为什么使用pow函数时输入的平方减少1? [英] Why do the square of input decreases by one when using pow function?

问题描述

这是一个简单的C程序，它接受用户输入的数字并得出平方.

Here is a simple C program, which accepts a number from the user and results it's square.

#include <stdio.h>
#include <math.h>
int main()
{
    int number;
    int result;
    printf("\nEnter the number\n");
    scanf("%d",&number);
    result=(pow(number,2));
    printf("\nThe result is %d\n",result);
    return 0;
}

问题是，每当我输入5,25,26等作为输入时，输出为24,624,675，即它减少1并且并非所有数字都这样.我正在使用CodeBlocks IDE.我想出了一个解决此问题的方法，但我想知道幕后正在发生什么，这是导致此错误的原因.

The problem is, whenever i enter 5,25,26 etc as input, the output is 24,624,675 i.e. it decreases by 1 and this does not happen with all numbers. I am using CodeBlocks IDE. I figured out a fix for this problem but I want to know what is happening behind the scene, which is causing this error.

推荐答案

来自这篇文章，苏拉夫告诉我们:

From this post, Sourav tells us:

pow()接受双精度参数并返回双精度值.

pow() takes double arguments and returns a double.

如果将返回值存储在int中并打印出来，则可能无法获得所需的结果.

If you store the return value into an int and print that, you may not get the desired result.

关于 为什么 的解释，pow()返回的值比您期望的少1，请参见

As for an explanation on why pow() is returning 1 less that you expect, see this post. Specifically:

pow()使用双数.这些表示形式为s * 2^e的数字​​，其中s是53位整数.因此，double可以存储2^53以下的所有整数，但只能存储2^53以上的一些整数.特别是，它只能表示偶数> 2^53，因为对于e > 0，该值始终是2的倍数.

pow() works with double numbers. These represent numbers of the form s * 2^e where s is a 53 bit integer. Therefore double can store all integers below 2^53, but only some integers above 2^53. In particular, it can only represent even numbers > 2^53, since for e > 0 the value is always a multiple of 2.

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