declaring a variable with only 2 decimal points in C - c

Question

I am trying to declare a variable eLon with only 1 decimal place so that if I have the code:
elon=359.8
printf("eLon = %f\n",eLon);
and the output will be
eLon = 359.8
However the output I get is:
eLon = 359.7999999.
How I know that I could modify the printf so that it would be:
printf(eLon is %0.1f\n",eLon);
to get the desired result. This is NOT what I want to do. I just want the variable itself to only have one decimal place so that it equals 359.8 not 359.7999999, since this is critical for any computations I make. Do you know how I should modify my code to get the desired result. I tried doing what was suggested in other inquiries but it did not work for example the code:
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
int main()
{
int endLonn;
float endLon,eLon;
endLon=359.8;
endLonn=359.8*10;
printf("%d is endLonn\n",endLonn);
eLon= endLonn / 10;
printf("elon is %f\n",eLon);
}
gives me the output:
elon is 359.00000000
Again this is also not what I am looking for. I want elon is 359.8. If you could help me tweak my code to get the desired result that would be great. Thank you for your time.

Answer 1

Instead of trying to cheat double or float and printf, you can use fixed point arithmetic. In the particular case of your example you will express variable in 0.1s of it's original value:
int in_10s;
in_10s=3598;
printf("My fixed point variable "
"represents %d.%01d value\n",
in_10s/10, abs(in_10s % 10));
In general case of n decimal places, the format to printf should be %d.%0nd, where n=log_10(scaling_factor^-1), and scaling factor is interpreted like here.

Answer 2

What you want is not a floating point number but a fixed point decimal, which is for example available in c# as decimal. c doesn't give you that. There are two "typical" approaches to roll your own:
Use a plain int and just have the decimal point at some position by convention. e.g. your value would be int elon=3598.
Use a struct with two ints, one for the whole number part and one for the amount of tenths (or hundredths, thousandths, ...)
In both cases, you will need to implement your own logic for output. The simple approach using a plain int at least lets you use basic arithmetics as usual.

Answer 3

You actually can't declare a variable to hold only one decimal figure. The only native types able to represent non-integer values are floating point types, on which any arithmetic operation which yield a value of the same nature. Floating point types cannot be limited to a fixed number of decimal figures.
In order to achieve what you are asking for, you have to either implement such behavior or use an existing implementation of what is called fixed-point airthmetic.
A pretty complete introduction can be found in Wikipedia's Fixed-Point arithmetic entry. At the end you will find a couple of libraries implementing it.

Answer 4

I just want the variable itself to only have one decimal place so that it equals 359.8 not 359.7999999,
You should use double:
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
int main(void){
int endLonn;
double endLon,eLon;
endLon=359.8;
endLonn=359.8*10;
printf("%d is endLonn\n",endLonn);
eLon = endLonn / 10;
printf("elon is %.1lf\n%.1lf",eLon,endLon);
return 0;
}
Output:
3598 is endLonn
elon is 359.0
359.8