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 that is multiplied by 1.5. Floating point values are not exact; they have a fixed, limited amount of precision.
The range of values for an integer depends on the machine. The minimum range is -134217728 to 134217727 (28 bits; i.e., to but some machines may provide a wider range. Many examples in this chapter assume an integer has 28 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. 268435457 ; 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 28-bit binary, the decimal integer 5 looks like this:
0000 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 1111
-1 is represented as 28 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 1111 1011
In this implementation, the largest 28-bit binary integer value is 134,217,727 in decimal. In binary, it looks like this:
0111 1111 1111 1111 1111 1111 1111
Since the arithmetic functions do not check whether integers go outside their range, when you add 1 to 134,217,727, the value is the negative integer -134,217,728:
(+ 1 134217727)
=> -134217728
=> 1000 0000 0000 0000 0000 0000 0000
Many of the functions described in this chapter accept markers for arguments in place of numbers. (See section Markers.) Since the actual arguments to such functions may be either numbers or markers, we often give these arguments the name number-or-marker. When the argument value is a marker, its position value is used and its buffer is ignored.
Floating point numbers are useful for representing numbers that
are not integral. 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 you are using.
The read-syntax 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 Emacs Lisp doesn't try to distinguish them.
Here are the read syntaxes for these special floating point
values:
In addition, the value -0.0 is distinguishable from
ordinary zero in IEEE floating point (although equal
and = consider them equal values).
You can use logb to extract the binary exponent of
a floating point number (or estimate the logarithm of an
integer):
(logb 10)
=> 3
(logb 10.0e20)
=> 69
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 to =
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.
Sometimes it is useful to compare numbers with
equal; it treats two numbers as equal if they have the
same data type (both integers, or both floating point) and the same
value. By contrast, = can treat an integer and a
floating point number as equal.
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)
(or (and (= x 0) (= y 0))
(< (/ (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 uses the kind
of division operation that corresponds to mod,
rounding downward. 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 Emacs Lisp, arithmetic functions
do not check for overflow. Thus (1+ 134217727) may
evaluate to -134217728, 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, if we continue,
foo
=> 4
If you want to increment the variable, you must use
setq, like this:
(setq foo (1+ foo))
=> 5
+ returns
0.
(+)
=> 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 more-numbers-or-markers from
number-or-marker, cumulatively. If there are no
arguments, the result is 0.
(- 10 1 2 3 4)
=> 0
(- 10)
=> -10
(-)
=> 0
* returns 1.
(*)
=> 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 an integer by 0,
an arith-error error is signaled. (See section Errors.) Floating point division by
zero returns either infinity or a NaN if your machine supports IEEE
floating point; otherwise, it signals an arith-error
error.
(/ 6 2)
=> 3
(/ 5 2)
=> 2
(/ 5.0 2)
=> 2.5
(/ 5 2.0)
=> 2.5
(/ 5.0 2.0)
=> 2.5
(/ 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. For floor, see
section Numeric Conversions.
The functions ffloor, fceiling,
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; fceiling, 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.
Shifting a 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 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 one place to the right divides the value of a positive integer by two, rounding downward.
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 134,217,727 produces -2 on a 28-bit
machine:
(lsh 134217727 1) ; left shift
=> -2
In binary, in the 28-bit implementation, the argument looks like this:
;; Decimal 134,217,727 0111 1111 1111 1111 1111 1111 1111
which becomes the following when left shifted:
;; Decimal -2 1111 1111 1111 1111 1111 1111 1110
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 ones in the empty bit
positions on the left, while lsh puts zeros in those
bit positions.
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 1111 1010
=>
1111 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) => 134217725
;; Decimal -6 becomes decimal 134,217,725.
1111 1111 1111 1111 1111 1111 1010
=>
0111 1111 1111 1111 1111 1111 1101
Here are other examples:
; 28-bit binary values
(lsh 5 2) ; 5 = 0000 0000 0000 0000 0000 0000 0101
=> 20 ; = 0000 0000 0000 0000 0000 0001 0100
(ash 5 2)
=> 20
(lsh -5 2) ; -5 = 1111 1111 1111 1111 1111 1111 1011
=> -20 ; = 1111 1111 1111 1111 1111 1110 1100
(ash -5 2)
=> -20
(lsh 5 -2) ; 5 = 0000 0000 0000 0000 0000 0000 0101
=> 1 ; = 0000 0000 0000 0000 0000 0000 0001
(ash 5 -2)
=> 1
(lsh -5 -2) ; -5 = 1111 1111 1111 1111 1111 1111 1011
=> 4194302 ; = 0011 1111 1111 1111 1111 1111 1110
(ash -5 -2) ; -5 = 1111 1111 1111 1111 1111 1111 1011
=> -2 ; = 1111 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.
; 28-bit binary values
(logand 14 13) ; 14 = 0000 0000 0000 0000 0000 0000 1110
; 13 = 0000 0000 0000 0000 0000 0000 1101
=> 12 ; 12 = 0000 0000 0000 0000 0000 0000 1100
(logand 14 13 4) ; 14 = 0000 0000 0000 0000 0000 0000 1110
; 13 = 0000 0000 0000 0000 0000 0000 1101
; 4 = 0000 0000 0000 0000 0000 0000 0100
=> 4 ; 4 = 0000 0000 0000 0000 0000 0000 0100
(logand)
=> -1 ; -1 = 1111 1111 1111 1111 1111 1111 1111
logior is passed just one argument, it returns that
argument.
; 28-bit binary values
(logior 12 5) ; 12 = 0000 0000 0000 0000 0000 0000 1100
; 5 = 0000 0000 0000 0000 0000 0000 0101
=> 13 ; 13 = 0000 0000 0000 0000 0000 0000 1101
(logior 12 5 7) ; 12 = 0000 0000 0000 0000 0000 0000 1100
; 5 = 0000 0000 0000 0000 0000 0000 0101
; 7 = 0000 0000 0000 0000 0000 0000 0111
=> 15 ; 15 = 0000 0000 0000 0000 0000 0000 1111
logxor is passed just one argument, it returns that
argument.
; 28-bit binary values
(logxor 12 5) ; 12 = 0000 0000 0000 0000 0000 0000 1100
; 5 = 0000 0000 0000 0000 0000 0000 0101
=> 9 ; 9 = 0000 0000 0000 0000 0000 0000 1001
(logxor 12 5 7) ; 12 = 0000 0000 0000 0000 0000 0000 1100
; 5 = 0000 0000 0000 0000 0000 0000 0101
; 7 = 0000 0000 0000 0000 0000 0000 0111
=> 14 ; 14 = 0000 0000 0000 0000 0000 0000 1110
(lognot 5)
=> -6
;; 5 = 0000 0000 0000 0000 0000 0000 0101
;; becomes
;; -6 = 1111 1111 1111 1111 1111 1111 1010
These mathematical functions 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 a positive integer, the value is chosen to be nonnegative and less than limit.
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.