Do while loop

**dmboasisu2** · 01-24-2002

Hi everyone.

I am writing a program. A snippet of the code is below. In particular, I am trying to do a data validation ( see ****** below). I'm asking the user for certain inputs, and then I get to the Do while statement below, and when I execute the program, and I get to the point where I ask the user for a Y/N, it prints out the printf twice. I don't know if I'm making any sense. I realize that this is happening because once I enter data for a certain statement, I hit enter and that triggers the "Do while". How do I stop it from printing the statement twice. I'm using the Microsoft C/C++ compiler...which I here is different than compilers like borland. I must use C/C++ though.

printf("job grade? (soap handler=1, water manager=2, team manager=3,supervisor=4 :");
scanf("%d",&grade);
{
if(grade != 2 && grade!= 3 && grade != 4)
bonus2 = 25;

else if(grade != 1 && grade!= 3 && grade != 4)
bonus2= 45;

else if(grade != 2 && grade!= 1 && grade != 4)
bonus2 = 75;

else if(grade != 2 && grade!= 3 && grade != 1)
bonus2 = 100;
}

*************************************************
//once I hit enter from the previous input, it triggers this do while and prints
Employee worked 3 overtimes?
You must enter a Y or a N

Employee worked 3 overtimes?
You must enter a Y or a N//

do
{
printf("\nEmployee worked 3 overtimes? : ");
printf("\nYou must enter a Y or a N : ");
scanf("%c",&ans);
*************************************************
}
while(!(ans == 'Y' || ans == 'N'));

{
if (ans =='Y')
bonus3 = 50;

else
bonus3 = 0;
}

**Prelude** · 01-25-2002

You're using scanf, a notoriously buggy function that can be easily avoided in this situation. Since you are only reading one character from the input stream you can either use ans = getchar() to replace the call to scanf, or clear the stream with a simple loop so that scanf will work properly.

It would also be a good idea to change the input to upper case in the event that the user enters a lower case letter. y and n should work as well as Y and N.

Code:

do 
{ 
    printf("\nEmployee worked 3 overtimes? : "); 
    printf("\nYou must enter a Y or a N : "); 
    while(getchar() != '\n')
        ;
    scanf("%c",&ans);
    toupper(ans);  
} 
while( (ans != 'Y' || ans != 'N') );

-Prelude

**clu82** · 01-25-2002

In this context, wouldn't the result of toupper need to be assigned back to the 'ans' variable?

Code:

ans = toupper(ans);

**Govtcheez** · 01-25-2002

No - toupper returns an integer and changes the string.

**clu82** · 01-25-2002

On my system this code

Code:

    
    char c = 'c';

    toupper(c);

    printf("(%c)\n", c);

prints this: (c)

whereas this code

Code:

    char c = 'c';

    c = toupper(c);

    printf("(%c)\n", c);

prints this: (C)

**Govtcheez** · 01-25-2002

You're right - when I actually compiled that, it wouldn't even let me...

It's Friday - my head's not totally here

**dmboasisu2** · 01-25-2002

Thanks everyone for your help. I'm still a beginner so I'm not sure about the code you gave me. BUt, I made some alterations since the post I made. This is what I have now (look below). When I run it, it works when I type "Y" (it displays the answers I want) but when I type "N", it does nothing...the program simple stops. Also, when I type in a lower case y or n, it does nothing. What is the simpliest thing to do to make this work. Thank you again. Code below (note: this is not the entire program)

{
if(grade != 2 && grade!= 3 && grade != 4)
bonus2 = 25;

else if(grade != 1 && grade!= 3 && grade != 4)
bonus2= 45;

else if(grade != 2 && grade!= 1 && grade != 4)
bonus2 = 75;

else if(grade != 2 && grade!= 3 && grade != 1)
bonus2 = 100;
}

printf("\nEmployee worked 3 overtimes? : ");
printf("\nYou must enter a Y or a N : ");

do

{ scanf("%c",&ans);
if (ans =='Y')
bonus3 = 50;

else if (ans =='N')
bonus3 = 0;

}

while(!(ans == 'Y') || (ans == 'N'));

printf("\nBonus 1 = %d, Bonus 2 = %d, Bonus 3 = %d\n",bonus1,bonus2,bonus3);

Barjor · 01-25-2002

This is one way you can check for Y and y

if (ans =='Y' || and =='y')

~Barjor

**Govtcheez** · 01-25-2002

> when I type in a lower case y or n,

That's because you're only checking for uppercase Y and N - see our discussion about toupper.

