On our machine
storm.cis.fordham.edu
,
a memory address is customarily written in hexadecimal notation.
Each place has 10 times the value of the previous one.
Depending on the number,
we might need to write ten decial digits:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
1000’s place | 100’s place | 10’s place | 1’s place |
---|---|---|---|
2 | 0 | 2 | 5 |
2 × 1000 = 2000 0 × 100 = 0 2 × 10 = 20 5 × 1 = 5 2025
Each place has 2 times the value of the previous one.
The only binary digits we will ever need to write are 0 and 1.
1024’s place | 512’s place | 256’s place | 128’s place | 64’s place | 32’s place | 16’s place | 8’s place | 4’s place | 2’s place | 1’s place |
---|---|---|---|---|---|---|---|---|---|---|
1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
1 × 1024 = 1024 1 × 512 = 512 1 × 256 = 256 1 × 128 = 128 1 × 64 = 64 1 × 32 = 32 0 × 16 = 0 1 × 8 = 8 0 × 4 = 0 0 × 2 = 0 1 × 1 = 1 2025
A binary digit (0 or 1) is called a bit.
So our number 2025 consists of a series of 11 bits when written in binary.
A series of 8 bits is called a
byte.
A series of 4 bits is called a
nibble.
We will usually consider bits in groups of 4 or 8 at a time,
so let’s expand our 11-bit number to 16 bits
by adding 5 leading 0’s.
Now our number 2025 consists of 2 complete bytes, because 2 = 16/8.
32768’s place | 16384’s place | 8192’s place | 4096’s place | 2048’s place | 1024’s place | 512’s place | 256’s place | 128’s place | 64’s place | 32’s place | 16’s place | 8’s place | 4’s place | 2’s place | 1’s place |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
0 × 32768 = 0 0 × 16384 = 0 0 × 8192 = 0 0 × 4096 = 0 0 × 2048 = 0 1 × 1024 = 1024 1 × 512 = 512 1 × 256 = 256 1 × 128 = 128 1 × 64 = 64 1 × 32 = 32 0 × 16 = 0 1 × 8 = 8 0 × 4 = 0 0 × 2 = 0 1 × 1 = 1 2025
As we just saw, writing the value of a number in binary usually takes many bits. That’s why they invented hexadecimal digits. Each hexadecimal digit (hex digit) stands for a series of 4 binary digits (bits). In other words, each hex digit stands for 1 nibble.
Here are the 16 nibbles that the 16 hex digits stand for.
hex digit | nibble (4 bits) |
---|---|
0 |
0000 |
1 |
0001 |
2 |
0010 |
3 |
0011 |
4 |
0100 |
5 |
0101 |
6 |
0110 |
7 |
0111 |
8 |
1000 |
9 |
1001 |
A |
1010 |
B |
1011 |
C |
1100 |
D |
1101 |
E |
1110 |
F |
1111 |
We can now write out 16-bit number
(2025 = 0000011111101001
)
with only 4 hex digits (07E9
):
32768’s place | 16384’s place | 8192’s place | 4096’s place | 2048’s place | 1024’s place | 512’s place | 256’s place | 128’s place | 64’s place | 32’s place | 16’s place | 8’s place | 4’s place | 2’s place | 1’s place | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
binary | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
hex | 0 | 7 | E | 9 |
Type control-d
into the first two programs
to signal that you’re done typing.
dectohex.C
:
convert base 10 to base 16.
hextodec.C
:
convert base 16 to base 10.
table.C
:
decimal (base 10) and hexadecimal (base 16), side by side.
address.C
:
output the address of a variable in hexadecimal.
The value of i is 10 The address of i is 0x7fff2757276c The number of bytes in i is 4
pointer.C
:
store the address of a variable in another variable,
called a pointer.
dereference.C
:
use the unary
*
operator to dereference a “pointer to an
int
”
to get the value of the original
int
.
*
operator means multiplication.)
struct.C
:
use the unary
*
operator to
dereference a “pointer to a structure”
to get the value of each field of the original structure.
increment
:
when you increment a pointer with ++
,
how much does the pointer increase?
loop.C
:
loop through an array of int
s
with a pointer to an int
.
begin
and end
really mean.
In
loop.C
,
change &a[0]
to begin(a)
,
and
change &a[n]
to end(a)
.
movingaverage.C
:
use a pointer to concentrate on three consecutive elements of an array.
reference2.C
:
a reference argument makes it hard to see that a function could change
the value of one of its actual arguments.
reference3.C
:
a pointer argument makes it obvious that a function could
change the value of one of its actual arguments.
minimum.C
:
pass the address of the first element of the array,
and the address that is just after the last element.
a
to a function,
let the formal arguments be the address of the first element,
and the address of just beyond the last element.
Then use
begin(a)
and
end(a)
as the actual arguments.
sort
that takes the same two formal arguments as the function in
minimum.C
.
The sort
function will sort the elements of the array into increasing order,
using the algorithm we saw in
bubblesortint.C
.
The end. Thank you.