Teaching myself C and finding that when I do an equation for a temp conversion it won't work unless I change the fraction to a decimal. ie,
tempC=(.555*(tempF-32)) will work but tempC=((5/9)*(tempF-32)) won't work.
Why?
According to the book "C Primer Plus" it should work as I'm using floats for both tempC and tempF.
It looks like you have integer division in the second case:
tempC=((5/9)*(tempF-32)) The 5 / 9 will get truncated to zero.
To fix that, you need to make one of them a floating-point type:
tempC=((5./9.)*(tempF-32)) 
float then the expression should be 5.0f / 9.0f. Otherwise the division might be performed on double precision, which is then rounded to float precision when its result is stored. Hopefully the compiler is smart enough to optimize that minor little issue away, but I wouldn't count on it.
When you do 5/9, 5 and 9 are both integers and integer division happens. The result of integer division is an integer and it is the quotient of the two operands. So, the quotient in case of 5/9 is 0 and since you multiply by 0, tempC comes out to be 0. In order to not have integer division, atleast one of the two operands must be float.
E.g. if you use 5.0/9 or 5/9.0 or 5.0/9.0, it will work as expected.
5/9 is an integer division not a floating point division. That's why you are getting wrong result.
Make 5 or 9 floating point variable and you will get correct answer.
Like 5.0/9 OR 5/9.0
f denoting floating point to constants. Likewise the u for unsigned integer constants5/9 is an integer expression, as such it gets truncated to 0. your compiler should warn you about this, else you should look into enabling warnings.
result = SOME_CONSTANT & MASK; may very well result in zero. Lets say you have similar bit masking operations all over the code, with different masks. You wouldn't want hundreds of warnings then, why should you be punished just because you write generic, consistent code without any magic numbers in it?SOME_CONSTANT / SOMETHING isn't really any different from my example. Lets say that the constant states the number of outputs this build of the program is using, and the division is used to calculate the nature of the output.If you put 5/9 in parenthesis, this will be calculated first, and since those are two integers, it will be done by integer division and the result will be 0, before the rest of the expression is evaluated.
You can rearrange your expression so that the conversion to float occurs first:
tempC=((5/9)*(tempF-32)); → tempC=(5*(tempF-32))/9;
or of course, as the others say, use floating point constants.
tempC= 5/9 * (tempF-32); is an example with the parenthesis removed. This example may or may not give the desired result, depending on whether the compiler evaluates left-to-right or right-to-left - it relies on unspecified behavior.