Want to make c program app for bank - c

Question

I have made the structure, but I'm stuck at the last point where I have to perform if/else coding. This is my very first program, using some tutorials on web.
The code is below:
#include<stdio.h>
#include<conio.h>
int main()
{
char fName;
char mName;
char lName;
int dob;
int branch;
int brcode;
char * pbprint;
printf("\tWELCOME TO BANK OF KANPUR :: ONE HAND ROLE TO YOUR SUCCESS");
printf("\n\n\n Customer Detail Form");
printf("\n ---------------------");
printf("\n\n\tCustomer Entity ");
printf("\n ");
printf("\n\n\n Customer's First Name : ");
scanf("%s", &fName);
printf("\n Customer's Middle Name : ");
scanf("%s", &mName);
printf("\n Customer's Last Name : ");
scanf("%s", &lName);
printf("\n\n\tDate of Birth ");
printf("\n ");
printf("\n\n\nDate[DD] : ");
scanf("%d", &dob);
printf("\nMonth[MM] : ");
scanf("%d", &dob);
printf("\nYear[YYYY] : ");
scanf("%d", &dob);
printf("\n\n\tBranch Details ");
printf("\n ");
printf("\n\n\nBranch Code : ");
scanf("%d", &brcode);
printf("\n\n\tPrinter ");
printf("\n ");
printf("\n\n\nWould You Like Us To Print Your Passbook [Y/N] : ");
scanf("%s", &pbprint);
// please help how to make if else statement here
// how to make the condition when if I enter yes it should show your
// printing is done else your printing is not done.
getchar();
return 0;
}

Answer 1

You are reading a string from console. If I understood, a simple way is to check only the first char from the string. You can do this:
if ( (pbprint[0] == 'Y') || (pbprint[0] == 'y') )
{
printf("your printing is done");
}
else
{
printf("your printing is not done");
}
The user is usually very creative, so you can choose how smart and versatile your input interface should be.

Answer 2

First of all change your variable declaration from char to char array, as you want strings to be entered. I am considering maximum you will enter 19 characters in strings so you variable declaration will look as below:
char fName[20];
char mName[20];
char lName[20];
char pbprint; // This is just character as you want single character input
You can change your taking character input as below:
printf("\n\n\nWould You Like Us To Print Your Passbook [Y/N] : ");
scanf(" %c",&pbprint); // Notice whitespace in the format string
if((pbprint == 'Y') || (pbprint == 'y')) // now you can do single character comparison
{
printf("Yes");
}
else
{
printf("No");
}
I hope this helps you to take single character input

Answer 3

Issue #1: You need to set aside space to store string data. The char datatype only allocates space for a single character value ('A', 'j', etc.). Strings are sequences of char values followed by a single 0-valued byte, so you need to set aside arrays of char large enough to hold a name plus the 0 terminator.
So, your name variables need to be declared as arrays of char:
#define NAME_LEN 30 // make this as big as you need
char fName[NAME_LEN + 1]; // Add a space for the 0 terminator
char mName[NAME_LEN + 1];
char lName[NAME_LEN + 1];
To read these using scanf, you would write
scanf( "%s", fName ); // note no & operator
The %s conversion specifier expects its corresponding argument to be a pointer to the first element of an array of char. Under most circumstances, an array expression (such as fname in the call above) will automatically be converted to a pointer, so you don't need to use the & operator in this case.
For safety's sake, you should check the result of the scanf call; it will return the number of successful conversions and assignments (0 if none of the conversions are successful), or EOF if it sees an end-of-file or error condition. You should also specify the maximum number of characters to be read within the conversion specifier; otherwise, if you type in more characters than your destination array is sized to hold, scanf will happily attempt to store those extra characters in the memory past the end of your array, leading to anything from corrupted data to a smashed stack to an outright crash:
if ( scanf( "%30s", fName ) == 1 ) // expect one successful conversion and assignment
{
// successfully read fName
}
else
{
// problem during read
}
Unfortunately, the length has to be "hard coded" into the conversion specifier. You can get around this by using some less-than-intuitive preprocessor macros:
#define S2(x) #x // surrounds x with double quotes
#define S1(x) S2(x) // causes x to be expanded before being passed to S2
#define E1(x) x // causes x to be expanded
#define FMT(x) S1(%E1(x)s) // writes a %xs conversion specifier, where
// x is the max length
Then you can write something like
if ( scanf( FMT(NAME_LEN), &fName ) == 1 )
which will expand out to
if ( scanf( "%30s", &fName ) == 1 )
The S2 and E1 macros look redundant, but they are necessary; you want to "expand" the argument x before applying the # operator to it. If we wrote
#define S1(x) #x
#define FMT(x) S1(%xs)
...
if ( scanf( FMT(NAME_LEN), &fName ) == 1 )
it would expand out to
if ( scanf( "%xs", &fName ) == 1 )
which is not what we want.
Issue #2: Simply declaring a pointer does not set aside storage for the thing you are pointing to. The declaration
char * pbprint;
creates the pointer variable pbprint, but it doesn't allocate any memory for the thing we want to point to. In this particular case, if all you want to store is a y or n value, then declare pbprint as a regular char and read using the %c conversion specifier, like so:
#include <ctype.h>
...
char pbprint;
...
if ( scanf( " %c", &pbprint ) == 1 ) // note leading blank in format string
{
if ( tolower( pbprint ) == 'y' ) // convert to lower case for comparison
{
// show print output
}
else
{
// don't show print output
}
}
else
{
// error on input
}
The leading blank before the %c conversion specifier is necessary to skip over any leading whitespace in the input stream; otherwise, you risk picking up a newline character from a previous entry, which isn't what you want here.
We also use the tolower library function (declared in ctype.h) to convert the input character to lower case for comparison. That way we don't have to write distinct branches for lowercase and uppercase letters.