C中的无限循环

发布于 2025-01-02 05:12:48 字数 116 浏览 0 评论 0原文

初学者在这里。为什么这是一个无限循环？

for (p = 0; p < 5; p += 0.5)
{
    printf("p=%2.2f\n",p);
}

原文

Beginner here.
Why is this an endless loop ?

for (p = 0; p < 5; p += 0.5)
{
    printf("p=%2.2f\n",p);
}

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昵称有卵用 2025-01-09 05:12:48

您会看到无限循环，因为您的 p 是整数类型（例如 int）。无论将 0.5 添加到 int 多少次，它都将保持为 0，因为 int 会截断分配给它的 double/fp 值。换句话说，它相当于在每个步骤上添加零的循环。

如果您将 p 设为 float 或 double，您的问题就会消失。

编辑（由 Oli Charlesworth 的评论建议）

值得注意的是，不鼓励使用浮点数和双精度数来控制循环，因为结果并不总是像示例中那样干净。将步长从 0.5（2 的负 1 次方）更改为 0.1（不是 2 的负整数次方）会改变您看到的结果以一种相当意想不到的方式。

如果您需要按非整数步骤进行迭代，您应该考虑使用这个简单的模式：

// Loop is controlled by an integer counter
for (int i = 0 ; i != 10 ; i++) {
    // FP value is calculated by multiplying the counter by the intended step:
    double p = i * 0.5;
    // p is between 0 and 4.5, inclusive
}

You see an endless loop because your p is of an integral type (e.g. an int). No matter how many times you add 0.5 to an int, it would remain 0, because int truncates double/fp values assigned to it. In other words, it is equivalent to a loop where you add zero on each step.

If you make p a float or a double, your problem would go away.

EDIT (Suggested by Oli Charlesworth's comment)

It is worth noting that using floats and doubles to control loops is discouraged, because the results are not always as clean as in your example. Changing the step from 0.5 (which is 2 to the negative power of 1) to 0.1 (which is not an integral negative power of 2) would change the results that you see in a rather unexpected way.

If you need to iterate by a non-integer step, you should consider using this simple pattern:

// Loop is controlled by an integer counter
for (int i = 0 ; i != 10 ; i++) {
    // FP value is calculated by multiplying the counter by the intended step:
    double p = i * 0.5;
    // p is between 0 and 4.5, inclusive
}

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