**dmboasisu2** · 01-25-2002

I understand what your saying about the toupper...the main issue right now is when I type "N", it doesn't assign the value I want to a certain variable. Once thats fixed, then I want to beable to do a validity check on the case of the letter or some other character that a user may enter. If someone could just throw this in a compiler, it may make more sense to you. The entire code is below. Sorry for taking up space. Thanks

#include <stdio.h>

int main ()

{

int seniority;
int employeenum;
int bonus1;
int bonus2;
int grade;
char ans;
int bonus3;

printf("Please enter the employee's number : ");
scanf("%d",&employeenum);
printf("\n");
printf("How many years has the person been employed with BMW? : ");
scanf("%d",&seniority);
printf("\n");

//Start of 1st Factor//

{
if(seniority>10 || seniority<0)
printf("You entered an incorrect value. Please enter a value between 0-10 : ");

if(seniority < 1)
bonus1 = 25;

else if (seniority >=1 && seniority <=3)
bonus1 = 50;

else if (seniority >=4 && seniority <=6)
bonus1 = 75;

else if (seniority >=7)
bonus1 = 100;

}
//End of 1st Factor//

printf("job grade? (soap handler=1, water manager=2, team manager=3,supervisor=4 :");
scanf("%d",&grade);

//Start of 2nd Factor//
{
if(grade != 2 && grade!= 3 && grade != 4)
bonus2 = 25;

else if(grade != 1 && grade!= 3 && grade != 4)
bonus2= 45;

else if(grade != 2 && grade!= 1 && grade != 4)
bonus2 = 75;

else if(grade != 2 && grade!= 3 && grade != 1)
bonus2 = 100;
}

printf("\nEmployee worked 3 overtimes? : ");
printf("\nYou must enter a Y or a N : ");

do

{ scanf("%c",&ans);
if (ans =='Y')
bonus3 = 50;

else if (ans =='N')
bonus3 = 0;

}

while(!(ans == 'Y') || (ans == 'N'));

printf("\nBonus 1 = %d, Bonus 2 = %d, Bonus 3 = %d\n",bonus1,bonus2,bonus3);

return 0;
}

Barjor · 01-25-2002

I don't have a compiler here put I would try to write it like this instead

{ scanf("%c",&ans);
if (ans =='Y')
bonus3 = 50;

if (ans =='N')
bonus3 = 0;

}

**dmboasisu2** · 01-25-2002

Thanks for your input. I tried that and it still doesn't work. Its quite strange that it doesn't...I am stumped.

Barjor · 01-25-2002

Me again..Still don't have a compiler...I don't get the logic in your last Do/While loop. Why do you have it there? What is the tought behind >> while(!(ans == 'Y' || ans == 'N')); << Not sure I see what you want it to do. Is the line >> printf("\nYou must enter a Y or a N : "); << not supposed to be inside the Do/While loop?. Am I confused? is it late friday and am I sick of work..YES.

**dmboasisu2** · 01-25-2002

Originally posted by Barjor
Me again..Still don't have a compiler...I don't get the logic in your last Do/While loop. Why do you have it there? What is the tought behind >> while(!(ans == 'Y' || ans == 'N')); << Not sure I see what you want it to do. Is the line >> printf("\nYou must enter a Y or a N : "); << not supposed to be inside the Do/While loop?. Am I confused? is it late friday and am I sick of work..YES.

Well, the logic behind [while(!(ans == 'Y' || ans == 'N'));] is that it should execute the loop only when it is NOT a capital Y or a capital N. AS for the printf in the Do while, I'm a beginner and don't completely understand how everything is executed. I just want it to do a validity check to see if the user types in Y or N. If they type a 'y' or a 'n', it should execute the loop and tell the user that it did something wrong and to do it again. Also, depending upon whether they say Y or N, a value must be attached to their answer. At this point, my program works when I type a 'Y'...but not when I type a 'N'. I am quite frustrated as well. Sorry for the confusion. I had a long day of work as well.

Unregistered · 01-25-2002

instead of while(!(ans == 'Y') || (ans == 'N'));

try while(ans != 'Y' && ans != 'N');

or while(!(ans == 'Y' || ans == 'N'));

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