Question

can someone help explain how this code works

Answer 1

#include<iostream>
using namespace std;
const int NUMVALUES=10; // this is the constant value
void unweighted(double, double, double, double);
double weighted(double, double, double, double);
int main()
{
double sum=0;
int i, j;
double number=0;
double s1,s2,s3,s4;
double t1=0, t2=0, t3=0, t4=0;
for(i=1;i<NUMVALUES;i++)
{
cout<<"enter the value of the test scores: "<<i<<endl;
cin>>s1;
t1+=s1;

cin>>s2;
t2+=s2;

cin>>s3;
t3+=s3;

cin>>s4;
t4+=s4;
cout<<"student "<<i<<" your weighted average is: "<<weighted(s1, s2, s3, s4)<<endl;//prints out the average score for the test scores
}
cout<<"the unweighted average of the test scores are"<<unweighted(t1, t2, t3, t4)<<endl;// prints the unweighted average for each student
return 0;
}
double weighted(double t1, double t2, double t3, double t4)//this function takes four doubles that are used used to calculate return the weighted average for an individual student
{
double weighted=0;
double avgT1, avgT2, avgT3, avgT4;
avgT1=t1*.10;
avgT2=t2*.25;
avgT3=t3*.30;
avgT4=t4*.35;
weighted=(avgT1+avgT2+avgT3+avgT4);
return weighted;
}
void unweighted(double sumT1, double sumT2, double sumT3, double sumT4)//this function takes in the sum of each test then calculates the average of each teset and prints them out
{
cout<<"the class average is: "<<endl;
cout<<"For test 1 is: "<<sumT1/NUMVALUES<<endl;
cout<<"For test 2 is: "<<sumT2/NUMVALUES<<endl;
cout<<"For test 3 is: "<<sumT3/NUMVALUES<<endl;
cout<<"For test 4 is: "<<sumT4/NUMVALUES<<endl;
cout<<endl;
}

Answer 2

Which parts do you have problems understanding?

Answer 3

@Recursing
The for loop and the 2 functions

Answer 4

The for loop looks to loop through 10 students asking for scores from four tests that they've taken.

The functions do what the comments say they do.

This code isn't technically difficult, it's just written poorly. The variable name stink. Commenting would help this code too. Take a copy of this code and do some global edits to make it more readable to you, and what it does will become apparent.