Answer:
for(String s:words)
if(s.endsWith("ing"))
System.out.println(s);
Explanation:
Create an enhanced for loop that iterates through the words array
Check if an element in words ends with "ing" using endsWith() method (Since it is said that strings are lowercase letters, we do not need to check it)
If you find one that ends with "ing", print the element
Answer:
Offset bits: 3-bits
Set number of cache: 12-bits
Tag bits: 7-bits
22-bit physical address
Explanation:
Since the system is 32K so,
=2⁵.2¹⁰
=2¹⁵
As we know that it is 8-way set associative so,
=2¹⁵/2³
=2¹⁵⁻³
=2¹²
2¹² are cache blocks
22-bit physical address
Off-set bits are 3 as they are calulated from 8-way set associative information.
Set number of cache : 12-bits
For tag-bits:
Add off-set bits and cache bits and subtract from the total bits of physical address.
=22 - (12+3)
=22 - 15
=7
To accomplish this without using a loop,
we can use math on a string.
Example:
print("apple" * 8)
Output:
appleappleappleappleappleappleappleapple
In this example,
the multiplication by 8 actually creates 8 copies of the string.
So that's the type of logic we want to apply to our problem.
<span>def powersOfTwo(number):
if number >= 0:
return print("*" * 2**number)
else:
<span>return
Hmm I can't make indentations in this box,
so it's doesn't format correctly.
Hopefully you get the idea though.
We're taking the string containing an asterisk and copying it 2^(number) times.
Beyond that you will need to call the function below.
Test it with some different values.
powersOfTwo(4) should print 2^4 asterisks: ****************</span></span>
Answer:
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