How to pass a structure variable within functions - c

Question

I created a simple C program to display about a simple bank transaction on withdrawals and deposits. I created the variables using structure. But there is a problem in passing structure variables within functions as they are not passing the values from one function to another as well as to main function.
#include <stdio.h>
#include <stdlib.h>
typedef struct Details{
char name[100];
char day[100];
int openingbalance;
int amount;
int closingbalance;
}detail;
void InputDetails(detail customer){
printf("Enter Customer name: ");
scanf("%s",customer.name);
printf("Enter the date: ");
scanf("%s",customer.day);
printf("Enter opening balance: ");
scanf("%d",&customer.openingbalance);
}
void transaction(detail customerr, int money){
int selection;
printf("Opening balance = %d\n",customerr.openingbalance);
printf("Select the transaction\n1.Withdrawal\n2.Deposit\n");
scanf("%d",&selection);
switch(selection){
case 1:
if(money > customerr.openingbalance){
printf("No sufficient balance in the account to process the request\n");
customerr.closingbalance = customerr.openingbalance;
}
else
customerr.closingbalance = customerr.openingbalance - money;
break;
case 2:
customerr.closingbalance = customerr.openingbalance + money;
}
}
void main()
{
detail client;
int amount;
InputDetails(client);
printf("Enter the transaction amount: ");
scanf("%d",&amount);
transaction(client, amount);
FILE *file1;
file1 = fopen("Bank.txt","a+");
printf("Name\tDate\tOpening Balance\tAmount Processed\tClosing Balance\n");
printf("%s \t%s \t%d \t%d \t%d\n",client.name,client.day,client.openingbalance,amount,client.closingbalance);
fprintf(file1,"%s \t%s \t%d \t%d \t%d\n",client.name,client.day,client.openingbalance,amount,client.closingbalance);
fclose(file1);
}
In the output it displays a wrong amount rather the entered opening balance and all the other variables including name and date.

Answer 1

You're passing the struct by value, which creates a copy of it. Operations on the struct in the function don't affect the original variables.
Instead of doing that, you may want to pass a pointer to the variable to your functions.
To do that, you need to change your function signatures to use a pointer to detail:
void InputDetails(detail *customer)
void transaction(detail *customerr, int money)
Since you're using a pointer, you need to dereference it to access the actual data pointed to by the pointer.
For other variable types you'd use the * indirection unary operator to access the value, which you can also do with structs:
(*customerr).closingbalance = (*customerr).openingbalance - money;
This is ugly. Instead of having to do this, you can use the -> operator, which is just a prettier way of doing the above:
customerr->closingbalance = customerr->openingbalance - money;
In your functions, you'd need to replace every use of the . operator on customerr with ->.
Doing this will save both time and memory since you don't have to make a local copy of the struct, and your program can function correctly.
When you call your functions, you'd use the & unary operator to take obtain the address of the struct variable and pass that, instead of the entire struct itself:
InputDetails(&client);
transaction(&client, amount);

Answer 2

You have to pass an object of a structure by reference through using a pointer to the object if you want that the original object would be changed in the function.
For example the function InputDetails should be defined like
void InputDetails(detail *customer){
printf("Enter Customer name: ");
scanf("%s",customer->name);
printf("Enter the date: ");
scanf("%s",customer->day);
printf("Enter opening balance: ");
scanf("%d",&customer->openingbalance);
}
and called like
InputDetails(&client);