小数表达式的n个项的和应为浮点数时的整数 [英] The sum of n terms of fractional expression is an integer when it should be a float

问题描述

在下面的代码中，为什么我没有得到有效的结果，除非我输入 term = 1.0/n ，而不是输入 term = 1/n .我已将term声明为float，这还不够吗?

In the following code why don't I get a valid result unless I put term = 1.0/n and not when term = 1/n. I have declared term as float, Shouldn't that be enough?

#include <stdio.h>

int main()
{    
     float sum = 0, term;
     int n, i;
     
     printf("enter the value of n:\n");
     scanf("%d", &n);
     term = 1.0 / n;
     for(i = 1; i <= n; i++)
     {  
        sum = term + sum; 
       
      }
     printf("Sum = %.3f\n", sum);   
     
     return 0; 
}

推荐答案

在下面的代码中，为什么我没有得到有效的结果，除非我输入 term = 1.0/n ，而不是输入 term = 1/n .我已经将 term 声明为 float .那还不够吗?

In the following code why don't I get a valid result unless I put term = 1.0/n and not when term = 1/n. I have declared term as float. Shouldn't that be enough?

不幸的是.

ISO/IEC 9899:2017 §6.5.5 6 states:

除以整数时，/运算符的结果为代数商，其中舍弃了任何小数部分. ¹⁰⁵⁾如果商 a/b 是可表示的，表达式(a/b)* b + a％b 应该等于 a [...]

When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.¹⁰⁵⁾ If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a [...]

¹⁰⁵⁾这通常被称为向零截断".

¹⁰⁵⁾ This is often called "truncation toward zero".

翻译:两个整数相除的结果是一个整数.

Translation: the result of this division of two integers is an integer.

即使将其分配给 float 或 double 变量也不够，它在分配之前将被截断，解决方案是将像您一样，在浮点数或双精度数中的分数表达式.

Even if you assign it to a float or double variable it's not enough, it will be truncated before it's assigned, the solution is turn one of the operands in the fractional expression in a float or a double, like you did.

这篇关于小数表达式的n个项的和应为浮点数时的整数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！