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Hi,I'm learning c and it's been interesting, I'm at the lowest of beginners if there's anything like that, without wasting much time this is my question.
I'm trying to print,the sum of a and b divided by c equals=? Given a,b and c are 10,2 and 100 respectively,I've tried but it's not giving me the right answer. I would appreciate if anyone could show me how it's done. Thanks

What I have tried:

#include <stdio.h>
int main(){
int a,b,c;
a=10;
b=2;
c=100;
printf("The sum of %d and %d divided by %d equals %d\n",a,b,c,(a+b)%10);
}
Posted
Updated 31-Mar-20 7:44am
v2

Try
C
#include <stdio.h>
int main()
{
  double a,b,c;
  a=10.0;
  b=2.0;
  c=100.0;
  printf("The sum of %g and %g divided by %g equals %g\n",a,b,c,(a+b)/c);
  return 0;
}
 
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Comments
Patrice T 31-Mar-20 13:58pm    
My 5 too.
I think the .0 are not needed since variables are floats.
CPallini 31-Mar-20 14:01pm    
Thank you.
Indeed they are not needed.
Member 14788459 31-Mar-20 18:08pm    
Thanks,it worked but I think I'll just have to go back to the textbook I'm reading because I have no idea what %g is or does or the double you used or why mine didn't give me the correct answer,what did I do wrong?
CPallini 1-Apr-20 2:15am    
Don't bother about '%g'. It is just a format specifier used to get 'nice output' from floating point numbers.
The important point is the usage of doubles (floating point numbers) instead of integers. See, for instance:
https://www.calebcurry.com/c-programming-tutorial-21-int-float-and-double-data-types/
Stefan_Lang 1-Apr-20 2:43am    
Actually, I always use .0 too, even when it's not needed. The reason is that sometimes the compiler might interpret the code differently based on the type of, e. g., a function parameter. Then it's important to specify the right type.

It also helps you detect inaccurate types: if you accidentally declare a variable as int (or similar), the compiler will point it out once you try to initialize or modify it with a .0 literal.
As a beginner, you have to read documentation (or at least do some search for anything new to you).
Where have you seen that % is the operator for division ?

Beware result of division of integers is an integer.
You will need to search on how to get floating point result on a division of integers.
 
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v2
Comments
CPallini 31-Mar-20 13:39pm    
My 5.
Patrice T 31-Mar-20 13:40pm    
Thank you
Maciej Los 1-Apr-20 1:52am    
5ed!
Patrice T 1-Apr-20 1:57am    
Thank you

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