Arithmetic operators and function calling in C
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by Robert Dalton
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Published on 2010-05-26T01:05:07Z
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2010/05/26
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I'm not quite sure why I can't do
double a = (float) my_Function(45) / 2048 / 2340 / 90;
printf("%.4",a); // prints out 0.00
But instead I have to use one more variable as:
double a = (float) my_Function(45);
double b = (float) a / 2048 / 2340 / 90;
printf("%.4",b); // prints out the correct value
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