Thread: Problem with some "while" and float stuff

I've been getting some strange problems with my code so far. You need to input hrs as an integer, rate as a float, and then compute salary as a float. If >40 hrs are eneterd it is time and a half. Make sense? Where am I going wrong? Instructor has not gone over float stuff, so I gathered what I could.

Code:

# include <stdio.h>

int main()

{

int hrs;
float rate;
float salary;


while(1){

	printf("Enter # of hours worked (-1 to end):");
	scanf("%d",&hrs);

	if(hrs == -1){
		return 0;
	}

	printf("Enter hourly rate of worker ($00.00):");
	scanf("%f",&rate);
	
	if(hrs<=40){
		salary = rate*hrs;
	}

	if(hrs>40){
		salary = (rate*40)+((hrs-40)*rate*1.5);
	}

	printf("Salary is %.2f \n", salary);
}

return 0;

}

If you want to use scanf() for reading in the floats, you'll probably want to use %f in place of %d. You will be told by members of this forum to not use scanf(), but in place use fgets() for controled keyboard input.

Edit: After seeing your post, I want to warn you to only use %f for the floats. . . maybe try the %i for the int.

Originally Posted by Kennedy

If you want to use scanf() for reading in the floats, you'll probably want to use %f in place of %d. You will be told by members of this forum to not use scanf(), but in place use fgets() for controled keyboard input.

Thanks for the advice, but I must use scanf for keyboard input due to class restrictions ATM. I will update the code above and put in %f in place of the %d

Right now the program half ass works. Inputting -1 will not terminate...

edit: it will terminate now, but not right when I enter -1. It will terminate after I put in the second value. Odd.

Also, I get this warning when compiling "warning C4244: '=' : conversion from 'double' to 'float', possible loss of data"

It does terminate actually, after you enter anything for rate. Try it.

edit: try this:

Code:

{
printf("Enter # of hours worked (-1 to end):");
	scanf("%d",&hrs);
	
	if(hrs == -1){
           return 0;
           }

edit: also, you need to get rid of

Code:

while(hrs != -1)

You'd have to do it like this:

Code:

printf("Enter # of hours worked (-1 to end):");
scanf("%d",&hrs);
while (hrs != -1){
	printf("Enter hourly rate of worker ($00.00):");
	scanf("%f",&rate);
	
	if(hrs<=40){
		salary = rate*hrs;
	}

	if(hrs>40){
		salary = (rate*40)+((hrs-40)*rate*1.5);
	}

	printf("Salary is %.2f \n", salary);

	printf("Enter # of hours worked (-1 to end):");
	scanf("%d",&hrs);
}

And you can get rid of the warning by changing the floats to doubles. But you'd also have to change the scanf for rate so it uses %lf.

>Could I just use all floats and keep things the way I have it?
Yes, and if you still have the warning, you could eliminate the warning by changing your constants to float. For example change 1.5 to 1.5f. The default for a floating point constant is double.

Or you can add a cast in front: (float) 1.5

Or you could use an uppercase F: 1.5F Uppercase is often preferred, simply because when you go to add an 'L' (for a long literal) a lowercase 'l' looks like a numeric 1 (one). (That being said, I usually use a lowercase 'f' myself.)