Question

我正在尝试制作一个计算几何平均值的程序。但是我得到了这个错误消息：左值作为赋值的左操作数

 #include<stdio.h>
 #include<conio.h>

int main() {
    float number1,number2,avarage;
    printf("enter the numbers which you want to calculate geometrical average of them > ");
    scanf("%f %f",&number1,&number2);

    average*average=number1*number1+number2*number2;

    printf("Geometrical avarage is:%f",sayi);

    getch();

    return 0;
}

（我尝试了＃34; ==＆＃34;也是运营商，但在这段时间分配给0.00平均值）

由于

Answer 1

这是一个数学问题，而不是编程问题。如果你知道

average*average = number1*number1+number2*number2

你必须首先评估平均值。解决方案是

average = sqrt(number1*number1+number2*number2)

EDIT1：但几何平均值应计算为

average = sqrt(number1 * number2)

根据http://en.wikipedia.org/wiki/Geometric_mean

Answer 2

使用此公式计算几何平均值：

Geometric average = square root of (number1 * number2)

#include<stdio.h>
#include<math.h>

int main()
{

    float number1,number2,average;

    printf("enter the numbers which you want to calculate geometrical average of them > ");
    scanf("%f %f",&number1,&number2);

    average=sqrt(number1*number2);

    printf("Geometrical average is:%f",average);

    return 0;
}

左值作为赋值的左操作数

2 个答案: