Problem with Scanf and if else statement

[bonett09]

Hi guys, I have a problem with this question, where the amount is constantly $0 and at the end if the user enters Y/N it just closes without saying "Transaction Completed/Cancelled".

Any tips on how to fix this?
Thanks

#include<stdio.h>
#include<iostream>
#include<math.h>


int main()
{
    int numofbees;
    float price = 5;
    char query;
    char Y;
    char N;


    printf("How many bees would you like to purchase?\n");
              scanf("%d", &numofbees);
        printf("That will be $%d, confirm? (Y/N)\n", price*numofbees);
              scanf("%c", &query);
        if(query == Y)
            printf("Transaction completed.");
        else if(query == N)
              printf("Transaction cancelled.");


        getchar();
        getchar();
        return 0;
    }

[jimblumberg]

There are a couple of things wrong with your program. Your first problem is being caused by the first scanf() leaving the end of line character in the input buffer, which your second scanf() accepts for it's entry. To fix this problem you need to ignore that character, the easiest way would be to add a space before your "%c" in your second scanf():

scanf(" %c", &query);

The next problem is with your if statements,

 if(query == Y)

You are comparing query to the value held in your Y variable. You have not initialized this variable, it could therefore hold anything. But you really don't need a variable, you can compare query to a const character like:

 if(query == 'Y')

Note the single quotes. You have the same problem with the second if statement. I myself would only use a simple else, not the else if. That way if the user does not enter Y you will get the second message. Also be aware of the fact that C is case sensitive, Y does not equal y.

Jim

[hk_mp5kpdw]

What are the values contained within your variables Y and N?

[Amberlampz]

In addition to what's being mentioned, you don't need math.h and you should get rid of the elseif() and just change it to a single if() statement.

[bonett09]

I see, I have managed to solve the problem regarding Transaction Accepted/Cancelled. However, I still can't find what is the problem which is withholding it from executing the proper price.

EDIT: In addition, regarding what jimblumberg said, would it be more suitable to use 1/2 instead of (Y/N) as && and || operators didn't work as planned.

[jimblumberg]

Why didn't the && and || operators work. Show the code where you tried using them.

Jim

[Tclausex]

I see, I have managed to solve the problem regarding Transaction Accepted/Cancelled. However, I still can't find what is the problem which is withholding it from executing the proper price.

EDIT: In addition, regarding what jimblumberg said, would it be more suitable to use 1/2 instead of (Y/N) as && and || operators didn't work as planned.

You declared price as float and numofbees as int. When you multiply them together numofbees will be cast 'up' as float and the result will be float. %d format specifier is therefore wrong.

[bonett09]

Why didn't the && and || operators work. Show the code where you tried using them.

Jim

#include<stdio.h>
#include <iostream>


int main()
{
    int numofbees;
    float price = 5;
    char query;


    printf("How many bees would you like to purchase?\n");
        scanf("%d", &numofbees);
    printf("That will be $%f, confirm? (Y/N)\n", price*numofbees);
        scanf(" %c", &query);


        if(query == 'y'&&'Y') {
            printf("Transaction completed.");
    }
        else {
            printf("Transaction cancelled.");
    }
        getchar();
        getchar();
        return 0;
    }

It accepts anything as a 'Yes'/Transaction Complete.

[jimblumberg]

In the following snippets both sides of the && must be complete. Also in this case you probably want to use the || (or) instead of and (&&).

if(query == 'y'&&'Y')

Like:

if(query == 'y' || guery == 'Y')

Jim

[Tclausex]

Your test is wrong on 2 counts.

if(query == 'y'&&'Y')

That if statement says: if query's value is 'y' AND character 'Y'. Character 'Y' is non-zero and is always true. So the only time that statement should evaluate to true is when you enter 'y'. So what you need is the equality test == on both sides of the logical operator &&. But, the second problem is that query can't hold the 'y' AND 'Y' at the same time. You should be using the OR operator.

if( query == 'y' || query == 'Y' )

[bonett09]

Thanks guys, that solved it

[inu11byte]

I just went ahead and altered the original code to provide a better result.

#include <stdio.h>


int main()
{
    int numofbees, price = 5;
    char query;
 
    printf("How many bees would you like to purchase?\n");
              scanf("%d", &numofbees);
    printf("That will be $%d.00, confirm? (Y/N)\n", price*numofbees);
              scanf(" %c", &query);
    query == 'Y' || query == 'y' ? printf("Transaction completed.") :
              printf("Transaction cancelled.");


        getchar();
        getchar();
        return 0;
}