Texas Instruments TI-89 Titanium User Manual page 840

Return
CATALOG
expression
Return [
Returns
expression
within a
Func
...
Prgm EndPrgm
Note: Use
program.
Note: Enter the text as one long line on the Home
screen (without line breaks).
right()
MATH/List menu
]) ⇒
list1 num
right( [,
Returns the rightmost
.
list1
If you omit
sourceString
right( [,
Returns the rightmost
character string
If you omit
comparison
right( )
Returns the right side of an equation or inequality .
rotate()
MATH/Base menu
integer1 [ #ofRotations ]
rotate( ,
Rotates the bits in a binary integer. You can enter
in any number base; it is converted
integer1
automatically to a signed, 32-bit binary form. If the
magnitude of
symmetric modulo operation brings it within the
range.
If
#of Rotations
#of Rotations
The default is ë 1 (rotate right one bit).
For example, in a right rotation:
Each bit rotates right.
0b00000000000001111010110000110101
Rightmost bit rotates to leftmost.
produces:
0b10000000000000111101011000011010
The result is displayed according to the
Appendix A: Functions and Instructions
]
as the result of the function. Use
... block, or
EndFunc
block.
without an argument to exit a
Return
list
elements contained in
num
, returns all of .
num list1
⇒ string
num
])
characters contained in
num
.
sourceString
, returns all of
num sourceString
⇒ expression
) ⇒
integer
is too large for this form, a
integer1
is positive, the rotation is to the left. If
is negative, the rotation is to the right.
Define factoral(nn)=Func
:local answer,count:1! answer
:For count,1,nn
:answerù count! answer:EndFor
:Return answer:EndFunc ¸ Done
factoral(3) ¸
right({1,3,ë 2,4},3) ¸
right("Hello",2) ¸
.
right(x<3) ¸
In Bin base mode:
rotate(0b1111010110000110101)
¸
0b10000000000000111101011000011010
rotate(256,1) ¸ 0b1000000000
In Hex base mode:
rotate(0h78E) ¸
rotate(0h78E,ë 2) ¸0h800001E3
rotate(0h78E,2) ¸
Important: To enter a binary or hexadecimal
number, always use the 0b or 0h prefix (zero,
not the letter O).
mode.
Base
6
{3 ë 2 4}
"lo"
3
0h3C7
0h1E38
837