使用数组计算 5 个整数的平均值

Question

I can't seem to get it right. The question is "Calculate the Average of 5 Integers using an array"

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int avg[5],i,total;
    int average;

    printf("Enter the marks entered in 5 subjects");
    for (i=0; i<5; ++i){
        scanf("%d",&avg[i]);
    }
    for(i=0; i<5; ++i){
        total = total + avg[i];
    }

    average= (float)total/5;
    printf("The average of 5 marks is %d",average);


    return 0;
}

Answer 1

1) Your answer CAN be a decimal number but you are storing it in an integer which ignores the decimal points. The variable average should be declared as float average; The line where you print the result should be changed to printf("The average of 5 marks is %f",average);

2) Initialize the variable total as int total = 0;

Answer 2

total应该初始化为0 ：

int avg[5],i,total=0;

Answer 3

In the for loop's first iteration, variable total is used uninitialized, so the result is wrong since it will automatically acquire a garbage value if not initialized explicitly. Take this whole thing:

#include <stdio.h>

int main() {
    int avg[5], i, total = 0, average;
    printf("Enter the marks entered in 5 subjects");
    for (i = 0; i < 5; ++i) {
        scanf("%d", &avg[i]);
        total += avg[i];
    }
    average = total / 5;
    printf("The average of 5 marks is %d", average);
    return 0;
}

Answer 4

You've just declared the variable total but didn't initialize it. So, total contains a garbage value. For that if you add anything to total, the value will not be the correct as expected. It will be added to the garbage value. So,initialize the variable total with 0.

int total=0;

You need not to use type casting. Just declare the average variable as double.

double average=0;
average = total/5.0;

And you should print as printf("The average of 5 marks is %lf", average);

Answer 5

total is a local variable, hence it needs to be initialised. Had it been within file scope (like, a global variable), or a static variable, you might not have to initialise it to 0 .

An initialisation could be simply like:

int avg[5], i, total = 0 ;

or

int avg[5], i, total;
total = 0 ;

Why do you need initialisation? Because of this statement:

 total = total + avg[i];

Here total is calculated using its previous value. What is total 's previous value the first time this statement is encountered? It could be anything, commonly referred to as garbage value, invoking undefined behaviour. Hence, you need initialisation to give this starting value to total . Note, you don't need to initialise average , because its value does not depend on its previous contents.

---

Another problem is with the concepts of typecasting . Here is the statement:

average= (float)total/5;

You are right about typecasting total to float (you may also have done total/5.0 instead). However, you are storing the result in an integer . This will result in a second typecasting, from the result in float to int .

Hence, you need to declare average as a float .

( Note : If having a float result is not your requirement, and you really need an integral answer, you may ignore this part).

Answer 6

You need to initialize variables before you use them in your code.

What goes WRONG in your code is,

int avg[5],i,total;

where you have not initialized the int variable "total" Hence this code will use a garbage value

for(i=0; i<5; ++i){
    total = total + avg[i];
}

You need this correction

int avg[5], i, total=0;