GNU Emacs supports two numeric data types: integers and floating point numbers. Integers are whole numbers such as -3, 0, 7, 13, and 511. Their values are exact. Floating point numbers are numbers with fractional parts, such as -4.5, 0.0, or 2.71828. They can also be expressed in exponential notation: 1.5e2 equals 150; in this example, `e2' stands for ten to the second power, and is multiplied by 1.5. Floating point values are not exact; they have a fixed, limited amount of precision.
Support for floating point numbers is a new feature in Emacs 19, and it is controlled by a separate compilation option, so you may encounter a site where Emacs does not support them.
The range of values for an integer depends on the machine. The range is -8388608 to 8388607 (24 bits; i.e., to on most machines, but on others it is -16777216 to 16777215 (25 bits), or -33554432 to 33554431 (26 bits). Many examples in this chapter assume an integer has 24 bits.
The Lisp reader reads an integer as a sequence of digits with optional initial sign and optional final period.
1 ; The integer 1. 1. ; The integer 1. +1 ; Also the integer 1. -1 ; The integer -1. 16777217 ; Also the integer 1, due to overflow. 0 ; The integer 0. -0 ; The integer 0.
To understand how various functions work on integers, especially the bitwise operators (see section Bitwise Operations on Integers), it is often helpful to view the numbers in their binary form.
In 24-bit binary, the decimal integer 5 looks like this:
0000 0000 0000 0000 0000 0101
(We have inserted spaces between groups of 4 bits, and two spaces between groups of 8 bits, to make the binary integer easier to read.)
The integer -1 looks like this:
1111 1111 1111 1111 1111 1111
-1 is represented as 24 ones. (This is called two's complement notation.)
The negative integer, -5, is creating by subtracting 4 from -1. In binary, the decimal integer 4 is 100. Consequently, -5 looks like this:
1111 1111 1111 1111 1111 1011
In this implementation, the largest 24-bit binary integer is the decimal integer 8,388,607. In binary, it looks like this:
0111 1111 1111 1111 1111 1111
Since the arithmetic functions do not check whether integers go outside their range, when you add 1 to 8,388,607, the value is the negative integer -8,388,608:
(+ 1 8388607) => -8388608 => 1000 0000 0000 0000 0000 0000
Many of the following functions accept markers for arguments as well as integers. (See section Markers.) More precisely, the actual arguments to such functions may be either integers or markers, which is why we often give these arguments the name int-or-marker. When the argument value is a marker, its position value is used and its buffer is ignored.
Emacs version 19 supports floating point numbers, if compiled with the
macro LISP_FLOAT_TYPE
defined. The precise range of floating
point numbers is machine-specific; it is the same as the range of the C
data type double
on the machine in question.
The printed representation for floating point numbers requires either a decimal point (with at least one digit following), an exponent, or both. For example, `1500.0', `15e2', `15.0e2', `1.5e3', and `.15e4' are five ways of writing a floating point number whose value is 1500. They are all equivalent. You can also use a minus sign to write negative floating point numbers, as in `-1.0'.
Most modern computers support the IEEE floating point standard, which
provides for positive infinity and negative infinity as floating point
values. It also provides for a class of values called NaN or
"not-a-number"; numerical functions return such values in cases where
there is no correct answer. For example, (sqrt -1.0)
returns a
NaN. For practical purposes, there's no significant difference between
different NaN values in Emacs Lisp, and there's no rule for precisely
which NaN value should be used in a particular case, so this manual
doesn't try to distinguish them. Emacs Lisp has no read syntax for NaNs
or infinities; perhaps we should create a syntax in the future.
You can use logb
to extract the binary exponent of a floating
point number (or estimate the logarithm of an integer):
The functions in this section test whether the argument is a number or
whether it is a certain sort of number. The functions integerp
and floatp
can take any type of Lisp object as argument (the
predicates would not be of much use otherwise); but the zerop
predicate requires a number as its argument. See also
integer-or-marker-p
and number-or-marker-p
, in
section Predicates on Markers.
t
if so, nil
otherwise.
floatp
does not exist in Emacs versions 18 and earlier.
t
if so, nil
otherwise.
t
if so, nil
otherwise.
wholenump
predicate (whose name comes from the phrase
"whole-number-p") tests to see whether its argument is a nonnegative
integer, and returns t
if so, nil
otherwise. 0 is
considered non-negative.
t
if so, nil
otherwise. The argument must be a number.
These two forms are equivalent: (zerop x)
== (= x 0)
.
