how to specify the input field.

**paradize** · 05-01-2006

Hye all. Im new to this forum. I would like to seek some help here.
Below is my source code for counting the net salaray of a worker. My question is, how for example, in the name area, I only want the user to input character, other than that, invalid data and for id only numbers ? And, how at the end of the program, to make the user input only y for continue the program, and n to exit the program.?

Code:

#include <stdio.h>

void main()
{	
	
	int id;																				
	char month[100],position[500],name[500],ex;											
	double basic_salary,EPF_deduction,gross_salary,total_OT_payment,net_salary,OT_rate; 
	double OT;																			

	OT_rate = 50; 
	ex = 'y';

	
		do	
	
	{
		printf("\n\nPlease enter month (Capital Letter): ");
		scanf("%s",month);

		
		printf("\n\n*******************************************************************");
		printf("\nPAYMENT SLIPS FOR THE MONTH %s 2005\nSMART CHOICE EDUCATION CENTRE\n",month);
		printf("*******************************************************************\n");
			
		
		printf("Staff name\t\t: ");
		scanf("%s",&name);
		printf("Staff ID\t\t: ");
		scanf("%d",&id);
		printf("Position\t\t: ");
		scanf("%s",&position);

		printf("*******************************************************************\n");
		
		
		printf("\nBasic Salary\t\t: RM ");
		scanf("%lf",&basic_salary);
		
		
		EPF_deduction = basic_salary * 0.1;

		printf("EPF Deduction\t\t: RM %.2lf\n",EPF_deduction);
		
		
		gross_salary = basic_salary - EPF_deduction;

		printf("Gross Salary\t\t: RM %.2lf\n",gross_salary);
		
		
		

		printf("*******************************************************************\n");
		
		
		printf("\nTotal OT hour\t\t: ");
		scanf("%lf",&OT);

		printf("OT Rate \t\t: RM %.2f\n",OT_rate);

		
		total_OT_payment = OT * OT_rate;

		printf("Total OT Payment \t: RM %.2lf\n",total_OT_payment);
		

		printf("*******************************************************************\n");

		
		net_salary = gross_salary + total_OT_payment;

		printf("Net Salary\t\t: RM %.2lf",net_salary);
		

		printf("\n*******************************************************************\n");
												
		
		printf("Thank you , for using this program and have a nice day.\n");
		
		
		printf("Do you want to continue?(y/n) : ");
		scanf("%s",&a);


	}
		
		while (ex =='y' || ex == 'Y');
}

I'll be glad to here from ya'll. Thank you

**00Sven** · 05-01-2006

You can use getch it will still have them enter whatever they want. It will only display the character if they enter a y or n. It is pretty complicated though.

Code:

char response;
printf("Continue? (y/n) :");
while(response != 'y' && response != 'n')
{
     response = getch();
}
putchar(response):
getchar();
if(response == 'y')
{
     /*Code Here*/
}
else if(response  == n')
{
     /*Code Here*/
}

getch gets a character without printing it onto the screen. The user does not need to hit enter either.
Here is the breakdown:
1. The Continue question is asked.
2. It assigns the character that the user enters to response without printing it onto the screen. When that character is either a y or n it breaks out of that loop.
3. Putchar puts the y or n out onto the screen.
4. Getchar waits for them to press enter to continue.
5. The if and else statements would contain your code for what to do.

This gives the appearance of just having it only get the character if they enter a y or n. I think that this is what you were looking for. Hope this helps.

Note: getch is not included on all compilers. If you state your compiler it can help. If using Dev-C++ you include <conio.h>
~Sven

**cdalten** · 05-01-2006

Originally Posted by 00Sven

You can use getch it will still have them enter whatever they want. It will only display the character if they enter a y or n. It is pretty complicated though.

Code:

char response;
printf("Continue? (y/n) :");
while(response != 'y' && response != 'n')
{
     response = getch();
}
putchar(response):
getchar();
if(response == 'y')
{
     /*Code Here*/
}
else if(response  == n')
{
     /*Code Here*/
}

getch gets a character without printing it onto the screen. The user does not need to hit enter either.
Here is the breakdown:
1. The Continue question is asked.
2. It assigns the character that the user enters to response without printing it onto the screen. When that character is either a y or n it breaks out of that loop.
3. Putchar puts the y or n out onto the screen.
4. Getchar waits for them to press enter to continue.
5. The if and else statements would contain your code for what to do.

This gives the appearance of just having it only get the character if they enter a y or n. I think that this is what you were looking for. Hope this helps.

Note: getch is not included on all compilers. If you state your compiler it can help. If using Dev-C++ you include <conio.h>
~Sven

The code is broken. The prototype for getchar() is
int getchar(void);

This means that the variable response should be declared as an int and not a char. This comes down to the fact char may be either signed or unsigned. If the machine is signed, getchar() will return -1 on error. However, if the machine it unsigned, the machine will return 255. If you mucky on this, I recommend getting a basic book on computer science that talks about hex to dec conversions.

Regards

**00Sven** · 05-01-2006

I do not refer to the return value of getchar. I simply use it to pause the program. You could take this out of the program but then as soon as they hit y or n it would do the code for that choice without them needing to hit enter. If you want it can be replaced with...

Code:

while( ( temp = getchar() ) != '\n' && temp != EOF );

This would keep on doing getch and placing the character into a temp which would need to be an int and then once the user hit enter it would continue with the program. This way the only thing that is output is the y or n and then nothing else. But it still waits until Enter is pressed.
~Sven

**cdalten** · 05-01-2006

And not testing for EOF leads to undefined behavior.

Code:

#include <stdio.h>

int main(void) {
    char response;
    printf("Continue? (y/n) :");
    while(response != 'y' && response != 'n')
    {
        response = getchar();
    }
    putchar(response);
        getchar();
    if(response == 'y')
    {
        /*Code Here*/
    }
    else if(response  == 'n')
{
     /*Code Here*/
}


}

$./asc
Continue? (y/n) :z

<-----here the program hangs.

using your revised code

Code:

while( ( temp = getchar() ) != '\n' && temp != EOF )

while having response declared as int would be a much sounder solution. With that, I need to cook some dinner.

**00Sven** · 05-02-2006

I thought that I had corrected that but I must have forgotten to. You are right response should be an int not a char.
~Sven

**dwks** · 05-02-2006

And don't use void main().

Use int main().

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