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basic operations task2
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elijah vasquez
basic operations task2
Commits
5b1e3d42
Commit
5b1e3d42
authored
Sep 28, 2022
by
elijah vasquez
🦍
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5b1e3d42
##include <string.h>
##include <stdio.h>
/* Print binary stored in plain 32 bit block */
void
intToBinary
(
unsigned
int
n
)
{
int
c
,
k
,
i
;
printf
(
"How the number is stored in memory "
);
for
(
c
=
31
;
c
>=
0
;
c
--
)
{
k
=
n
>>
c
;
// k = n dividedby 2 c times
// or
// take the binary representation of k and move all the digits to the right c times
// since the loop c starts at 31 the first round of the loop moves the left most bit to the units column ( and rottes the rest round)
// we can then use a bitwise and (&) command to compare the righmost digit (previously the leftmost) to 000000000001
// if (k & 1) is true then that bit is 1 otherwise it is zero
// we can then move on to the next loop and look at the second leftmost and so on.
if
(
k
&
1
)
printf
(
"1"
);
else
printf
(
"0"
);
}
printf
(
" X * 2 ^ Y
\n
"
);
}
int
main
(
void
)
{
unsigned
int
m
;
float
f
=
3
.
14
;
printf
(
"
\n\n\n
f = %f
\n
"
,
f
);
/* See hex representation */
printf
(
"Float Point Representation of f = %a"
,
f
);
printf
(
" hexadecimal X * 16 ^ Y
\n
"
);
/* Copy memory representation of float to plain 32 bit block */
memcpy
(
&
m
,
&
f
,
sizeof
(
m
));
intToBinary
(
m
);
printf
(
"
\n\n\n
"
);
return
0
;
}
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