Advanced Fortran Programming 007 Binary; Octal and Hexadecimal interpretations

hello guys my name is Aaron and welcome
to my channel this series is a series of
tutorials on advanced Fortran
programming now in this tutorial I just
thought of mentioning one little one
little thing I forgot to mention the
last time and after that we'll go into a
new program okay now suppose if you have
two values let's say and you want to
combine this to form a complex number
there is an inbuilt intrinsic function
available Fortran and that is given by C
and plx okay now this function reality
match we put in complex number and from
which the real part is extracted and the
imaginary part is extracted and they
unused okay this function is the exact
opposite where and you put plug in the
real part and the imaginary part values
for the a real and imaginary part and
this complex function this.kc MPX
functional complex function majesties to
to produce complex number out of it and
then saves it to see now what I did is
there exactly the same program that we
saw last time except that I added an I
added that missing variable C and if you
guys notice it was initially a b and b I
needed you see here okay and then what I
did is there using this complex function
I passed a very available size seven ten
to eight point nine so C is actually
complex of five point seven and eight
point and so what would be is that C
will be printed as five point seven plus
eight point nine okay so to show to show
that I just work properly I'm come
printing it here so I compiled it build
it and executing it just ask me for
values like that there's enter these two
okay that's not a part that's not very
important here the important thing is
over here then we entered CM plx five
point seven comma eight point nine it C
is getting printed here's five point six
nine and something in eight point eight
point eight nine and something
apparently this is some kind of a
storing issue I suppose I have no idea
why it's kind of mess going up messed up
like this it's something I have to know
maybe I have to consider I have no I
don't know why it is and I'll let me
think about it if I if I find an answer
for you guys as to why this kind of
stuff happens I let you guys do not in
one of the upcoming tutorials okay
and that okay that being said that's
good let's go to the main program that
main power program that I thought of
explaining guys today okay and this
programs written over here what this
program does is that this program that
helps you to work with octal binary and
hexadecimal numbers okay now I'll be all
this time you are working at the decimal
decimal system man and we have our where
the number digits are between 0 to 9 and
the numerals are 0 tonight and we'll and
we were using them and we were using
them to express integers and the real
numbers and so on and so forth okay now
what you can do is that in Fortran just
like see you can use the formatting
features in Fortran to print to print
out display the real number real numbers
and the binary number binary equivalence
of the numbers of the real numbers and
integers and they use them accordingly
now what if they what I've done is that
this is that I have introduced to way I
already in since I mean
initialize two variables I equal 67 and
J equals forty five point six eight just
random numbers nothing much there's no
special dinner just they're just random
numbers I is an integer and J is real
tight and what I did is that I'm just
using the print statement and
right statement to get there Alicia what
it what happens is that if you want to
get the real of real parts and imaginary
parts and uh you said you just want to
get them printed with didn't want to get
them printed in binary formats or other
octal formats or something of that sort
what do you need is I already need is
actually the format the format specifier
if I had to write this in sup write this
in print it's a little bit it's a little
annoying so what I did is that I used
this write function how does what you do
is that yourself to use what do you use
what do you call
yes I'd write this far matting within
the print statement and go about with it
are you sir yet write a separate formal
statement again and again and go about
with it so I thought you know this is
easy because this is easy for easy
business or thought we'll go ahead with
this right statement but doesn't matter
okay this first line we are printing the
real J in realice okay and a printing J
in the field of 14 characters in the
feel of 14 spaces with H various places
does designed for I mean I like
allocated or assigned for the via in
dimension in decimal part of it and here
when I write that I write this number in
binary I'm using this far I'm using this
specifier be okay and fortify so far the
using 45 spaces I'm printing the outer
value here using this oh and the
hexadecimal with this character Z or Z
okay and this 20 these are just numbers
this spaces or stuff I'll explain you
guys why it does and similarly for
integers I'm using the same formatting
okay and I'm printing the I if instead
of the real part I'm just writing it as
I four because this number is an integer
so the real the real formatting will
should not be done if it should not be
done nothing like nothing more there
were two compile this builders and
execute this okay let me move this to
the screen over here okay the real J is
forty five point six eight something
fine the real Jane binary is this one
a hell of a number one big hell of a
number okay and if you guys noticed
let me just count this as 4 8 12 16 20
24 24 28 31 that 24 28 31 there are like
31 value 31 digits over here and and
they actually if they're not as s it's
not actually 31 it's actually 32 because
there there is a small value over here
which is not printed it's actually this
number is actually 32 did 32 digits long
so the binary number corresponding to
forty five point six head is very two
digits long but you should be asking me
so this is the binary where is the
rematch as a real number part of it
exactly where is the real number part of
it I mean the point of the decimal point
now you have to keep in mind that then
while number while floating point
numbers are actually stored and computer
they do not they get stored in a
peculiar manner okay what what I -
please stand it says another stand
that's okay okay I'll do one thing let
me just keep this aside okay I will I
started not over here okay there is this
I Triple E format okay like something
with all the I Triple E or I mean there
might be sadder farmers to store real
numbers okay how do they do is that they
calculate the number of bytes so let's
say for let's say we take a four byte
let's say we take a four byte float okay
you have 32 bits since it's a four by
four by float you have a 32 bits and now
among this 32 bit something like say
just giving just giving a wild number
just for you guys to understand okay
something like once one bit
does used for sign convention okay
something like let's say 15 bit 15 bits
for mantissa mantissa rest equals
exponent this is how the order it goes
now I will explain you guys this is
experiment in power of 2 exactly this is
how this is how it goes ok for an 8-bit
