Programming in c dodgy if statements? - c

Question

I'm writing code that reads the input for water consumption and displays the bill. the fixed price is $10. for the first 30 meters it costs $0.6 per meter, for the next 20 meters it costs $0.8, then the next 10 meters it cost $1.2 per meter and additional meters are $1.2 but for some reason my code asks the user to input a value and does nothing else. what am I doing wrong? Is it my if statements? I feel as though they are okay and my calculations may be the problem. here is the code so far:
#include <stdio.h>
int main()
{
int watercom, fixedrate, first30, next20, next10, additional, firstcal, secondcal,thirdcal, fourthcal;
printf("what is our water consumption?\n");
scanf("%i\n",&watercom);
fixedrate = 10;
first30 = 0.6;
next20 = 0.8;
next10 = 1.0;
additional = 1.2;
firstcal = fixedrate * first30;
secondcal = fixedrate * next20;
thirdcal = fixedrate * next10;
fourthcal = fixedrate * additional;
if ( watercom <= 30)
printf("your bill is %i", &firstcal);
else if (watercom >= 31 && watercom <= 50)
printf("your bil is %i", &secondcal);
else if (watercom >= 51 && watercom >= 60)
printf("your bill is %i", &thirdcal);
else if (watercom >= 61)
printf("your water bill is %i", fourthcal);
return 0;
}

Answer 1

if ( watercom <= 30)
printf("your bill is %i", &firstcal);
else if (watercom >= 31 && watercom <= 50)
printf("your bil is %i", &secondcal);
else if (watercom >= 51 && watercom >= 60)
printf("your bill is %i", &thirdcal);
else if (watercom >= 61)
printf("your water bill is %i", fourthcal);
There are a number of simplifications that can be applied:
You don't need the >= part in each if statement as if watercom <= 30 then it is implicitly >= 31.
Also you don't pass in the address-of (&) values to printf, you pass in the values directly.
You also don't have a default case for when watercom is outside the range of expected values:
if ( watercom <= 30 )
printf("your bill is %i", firstcal);
else if ( watercom <= 50)
printf("your bill is %i", secondcal);
else if ( watercom <= 60)
printf("your bill is %i", thirdcal);
else
printf("your water bill is %i", fourthcal);

Answer 2

In addition to what others have pointed out, all your variables are int but some need to be double [because you're setting them to 0.6, etc.]:
int watercom;
double fixedrate;
double first30;
double next20;
double next10;
double additional;
int firstcal;
int secondcal;
int thirdcal;
int fourthcal;

Answer 3

Yeah your only int is fixedRate. The rest are floats.
Also by what you say in your question that is not what is in your program. If the fixedRate is 10 and the first 30 is at a rate of .6 then if I input 30 I should get fixedRate + (30*.6).
You need to add the fixedRate to all calculations and add the calculations for each tier together for your final amount.
Work the problem on paper before you try to write the program. You should know what output is expected before you test your program.
😊Happy Day!

Answer 4

The algebra of the calculations, as they appeared in the first version of this post, seems all wrong to me. To get you started on that: as well as these amount * rate multiplications, you need to be performing an addition --- a fixed rate term, plus another amount if you go over the initial allowance, plus another amount when you enter the next price bracket, and so on. To calculate each of these terms, you need to be subtracting the amount of water already-paid-for in previous terms as you go along.
But that's not the question. You asked why the program was not printing anything at all. For me, after making a command-line executable out of this using Visual Studio on Windows, I did replicate that problem: I input the value, then nothing seems to happen. But then the answer is finally printed when I press control-C to halt the program.
To fix this, remove the \n from the scanf command.
Then proceed to your other problems, which include (a) getting the algebra right, and (b) printing the value (e.g. firstCal) not the address (&firstCal) of your variables.