To test numbers for numerical equality, you should normally use
=
, not eq
. There can be many distinct floating point
number objects with the same numeric value. If you use eq
to
compare them, then you test whether two values are the same
object. By contrast, =
compares only the numeric values
of the objects.
At present, each integer value has a unique Lisp object in Emacs Lisp.
Therefore, eq
is equivalent =
where integers are
concerned. It is sometimes convenient to use eq
for comparing an
unknown value with an integer, because eq
does not report an
error if the unknown value is not a number--it accepts arguments of any
type. By contrast, =
signals an error if the arguments are not
numbers or markers. However, it is a good idea to use =
if you
can, even for comparing integers, just in case we change the
representation of integers in a future Emacs version.
There is another wrinkle: because floating point arithmetic is not exact, it is often a bad idea to check for equality of two floating point values. Usually it is better to test for approximate equality. Here's a function to do this:
(defvar fuzz-factor 1.0e-6) (defun approx-equal (x y) (< (/ (abs (- x y)) (max (abs x) (abs y))) fuzz-factor))
Common Lisp note: Comparing numbers in Common Lisp always requires
=
because Common Lisp implements multi-word integers, and two distinct integer objects can have the same numeric value. Emacs Lisp can have just one integer object for any given value because it has a limited range of integer values.
t
if so, nil
otherwise.
t
if they are not, and nil
if they are.
t
if so, nil
otherwise.
t
if so, nil
otherwise.
t
if so, nil
otherwise.
t
if so, nil
otherwise.
(max 20) => 20 (max 1 2.5) => 2.5 (max 1 3 2.5) => 3
(min -4 1) => -4
To convert an integer to floating point, use the function float
.
float
returns
it unchanged.
There are four functions to convert floating point numbers to integers; they differ in how they round. These functions accept integer arguments also, and return such arguments unchanged.
If divisor is specified, number is divided by divisor
before the floor is taken; this is the division operation that
corresponds to mod
. An arith-error
results if
divisor is 0.
Emacs Lisp provides the traditional four arithmetic operations: addition, subtraction, multiplication, and division. Remainder and modulus functions supplement the division functions. The functions to add or subtract 1 are provided because they are traditional in Lisp and commonly used.
All of these functions except %
return a floating point value
if any argument is floating.
It is important to note that in GNU Emacs Lisp, arithmetic functions
do not check for overflow. Thus (1+ 8388607)
may evaluate to
-8388608, depending on your hardware.
(setq foo 4) => 4 (1+ foo) => 5
This function is not analogous to the C operator ++
---it does
not increment a variable. It just computes a sum. Thus,
foo => 4
If you want to increment the variable, you must use setq
,
like this:
(setq foo (1+ foo)) => 5
+
returns 0. It does not check for overflow.
(+) => 0 (+ 1) => 1 (+ 1 2 3 4) => 10
-
function serves two purposes: negation and subtraction.
When -
has a single argument, the value is the negative of the
argument. When there are multiple arguments, -
subtracts each of
the other-numbers-or-markers from number-or-marker,
cumulatively. If there are no arguments, the result is 0. This
function does not check for overflow.
(- 10 1 2 3 4) => 0 (- 10) => -10 (-) => 0
*
returns 1. It does
not check for overflow.
(*) => 1 (* 1) => 1 (* 1 2 3 4) => 24
If all the arguments are integers, then the result is an integer too.
This means the result has to be rounded. On most machines, the result
is rounded towards zero after each division, but some machines may round
differently with negative arguments. This is because the Lisp function
/
is implemented using the C division operator, which also
permits machine-dependent rounding. As a practical matter, all known
machines round in the standard fashion.
If you divide by 0, an arith-error
error is signaled.
(See section Errors.)
(/ 6 2) => 3 (/ 5 2) => 2 (/ 25 3 2) => 4 (/ -17 6) => -2
The result of (/ -17 6)
could in principle be -3 on some
machines.
For negative arguments, the remainder is in principle machine-dependent since the quotient is; but in practice, all known machines behave alike.
An arith-error
results if divisor is 0.
(% 9 4) => 1 (% -9 4) => -1 (% 9 -4) => 1 (% -9 -4) => -1
For any two integers dividend and divisor,
(+ (% dividend divisor) (* (/ dividend divisor) divisor))
always equals dividend.
Unlike %
, mod
returns a well-defined result for negative
arguments. It also permits floating point arguments; it rounds the
quotient downward (towards minus infinity) to an integer, and uses that
quotient to compute the remainder.
