c primer plus 第 6 章 ex12

Question

每个人，c primer 加上第 6 章 ex12:

**Consider these two infinite series:

1.0 + 1.0/2.0 + 1.0/3.0 + 1.0/4.0 + ...

1.0 - 1.0/2.0 + 1.0/3.0 - 1.0/4.0 + ...

Write a program that evaluates running totals of these two series up to some limit of number of terms. Hint: –1 times itself an odd number of times is –1, and –1 times itself an even number of times is 1. Have the user enter the limit interactively; let a zero or negative value terminate input. Look at the running totals after 100 terms, 1000 terms, 10,000 terms. Does either series appear to be converging to some value?**

#include <stdio.h>
int main(void)
{
    int times, a, b, d;
    float sum1, sum2, c;
    printf("Enter the times: ");
    scanf("%d", &times);
    while (times > 0)
    {
        sum1 = 0;
        sum2 = 0;

        for (a = times; a >= 1; a--)
            sum1 += 1.0 / (float)a;
        printf("The sum1 is %f\n", sum1);


        for (b = times; b >= 1; b--)
        {
            c = -1.0;
            while ((d = b) % 2 == 1)
            { 
                c = 1.0;
                d++;
            }
            sum2 += (c / (float)b);
        }
        printf("The sum2 is %f\n", sum2);

        printf("Enter the times again: ");
        scanf("%d", &times);
    }
    return 0;
}

我的代码有什么问题？

Answer 1

这里:

while ((d = b) % 2 == 1)
{ 
  c = 1.0;
  d++;
}

您将 b 的值赋给 d(通过 d = b)，然后，您检查此值对 2 的模数是否相等到 1。如果是这种情况，您将永远进入循环，因为 b 的值永远不会改变。当然，您在循环内递增 d，但在您的检查中，您将其值重置为 b，从而导致无限循环。

关于你的练习，如果 b 是偶数，你试图将 c 设置为 -1，如果 b 是奇数，则设置为 1。这可以通过条件赋值轻松完成:

c = (b % 2 == 0) ? -1.0 : 1.0;

或者，正如您问题中的提示所暗示的，您可以在启动循环之前将 c 初始化为 1(或 -1)并在里面做 c = -1.0 * c