C-为什么当我尝试输入3个数字作为总和和平均值时，答案为0.00000？

Question

#include <stdio.h>

int main(int argc, char *argv[]) 
{

float num1 = 0;
float num2 = 0;
float num3 = 0;
float sum = num1 + num2 + num3;
float average = sum / 3;


printf("Enter three numbers:\n"); //Enter three floats
scanf("%f %f %f", &num1, &num2, &num3); //Assign to num1, num2, num3 

printf("The sum of these three numbers is %f.\n", sum);//Print sum
printf("The average of these numbers is %f.\n", average);//Print average

}

这就是显示的内容。

Enter three numbers:

12.0

10.0

12.0

The sum of these three numbers is 0.000000.

The average of these numbers is 0.000000.

Answer 1

C程序一次从上至下执行一条指令。 在接受数字之前，您已经计算了sum和average 。 由于所有三个数字均为0，因此sum=0为sum=0和average=0 0。

main(int argc, char *argv[]) {
    float num1,num2,num3,sum,average;

    printf("Enter three numbers:\n"); //Enter three floats
    scanf("%f %f %f", &num1, &num2, &num3); //Assign to num1, num2, num3 

    sum = num1 + num2 + num3;
    average = sum / 3;

    printf("The sum of these three numbers is %f.\n", sum);//Print sum
    printf("The average of these numbers is %f.\n", average);//Print average
    return 0;
}

Answer 2

请记住，C在没有循环或条件的函数中自上而下运行。

您创建了num1，num2和num3，分别为0，并在输入数字之前找到了它们的总和和平均值。

进行如下操作：

float num1 = 0;
float num2 = 0;
float num3 = 0;
float sum = 0;
float average = 0;   

printf("Enter three numbers:\n"); //Enter three floats
scanf("%f %f %f", &num1, &num2, &num3); //Assign to num1, num2, num3 

sum = num1 + num2 + num3; //calculate
average = sum / 3;

printf("The sum of these three numbers is %f.\n", sum);//Print sum

printf("The average of these numbers is %f.\n", average);//Print average

Answer 3

您似乎误解了变量定义的用法。 这些：

float num1 = 0;
float num2 = 0;
float num3 = 0;
float sum = num1 + num2 + num3;
float average = sum / 3;

在读取完成后，并没有定义sum计算方式，而是实际使用这些值并在程序调用first scanf之前将sum和average计算为0 。

scanf("%f %f %f", &num1, &num2, &num3);
sum = num1 + num2 + num3;                     // <-- place it here
average = sum / 3;