integer 8 byte in the eight byte float
ok there it will be like 64 bits right
okay if I remember properly one bit is
for the sign convention and this bit is
actually the leftmost bit okay so this
here again leftmost bit same here and
then something like 52 by bitch 52 bits
from antis I suppose and other 11 bits
I'm not sure as to what I mean these are
okay let me just try this that means I
write it here my name is guessing
guessing so I mean I might be wrong here
so I'll just using this as guessing here
and then I'm like this as exponent again
this is also guessing okay so don't take
this too literally
but this but I know the values might
slightly changed if you guys look take
this out of Wikipedia or any other stuff
any other way any other place in the
internet well - I know for to know how
decimal numbers means how floaty numbers
are stored what is what did get stored
is that each part each part of the
mantissa stored separately an extra on
the store separately okay in this kind
of a notation 50 in services 64-bit
integer right eight
a bit flawed number what happens is that
among all the 64-bit RIA which are
available the leftmost bit of the 8 8
bytes of the 8 bytes is used for storing
the sign 0 for positive and the 1 for
negative something I'm like something of
that sort
and then subsequently the next
subsequent 52 bits might be used for
storing the X mantissa 52 or something
of that sort
this man decides something like point 2
3 something like that ok and the
remaining 11 bits is used for storing
and storing the exponent but this X the
power of their exponent but this power
of the experiment is is that in power in
power of 2 so what you have is that
something like example example to let
something like let's say zero point 2 5
6 7 8 9 something of that sort and this
will be this will be something like 26
or 27 or with the night with with a plus
or minus negative sense this will be 26
or 27 or 22 or something okay so this is
27 this is actually 2 power 27 now so
the number is actually here it is 2.5
2.5 6 3 9 8 times the 2 power 27 which
will be approximated to based it which
will approximate to a base then value
okay and this man this number stored as
the in binary and this part this int
exponent is stored as a binary and
that's our job that's how it goes so
that's the reason why then you print
this J with binary for with the real
number formal gets in a printed nicely
but when you print J in a decimal form
and if you get to print the J I mean
real numbers printed I see but if you
print a in binary format okay it just
prints only the only a proper bind
without any decimals okay no you mind by
asking me could this be because of the
because the fact that I didn't use any
decimal points here well let's try it
out I'll just use just for this mother
that'll mean you seven over here and
then something like seven over here for
octave and seven over here for except
decimal but guys not this B for binary
or for octal and zz4 except the symbol
x decimal and I let me cart it over here
okay hexadecimal okay now if I compile
this build this and execute this okay I
think I already have sorry
we're checking a few stuff yeah I think
there was a small issue but even when
even when the decimal points are adders
they're just working the same they're
like they're like exactly the same way
it's because this binary number is being
converted into an this entire binary
number when you express this as a real
number okay that real number conventions
of using the certain pass for mantissa
certain paths for exponent and all takes
into consideration and then that number
is print that's kind of inconvenient and
it's done over here where this where as
the octal number the same number is
actually converted into its equal and
octal format and gets printed over this
binary number gets converted into its
equal and octal format and gets printed
here and similarly this binary number
gets converted will equal equal to xn xn
decimal number and gets printed here
simple as that now for integers it's
simple if I were to run this again okay
fine digits it's very fairly
straightforward what it does is that it
just kind of since there aren't any
complicated not complicated notions like
what you see on what you see for
integers oh sorry Odyssey for floors
it's fairly simple
fairly simple it just takes the number
and converts it converts into equal and
equal binary notation and then prints it
the only thing is that if you guys
notice it will just print the four five
six seven digits here this last digit
there is actually one more digit this
digit takes care of the sign convention
sign convention and nothing more okay
suppose if I were to know show you guys
how this actually works they didn't put
a negative sign over here and then I put
a negative sign over here if I were to
compile build and execute this yeah
perfect perfect
now now you guys notice the now you guys
notice a discrepancy over here since I
put a negative sign over here the
numbers if you just compare this value
and the previous value which are shown
this octal number and this exodus number
will be different entirely different and
if you guys notice there will be like
for 816 for 812 sorry now it's 16 20 24
28 and 32 digits properly here the
output value in decimal value are fine
different whereas in the binary value
you have you're having some trouble here
okay it's not printing it and properly
will digest it accordingly now the octal
number gets different there because I
put a one on the top the entire value
gets different so it needs lot more
space little more than 20 pitches to
explain what it is and the octal number
is different and the hexadecimal is
actually different all right
that's that's how it is so what did what
fortran does is that it does not
literally convert the number into its
binary format and print binary format or
exodus well former an octal formatted
printed rather it just takes the binary
format in the memory binary format in
the memory uses the standard conventions
used for converting it to an integer or
real and does just that job and prints
it whereas for the octal and decimal
octal and hexadecimal pad it literally
takes the
after a binary value and excessive value
on an ASUS basis and prints them and
prints them and actually prints them
accordingly so I said no for I think
Fortran does not convert does not
convert smartly converters but them I
think there might be options to do that
do that but it's worth but it's worth
experimenting these this is actually a
very less important but this actually
explains one important concept as hot as
to how the valleys are being stored in
the computer well at least have been
stored and the memory of the memory of
Fortran this is something important
something important over here that's why
that's one of the reasons why I wanted
to explain this program okay now that's
alright that's all I have for you guys
to mr. Durell thank you guys for
watching and see you guys next tutorial
and in the next tutorial we'll be
working on some new more advanced
concepts like for all for all statement
and some process some procedures or no
okay thank you guys for watching and see
you guys next time