显示所有数据链表的单独总和

Question

I am writing a data structure program using c++ . This is my code to display sum of quiz I and II of all students. But it is only display the sum of the first student.

int totalscore(){
    double sum;
    node *pcurrent;
    pcurrent=pfirst;
    while(pcurrent!=NULL)
    {
        sum= (pcurrent->quizI + pcurrent->quizII);


        pcurrent =pcurrent->pnext++;//move  
    }

    cout<<"The  individual sum are " <<sum ;
};

int main() {
    linkedlist apt;
    apt.insert("643431 ", 50, 40);
    apt.insert( "655222",100, 20);
    apt.insert( "655444",70, 30 );
    apt.insert( "651515",10, 30);
    apt.insert(" 644444", 70, 32);
    apt.displaylist();
    apt.totalscore();


    return 0;
}

Thank for helping

Answer 1

You're not moving to the next node correctly.

Change this line

pcurrent = pcurrent->pnext++;

to

pcurrent = pcurrent->pnext;

And, for the individual sum of each student in the linked list separately, you have to calculate the sum and print it in the loop for all the nodes in the loop like this:

void totalscore() // no return value
{
    node *pcurrent = pfirst;
    while ( pcurrent != NULL )
    {
        double sum = (pcurrent->quizI + pcurrent->quizII);
        cout << "Sum: " << sum << '\n';

        pcurrent = pcurrent->pnext;
    }
}

The method name totalscore() with the return type implies that you want to return the sum of all the students' score in the likned list eg:

double totalscore() // sum is double type so return the same
{
    double sum = 0.0;

    node *pcurrent = pfirst;
    while ( pcurrent != NULL )
    {
        sum += (pcurrent->quizI + pcurrent->quizII);
        pcurrent = pcurrent->pnext;
    }

    cout << "Sum " << sum << '\n';
    return sum; // return the sum when calculated
}

If you want to return the individual sum of all the students from the function separately:

1. you can calculate the sum and store it in a container like std::vector ; or,
2. add another element sum in the node itself and calculate the sum as you store data and use that whenever you need the sum.