An arith-error
results if divisor is 0.
(mod 9 4) => 1 (mod -9 4) => 3 (mod 9 -4) => -3 (mod -9 -4) => -1 (mod 5.5 2.5) => .5
For any two numbers dividend and divisor,
(+ (mod dividend divisor) (* (floor dividend divisor) divisor))
always equals dividend, subject to rounding error if either argument is floating point.
The functions ffloor
, fceil
, fround
and
ftruncate
take a floating point argument and return a floating
point result whose value is a nearby integer. ffloor
returns the
nearest integer below; fceil
, the nearest integer above;
ftruncate
, the nearest integer in the direction towards zero;
fround
, the nearest integer.
In a computer, an integer is represented as a binary number, a sequence of bits (digits which are either zero or one). A bitwise operation acts on the individual bits of such a sequence. For example, shifting moves the whole sequence left or right one or more places, reproducing the same pattern "moved over".
The bitwise operations in Emacs Lisp apply only to integers.
lsh
, which is an abbreviation for logical shift, shifts the
bits in integer1 to the left count places, or to the right
if count is negative, bringing zeros into the vacated bits. If
count is negative, lsh
shifts zeros into the leftmost
(most-significant) bit, producing a positive result even if
integer1 is negative. Contrast this with ash
, below.
Here are two examples of lsh
, shifting a pattern of bits one
place to the left. We show only the low-order eight bits of the binary
pattern; the rest are all zero.
(lsh 5 1) => 10 ;; Decimal 5 becomes decimal 10. 00000101 => 00001010 (lsh 7 1) => 14 ;; Decimal 7 becomes decimal 14. 00000111 => 00001110
As the examples illustrate, shifting the pattern of bits one place to the left produces a number that is twice the value of the previous number.
The function lsh
, like all Emacs Lisp arithmetic functions, does
not check for overflow, so shifting left can discard significant bits
and change the sign of the number. For example, left shifting 8,388,607
produces -2 on a 24-bit machine:
(lsh 8388607 1) ; left shift => -2
In binary, in the 24-bit implementation, the argument looks like this:
;; Decimal 8,388,607 0111 1111 1111 1111 1111 1111
which becomes the following when left shifted:
;; Decimal -2 1111 1111 1111 1111 1111 1110
Shifting the pattern of bits two places to the left produces results like this (with 8-bit binary numbers):
(lsh 3 2) => 12 ;; Decimal 3 becomes decimal 12. 00000011 => 00001100
On the other hand, shifting the pattern of bits one place to the right looks like this:
(lsh 6 -1) => 3 ;; Decimal 6 becomes decimal 3. 00000110 => 00000011 (lsh 5 -1) => 2 ;; Decimal 5 becomes decimal 2. 00000101 => 00000010
As the example illustrates, shifting the pattern of bits one place to the right divides the value of the binary number by two, rounding downward.
ash
(arithmetic shift) shifts the bits in integer1
to the left count places, or to the right if count
is negative.
ash
gives the same results as lsh
except when
integer1 and count are both negative. In that case,
ash
puts a one in the leftmost position, while lsh
puts
a zero in the leftmost position.
Thus, with ash
, shifting the pattern of bits one place to the right
looks like this:
(ash -6 -1) => -3 ;; Decimal -6 becomes decimal -3. 1111 1111 1111 1111 1111 1010 => 1111 1111 1111 1111 1111 1101
In contrast, shifting the pattern of bits one place to the right with
lsh
looks like this:
(lsh -6 -1) => 8388605 ;; Decimal -6 becomes decimal 8,388,605. 1111 1111 1111 1111 1111 1010 => 0111 1111 1111 1111 1111 1101
Here are other examples:
; 24-bit binary values (lsh 5 2) ; 5 = 0000 0000 0000 0000 0000 0101 => 20 ; = 0000 0000 0000 0000 0001 0100 (ash 5 2) => 20 (lsh -5 2) ; -5 = 1111 1111 1111 1111 1111 1011 => -20 ; = 1111 1111 1111 1111 1110 1100 (ash -5 2) => -20 (lsh 5 -2) ; 5 = 0000 0000 0000 0000 0000 0101 => 1 ; = 0000 0000 0000 0000 0000 0001 (ash 5 -2) => 1 (lsh -5 -2) ; -5 = 1111 1111 1111 1111 1111 1011 => 4194302 ; = 0011 1111 1111 1111 1111 1110 (ash -5 -2) ; -5 = 1111 1111 1111 1111 1111 1011 => -2 ; = 1111 1111 1111 1111 1111 1110
For example, using 4-bit binary numbers, the "logical and" of 13 and 12 is 12: 1101 combined with 1100 produces 1100. In both the binary numbers, the leftmost two bits are set (i.e., they are 1's), so the leftmost two bits of the returned value are set. However, for the rightmost two bits, each is zero in at least one of the arguments, so the rightmost two bits of the returned value are 0's.
