即使退出条件满足并进入，错误消息仍会打印

Question

I'm having quite the issue trying to figure out why these code segments are printing the error message even when I have done cout statements within the block that should return true / not print that error message. Any ideas? I'm new here so please let me know if this isn't allowed. Thanks!

Use of function:

case 'a':
        {
            // Format: a ID credits GPA
            // Adds a student with the given student ID (ID), number of 
            // credits (credits), and overall GPA (GPA) to the database. 
            // If the student is already in the database, an error 
            // message should be printed indicating this.

            int credits = 0;
            double gpa = 0;
            cin >> studentID;
            cin >> credits;  
            cin >> gpa;

            // Adds the student and checks to see if the student was actually added
            // or if there was an existing student with the specified ID

            bool added = addStudent(studentID, credits, gpa);
            if(added == false);
            {
                cout << "Student already exists in database, nothing changed." << endl;
                // Still prints this when executed with valid 
            }
            break;
        }

Function for adding student to array:

bool addStudent (int id, int numCredits, double gpa) {

// Check to see if student exists

if (nextEntry != 0)
{
    for (int x = 0; x < 7000; x++) {
        Student tmp = studentRecords[x];
        if (tmp.studentId == id)
        {
            return false;
            cout << "hey" << endl;
        }
    }
}

// If student does not exist, add to records database
if (nextEntry != 7000)
{
    studentRecords[nextEntry].studentId = id;
    studentRecords[nextEntry].numCredits = numCredits;
    studentRecords[nextEntry].gpa = gpa;
    nextEntry++;
    return true;
    // confirmed I can get here
}

return false;
}

Answer 1

You have an extra ; after your if (added==false) statement. This will terminate the if statement and the code after will run regardless of the check.