#include <stdio.h> #include <stdlib.h> #define calc(a,b) (a*b)/(a-b) void calculate(){ int a = 20, b = 10; printf("%f\n", calc(a+4,b-2));//output 0.00000 } what to do to print the actual answer, 4.83.
3 Answers
#define calc(a,b) ((a)*(b))/((a)-(b)) Can you spot the extra parentheses?
--> calc(a+4,b-2) resolves to ((a+4)*(b-2))/((a+4)-(b-2)). Correct.
Your solution without the extra parentheses:
--> calc(a+4,b-2) resolves to (a+4*b-2)/(a+4-b-2). Which is very different!
1 Comment
James O'Doherty
Very good catch! While the integer division may cause rounding issues, the macro is definitely the real problem here. The expanded (a+4*b-2)/(a+4-b-2) isn't the same as the intended ((a+4)*(b-2))/((a+4)-(b-2))
The problem here is with your datatypes which are ints, not with format-specifiers. Integer division is always truncated to the whole numbers. You should consider changing your variables to float instead of int.
Try this:
#include <stdio.h> #include <stdlib.h> #define calc(a,b) (a*b)/(a-b) void calculate(){ float a = 20.0f, b = 10.0f; printf("%f\n", calc(a+4,b-2));//output 0.00000 } 
1 Comment
Keith Thompson
Or double (which, on many systems, is just as fast as float).
You need to fix the expression first. calc(a+4,b-2) is of type int, and integer division truncates.
For example, you could change the declarations to:
double a = 20.0, b = 10.0; and then change "%f\n" to "%.2f\n" to get two decimal places.
(24 * 8) / (24 - 8) = 192 / 16 = 12