Why it keeps giving me a garbage value

Question

1.00/5 (1 vote)

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I want to pass the struct by ref

What I have tried:

#include<iostream>
#include<string>
#include<conio.h>	
using namespace std;

struct Bankinfo{
	string name,location,manager_name;
	int bank_id,emp_numbers;
};

Bankinfo getBankinfo(){

	Bankinfo n;
	cout<<"enter name,location,manager name,bank id,employees number"<<endl;
	getline(cin,n.name);
	getline(cin,n.location);
	getline(cin,n.manager_name);
	cin>>n.bank_id;
	cin>>n.emp_numbers;
	return n;
}


void displayBankinfo(Bankinfo *){
	Bankinfo n;
	cout<<n.name<<endl;
	cout<<n.location<<endl;
	cout<<n.manager_name<<endl;
	cout<<n.bank_id<<endl;
	cout<<n.emp_numbers<<endl;
}



int main(){

Bankinfo n;



getBankinfo();
displayBankinfo(&n);
	return 0;

}

Posted 5-Nov-21 11:15am

Member 15420232

Updated 6-Nov-21 3:41am

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merano99 · Accepted Answer · 2021-11-06T03:41:00

1. In C ++, parameters can be passed to a function either by pointers or by references.
Here it was subsequently requested to implement the parameter transfer via pointer.
In C, Pass-by-reference is simulated by passing the address of a variable.
This is referred to as "C style pass-by-reference. (Source: www-cs-students.stanford.edu)
2. The questioner wants a solution with a structure instead of a class.

Unfortunately, the result is not good C ++ code, but rather a mixture of C ++ and old C code.

So here is the result:

C++

typedef struct {
	string name, location, manager_name;
	int bank_id, emp_numbers;
} Bankinfo;

void getBankinfo(Bankinfo* b)
{
	cout << "Name: "; 
	getline(cin, b->name);
	cout << "Location: ";
	getline(cin, b->location); 
	cout << "Manager-Name: ";
	getline(cin, b->manager_name);
	cout << "Bank ID: ";
	cin >> b->bank_id;
	cout << "Employees-Number: ";
	cin >> b->emp_numbers;
}

void displayBankinfo(const Bankinfo *b) {
	cout << b->name << endl;
	cout << b->location << endl;
	cout << b->manager_name << endl;
	cout << b->bank_id << endl;
	cout << b->emp_numbers << endl;
}

int main() 
{
	Bankinfo b;
	getBankinfo(&b);
    cout << "-------------------------------\n";
	displayBankinfo(&b);
	return 0;
}

merano99 · Accepted Answer · 2021-11-05T12:18:00

Solution 2

At first glance I see several problems. On the one hand, no reference is passed to the subroutine. It should look like this:

C++

void displayBankinfo(const Bankinfo &n) {
	// Bankinfo n;
	cout << n.name << endl;
	cout << n.location << endl;
	cout << n.manager_name << endl;
	cout << n.bank_id << endl;
	cout << n.emp_numbers << endl;
}

Since the values are only output and should not change, I would declare them as const.

The second point is that the structure is created in the first subprogram, but then not assigned in the main program.

C++

int main() 
{
	Bankinfo n;

	n = getBankinfo();  // assign here!
	displayBankinfo(n);
	return 0;
}

Remarks:
1. I would already use the reference from the main program when reading in the values.
2. Is there a good reason to use a structure here instead of a class?
3. Reading in the values seems very error-prone and is also not secured.

Posted 5-Nov-21 12:18pm

merano99

Comments

Member 15420232 5-Nov-21 18:38pm

1-me too
2-the question wants it with structure
3-also wants the function void displayBankinfo to take (*) as a parameter
is there other solution to display the output using (*) as a parameter?

phil.o 5-Nov-21 20:24pm

Yes, you have to use the Member Access Operator ->[^]
When you work with a pointer, you cannot use . as member accessor; you have to use -> operator:

struct thing { int value; };
thing* pointerToThing = new thing;
pointerToThing.value = 0; // Will not compile. Wrong operator.
pointerToThing->value = 0; // OK

Member 15420232 5-Nov-21 20:35pm

you mean like
void displayBankinfo(Bankinfo *n){

cout<<n->name<<endl;
cout<<n-="">location<<endl;
cout<<n-="">manager_name<<endl;
cout<<n-="">bank_id<<endl;
cout<<n-="">emp_numbers<

merano99 5-Nov-21 21:21pm

If you had said from the beginning what exactly you are allowed to and what not you could have given better answers immediately.

When you have to use * as reference you have to write:

void displayBankinfo(Bankinfo *n) {
cout << n->name << endl;
...

Member 15420232 5-Nov-21 22:14pm

I tried that but it still gives me a garbage value ):

merano99 6-Nov-21 9:05am

It definitely works.
Did you assign the return value in main as I suggested?

n = getBankinfo();
displayBankinfo(&n);

Member 15420232 6-Nov-21 9:14am

it worked!!

I forgot to assign it my bad
thanks:)

phil.o · Accepted Answer · 2021-11-05T11:23:00

Solution 1

Because you do not use the parameter of the displayBankinfo function; you did not even give it a name. Instead, you create a brand new Bankinfo instance (which contains garbage since it is unitialized), and display its values. You may be searching for something like

C++

void displayBankinfo(Bankinfo *n){
	// display n's values here. no need to create a new instance, just use the parameter
}

Posted 5-Nov-21 11:23am

phil.o

v2

Comments

Member 15420232 5-Nov-21 18:43pm

did not work

phil.o 5-Nov-21 20:01pm

not specific enough as an issue description...

Member 15420232 5-Nov-21 20:40pm

void displayBankinfo(const Bankinfo *n){

cout<<n->name<<endl;
cout<<n-="">location<<endl;
cout<<n-="">manager_name<<endl;
cout<<n-="">bank_id<<endl;
cout<<n-="">emp_numbers<

phil.o 5-Nov-21 20:52pm

This is still not an issue description.
You should also write

Bankinfo n = getBankinfo();

in your main method. Right now you create the variable but you do not affect it the result returned by the getBankInfo() method.

Member 15420232 6-Nov-21 9:24am

I finally did it

all what I had to do is:
1- void displayBankinfo(BANKinfo *n)
2-use the -> operator
3- assign n== getBankinfo()

thank you for helping me
:)