Literals

A literal is a value written straight into the source. Integers can be written in four bases, floating-point numbers with a point or an exponent, and a suffix can pin a literal to a particular type.

'' Integers in four bases. The prefix names the base.
print "decimal 255        ="; 255
print "hex     &hFF       ="; &hFF
print "octal   &o377      ="; &o377
print "binary  &b11111111 ="; &b11111111

'' Floating-point literals: a point, an exponent, or both.
print "1.5     ="; 1.5
print "2.5e3   ="; 2.5e3
print "1e-4    ="; 1e-4
print ".25     ="; .25

'' A suffix fixes the literal's type; sizeof shows the width it chose.
print "sizeof( 1 )    ="; sizeof( 1 )
print "sizeof( 1L )   ="; sizeof( 1L )
print "sizeof( 1LL )  ="; sizeof( 1LL )
print "sizeof( 1UL )  ="; sizeof( 1UL )
print "sizeof( 1.5 )  ="; sizeof( 1.5 )
print "sizeof( 1.5F ) ="; sizeof( 1.5F )

'' Booleans and text.
print "true and false: "; true; " "; false
print "a ustring literal: "; "in double quotes"
print !"an escaped literal:\tafter a tab"

Output

decimal 255        = 255
hex     &hFF       = 255
octal   &o377      = 255
binary  &b11111111 = 255
1.5     = 1.5
2.5e3   = 2500
1e-4    = 0.0001
.25     = 0.25
sizeof( 1 )    = 8
sizeof( 1L )   = 4
sizeof( 1LL )  = 8
sizeof( 1UL )  = 4
sizeof( 1.5 )  = 8
sizeof( 1.5F ) = 4
true and false: true false
a ustring literal: in double quotes
an escaped literal:	after a tab

A literal with a point or an exponent (1.5, 2.5e3, .25) is floating point. tiko has no digit separator: 1_000_000 is error TKO0420: expected the end of the statement.

A plain whole-number literal is an integer, which is 8 bytes on the 64-bit targets; a plain floating-point literal is a float64. The suffixes change that:

Literal types shows how to see the type a literal gets, including one case that surprises people.

true and false are values of type boolean. Text goes in double quotes; with a ! in front, backslash escapes such as \t are read, so the last line prints a real tab. Strings has the full set.

Activities

1. Print &h7FFFFFFF and &b10000000 and check the values by hand.
2. Print cbool( &o17 = &b1111 ) and work out why the answer is what it is.