Therefore,
(logand 13 12) => 12
If logand
is not passed any argument, it returns a value of
-1. This number is an identity element for logand
because its binary representation consists entirely of ones. If
logand
is passed just one argument, it returns that argument.
; 24-bit binary values (logand 14 13) ; 14 = 0000 0000 0000 0000 0000 1110 ; 13 = 0000 0000 0000 0000 0000 1101 => 12 ; 12 = 0000 0000 0000 0000 0000 1100 (logand 14 13 4) ; 14 = 0000 0000 0000 0000 0000 1110 ; 13 = 0000 0000 0000 0000 0000 1101 ; 4 = 0000 0000 0000 0000 0000 0100 => 4 ; 4 = 0000 0000 0000 0000 0000 0100 (logand) => -1 ; -1 = 1111 1111 1111 1111 1111 1111
logior
is
passed just one argument, it returns that argument.
; 24-bit binary values (logior 12 5) ; 12 = 0000 0000 0000 0000 0000 1100 ; 5 = 0000 0000 0000 0000 0000 0101 => 13 ; 13 = 0000 0000 0000 0000 0000 1101 (logior 12 5 7) ; 12 = 0000 0000 0000 0000 0000 1100 ; 5 = 0000 0000 0000 0000 0000 0101 ; 7 = 0000 0000 0000 0000 0000 0111 => 15 ; 15 = 0000 0000 0000 0000 0000 1111
logxor
is passed just one argument, it returns that argument.
; 24-bit binary values (logxor 12 5) ; 12 = 0000 0000 0000 0000 0000 1100 ; 5 = 0000 0000 0000 0000 0000 0101 => 9 ; 9 = 0000 0000 0000 0000 0000 1001 (logxor 12 5 7) ; 12 = 0000 0000 0000 0000 0000 1100 ; 5 = 0000 0000 0000 0000 0000 0101 ; 7 = 0000 0000 0000 0000 0000 0111 => 14 ; 14 = 0000 0000 0000 0000 0000 1110
(lognot 5) => -6 ;; 5 = 0000 0000 0000 0000 0000 0101 ;; becomes ;; -6 = 1111 1111 1111 1111 1111 1010
These mathematical functions are available if floating point is supported. They allow integers as well as floating point numbers as arguments.
(asin arg)
is a number between -pi/2
and pi/2 (inclusive) whose sine is arg; if, however, arg
is out of range (outside [-1, 1]), then the result is a NaN.
(acos arg)
is a number between 0 and pi
(inclusive) whose cosine is arg; if, however, arg
is out of range (outside [-1, 1]), then the result is a NaN.
(atan arg)
is a number between -pi/2
and pi/2 (exclusive) whose tangent is arg.
(log10 x)
== (log x 10)
, at least approximately.
A deterministic computer program cannot generate true random numbers. For most purposes, pseudo-random numbers suffice. A series of pseudo-random numbers is generated in a deterministic fashion. The numbers are not truly random, but they have certain properties that mimic a random series. For example, all possible values occur equally often in a pseudo-random series.
In Emacs, pseudo-random numbers are generated from a "seed" number.
Starting from any given seed, the random
function always
generates the same sequence of numbers. Emacs always starts with the
same seed value, so the sequence of values of random
is actually
the same in each Emacs run! For example, in one operating system, the
first call to (random)
after you start Emacs always returns
-1457731, and the second one always returns -7692030. This
repeatability is helpful for debugging.
If you want truly unpredictable random numbers, execute (random
t)
. This chooses a new seed based on the current time of day and on
Emacs's process ID number.
If limit is nil
, then the value may in principle be any
integer. If limit is a positive integer, the value is chosen to
be nonnegative and less than limit (only in Emacs 19).
If limit is t
, it means to choose a new seed based on the
current time of day and on Emacs's process ID number.
On some machines, any integer representable in Lisp may be the result
of random
. On other machines, the result can never be larger
than a certain maximum or less than a certain (negative) minimum.