does emacs has rational number data type

Question

I eval a lisp expression in scratch

(+ (/ 1 2) (/ 1 2))

I got a 0.

normally it should be 1.

Mildamilde

Didn't you just answer the question yourself? – Congdon

Answer 1

As Oleg points out, operators usually default to integer arithmetic unless you include floating point arguments (like 1.0).

With respect to your question about rational number support, emacs-calc (which is part of emacs) supports many number types including fractions (i.e. rational numbers), complex numbers, infinite precision integers, etc. Your code must call emacs-calc functions (instead of /, etc.) in order to use calc's arithmetic.

GNU Emacs Calc Manual:

• Fractions
• Index of Lisp Math Functions

Colorcast

Ah, calc to the rescue yet again. I should have thought of that. – Blepharitis

Answer 2

Try this way

(+ (/ 1.0 2) (/ 1.0 2))

According to emacs doc

Function: / dividend divisor &rest divisors

if all the arguments are integers, then the result is an integer too.

Kalevala

That fails to answer the question, although between the lines, you can infer that because Emacs uses floats, it doesn't have a separate type for real numbers. (Many Lisp implementations have one; that is, 1/3 is represented precisely, rather than as a floating-point approximation.) – Congdon

Answer 3

You can read all about numbers in elisp here:

C-hig (elisp) Numbers RET

As already indicated by tripleee, it is apparent that the answer is "no".

Blepharitis

...except that, as per Juancho's answer, the calc library makes that a "yes". – Blepharitis

Answer 4

Emacs calc has rational data type: use colon, like 1:2 == 0.5 or 5:3 == 1 + 2:3 == 1:2:3.

This way Emacs calc simplifies expressions, for example if you deal with display resolutions for 1920:1080 it prints 16:9! If you want 1440p with the 16:9 ratio: 1440 * 16:9 ⇒ 2560.

Guanabana