exponent then adds it to other modified digits
                //Adds 2 to current exponent to increase digit multiplier by 100
                int digit = temp % 10;
                temp = temp / 10;
                int timesTwo = digit * 11;
                sum = (timesTwo * pow(10, exponent)) + sum;
                exponent = exponent + 2;
                cout << sum << endl;
                cout << endl;
            }

            cout << "Number is even, doubling each digit in the integer ..." << endl;
            cout << sum << endl;
        }

        //If integer is odd this runs the same as the above function except it triples the digit and adds 3 to the multiplier
        else
        {
            for (int i = mycount; i > 0; i--)
            {
                int digit = temp % 10;
                temp = temp / 10;
                int timesThree = digit * 111;
                sum = (timesThree * pow(10, exponent)) + sum;
                exponent = exponent + 3;
                cout << sum << endl;
            }
            cout << "Number is odd, tripling each digit in the integer ..." << endl;
            cout << "Result: " << sum << endl;
        }

        cout << "Would you like to enter another integer? (y/n): ";
        cin >> restart;
    }

    return 0;
}

Answer 1

为了调试这个，我打印第一个循环中的所有输入，而不仅仅是输出。我能看到的唯一可疑的事情是pow()返回一个浮点数，这可能会在转换中向下舍入到整数。

为什么不完全避免pow()和隐式转换，而是使用每轮乘以的整数因子呢？

sum = (timesTwo * factor) + sum;
factor += 100;

顺便说一下：对于两种情况都没有必要计算数字和不同的循环 - 你可以将程序的核心简化为类似

bool even = (integer & 1) == 0;
int digitFactor = even ? 11 : 111;
int stepFactor = even ? 100 : 1000;
int result = 0;
while(integer != 0) {
  int digit = integer % 10;
  result += digit * digitFactor;
  integer /= 10;
  digitFactor *= stepFactor;
}

程序仅从偶数值输出中减去1

1 个答案: