Answer:
nope u gonna fail ahhahaha
Explanation:
my dead cat ran across my keyboard
Answer:
65535 x 32768 =214 7450880
which is in the range of 32 bit unsigned integers, so there will be no overflow and we wont be in a need of more bits than 32 bit.
Explanation:
An overflow error indicates that software attempted to write down data beyond the bounds of memory.
Each program features a section of memory allocated for a stack. The stack is employed to store internal data for the program and is extremely fast and keep track of return addressing. A program may jump to a neighborhood that reads some data from the disk drive, then it returns from that routine to continue processing the info. The stack keeps track of the originating address, and therefore the program uses that for the return. Kind of like leaving breadcrumbs to seek out your way back. The stack features a limited amount of space for storing. If software attempts to access a neighborhood of the stack beyond its limits, an overflow error occurs.
• signed 32-bit integers support the range [-2147483648,2147483647]
• unsigned 32-bit integers support the range [0,4294967295]
And if you go outside this range, even temporarily, you would like to be very careful. Most cases handle the overflow technically and provide you "wraparound" or "modulo" behavior where carry bits outside the range just disappear and you're left with the low-order bits like the precise result.
In computer architecture of 32 bit, memory addresses, or other data units are those who are 32 bits (4 octets) wide. Also, 32-bit CPU and ALU architectures are those who have supported registers, address buses, or data buses of that size. 32-bit microcomputers are computers during which 32-bit microprocessors are the norm.
Answer:
Here is the Python program:
def exec_2(f):
f()
f()
def exec_4(f):
exec_2(f)
exec_2(f)
def print_koala():
print("koala")
exec_4(print_koala)
Explanation:
The program works as follows:
First method:
def exec_2(f):
f()
f()
any function that is passed into exec_2() will be executed twice without parameters.
Second method:
def exec_4(f):
exec_2(f)
exec_2(f)
function exec_4(f) takes in a function f as a parameter and executes the function f 4 times using the function exec_2. At first exec_2(f) call the function f is executed twice and and second exec_2(f) call the function f is again executed twice so the function f executes 4 times in total.
Third method:
def print_koala():
print("koala")
a function that prints out the koala once
This function has a print statement which is passed the string koala to print on output screen.
exec_4(print_koala)
function exec_4 is used to print koala four times. This function exec_4 takes as argument the function print_koala() So this works as follows:
exec_4(print_koala)
The body of exec_4 executes as:
exec_2(print_koala)
exec_2(print_koala)
Now at first call to exec_2(print_koala) the program moves to the function exec_2 which executes print_koala() method twice. Since print_koala() method has a print statement print("koala") so this is executed twice:
koala
koala
Now at second call exec_2(print_koala) the program moves to the function exec_2 which executes print_koala() method twice again. Since print_koala() method has a print statement print("koala") so this is executed again twice:
koala
koala
So the output of the entire above program is:
koala
koala
koala
koala
The screenshot of program along with its output is attached.