C - 为什么答案为0.00000,当我尝试输入3个数字作为总和和平均值？

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程序一次从上到下执行一条指令.您已计算sum并average在接受数字之前.这评估为sum=0并且average=0因为所有三个数字都是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从上到下运行.

您创建了num1,num2和num3,每个都为0,并在输入数字之前找到它们的总和和平均值.

做如下:

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;

没有定义的方式sum读取完成后,将被计算,但实际上使用这些值和计算sum,并average为0程序调用之前先scanf.

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