Thread: Function problem - too few args

Notice:

Code:

    if(choice=1){
         inputDB();
    }

Clearly you call inputDB() with no arguments, but you declared it to take three arguments.

Incidentally, choice=1 in that context is probably a mistake.

Ah yup, ive changed it a bit now:

Code:

#include <cstdlib>
#include <iostream>

using namespace std;

struct DB {
       int age;
       char name[20];
       char location[20];
};

void inputDB(int a, char n[20], char l[20]);
void viewDB(DB dbstats);

int main(int argc, char *argv[])
{
    int age;
    char name[20];
    char location[20];
    
    int choice;
    cout << "Welcome\n";
    cout << "1. Input Data\n";
    cout << "2. View Data\n";
    cout << "3. Quit\n";
    cin >> choice;
    
    if(choice==1){
         inputDB(age, name, location);
    }
    else if(choice==2){
         viewDB(age, name, location);
    }   
    else if(choice=3){
         return EXIT_SUCCESS;
    }
    else{
         cout << "Not an option\n";
    }      
    
    
    system("PAUSE");
    return EXIT_SUCCESS;
}

void viewDB(DB dbstats)
{
     cout << "Age: " << dbstats.age << endl;
     cout << "Name: " << dbstats.name << endl;
     cout << "Location: " << dbstats.location << endl;
}
void inputDB(int a, char n[20], char l[20])
{
     cout << "Enter Age: ";
     cin >> a;
     cout << "\nEnter Name: ";
     cin >> n;
     cout << "\nEnter Location: ";
     cin >> l;
     DB input = {a, n[20], l[20]};
}

but i get this error: `int' to non-scalar type `DB' requested

but i get this error: `int' to non-scalar type `DB' requested

Compare:

Code:

void viewDB(DB dbstats);

with its attempted use:

Code:

viewDB(age, name, location);

By the way, this line accesses two arrays out of their bounds:

Code:

DB input = {a, n[20], l[20]};

Ok heres what i got now, but this bit has me really stumped :S

Error: too few arguments to function `void viewDB(DB)'

Code:

#include <cstdlib>
#include <iostream>

using namespace std;

struct DB {
       int age;
       char name[20];
       char location[20];
};

void inputDB(int a, char n[20], char l[20]);
void viewDB(DB dbstats);

int main(int argc, char *argv[])
{
    int age;
    char name[20];
    char location[20];
    
    int choice;
    cout << "Welcome\n";
    cout << "1. Input Data\n";
    cout << "2. View Data\n";
    cout << "3. Quit\n";
    cin >> choice;
    
    if(choice==1){
         inputDB(age, name, location);
    }
    else if(choice==2){
         viewDB();
    }   
    else if(choice==3){
         return EXIT_SUCCESS;
    }
    else{
         cout << "Not an option\n";
    }      
    
    
    system("PAUSE");
    return EXIT_SUCCESS;
}

void viewDB(DB dbstats)
{
     cout << "Age: " << dbstats.age << endl;
     cout << "Name: " << dbstats.name << endl;
     cout << "Location: " << dbstats.location << endl;
}
void inputDB(int a, char n[20], char l[20])
{
     cout << "Enter Age: ";
     cin >> a;
     cout << "\nEnter Name: ";
     cin >> n;
     cout << "\nEnter Location: ";
     cin >> l;
     DB dbstats = {a, n[20], l[20]};
}

Ok heres what i got now, but this bit has me really stumped :S

Now compare:

Code:

void viewDB(DB dbstats);

with its attempted use:

Code:

viewDB();

Error: too few arguments to function `void viewDB(DB)'

Do you understand the error message?

Yes sorry for this, im overlooking some pretty simple errors, I'm really bad at coding on my own, cause usually if im with someone else they'll see something thats really obvious that i cant

Code:

#include <cstdlib>
#include <iostream>

using namespace std;

struct DB {
       int age;
       char name[20];
       char location[20];
};

void inputDB(int a, char n[20], char l[20]);
void viewDB(DB dbstats);

int main(int argc, char *argv[])
{
    int age;
    char name[20];
    char location[20];
    
    int choice;
    cout << "Welcome\n";
    cout << "1. Input Data\n";
    cout << "2. View Data\n";
    cout << "3. Quit\n";
    cin >> choice;
    
    if(choice==1){
         inputDB(age, name, location);
    }
    else if(choice==2){
         viewDB(DB dbstats);
    }   
    else if(choice==3){
         return EXIT_SUCCESS;
    }
    else{
         cout << "Not an option\n";
    }      
    
    
    system("PAUSE");
    return EXIT_SUCCESS;
}

void viewDB(DB dbstats)
{
     cout << "Age: " << dbstats.age << endl;
     cout << "Name: " << dbstats.name << endl;
     cout << "Location: " << dbstats.location << endl;
}
void inputDB(int a, char n[20], char l[20])
{
     cout << "Enter Age: ";
     cin >> a;
     cout << "\nEnter Name: ";
     cin >> n;
     cout << "\nEnter Location: ";
     cin >> l;
     DB dbstats = {a, n[20], l[20]};
}

Thats the changes, cleared up the previous error but i get:

Code:

Line 32: expected primary-expression before "dbstats"

for where i declare this:

Code:

void viewDB(DB dbstats);

Yes sorry for this, im overlooking some pretty simple errors

The error message you quoted left out the number of the line at which the error was detected. Use that line number to help you figure out the error.

Line 32 is not where you declare the prototype. It's where you are calling the function.