Arithmetic operators and function calling in C

Question

I'm not quite sure why I can't do

double a = (double) my_Function(45) / 2048 / 2340 / 90;
printf("%.4f",a); // prints out 0.00

But instead I have to use one more variable as:

double a = (double) my_Function(45);
double b = a  / 2048 / 2340 / 90;
printf("%.4f",b);  // prints out the correct value

Answer 1

It comes down to order of operations.

You basically are doing this in the first line

double a = (double)(my_Function(45) / 2048 / 2340 / 90);

Answer 2

What does my_Function return? Perhaps your answer is less than 10^(-4).