为什么除法的结果是零而不是小数？

Question

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.

Answer 1

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))

Answer 2

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 .

Eg if you use 5.0/9 or 5/9.0 or 5.0/9.0, it will work as expected.

Answer 3

5\/9 is an integer division not a floating point division. That's why you are getting wrong result.

Answer 4

5/9是一个整数表达式，因此它会被截断为 0。你的编译器应该警告你，否则你应该考虑启用警告。

Answer 5

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.