# 14 ^ 23

Yogesh Chhawasaria

Ranch Hand

Posts: 53

posted 12 years ago

Hi This was at the javaprepare test

14 ^ 23 actual ans was 25

but i reached 37 as follows

1 - 0001

4 - 0100

2 - 0010

3 - 0011

So 14 0001 0100

^

23 0010 0011

-------------------------

0011 0111

3 7

Can anyone do a postmortem of this.

14 ^ 23 actual ans was 25

but i reached 37 as follows

1 - 0001

4 - 0100

2 - 0010

3 - 0011

So 14 0001 0100

^

23 0010 0011

-------------------------

0011 0111

3 7

Can anyone do a postmortem of this.

When you have eliminated all which is impossible, then whatever remains, however improbable, must be the truth.

Corey McGlone

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Posts: 3271

posted 12 years ago

A single hexadecimal digit is equivalent to 4 binary digits. That's because 16 is equivalent to 2^4. However, one binary digit

14 decimal = 1110 binary

23 deimcal = 10111 binary

14 ^ 23 = 1110 ^ 10111 = 11001 = 25

I hope that helps,

Corey

**is not**equivalent to 4 binary digits because 10 is not equivalent to 2^4. Therefore, your method of converting decimal digits to binary digits is flawed.14 decimal = 1110 binary

23 deimcal = 10111 binary

14 ^ 23 = 1110 ^ 10111 = 11001 = 25

I hope that helps,

Corey

posted 12 years ago

In other words, values are not stored how you apparently think they are. It is not true that each digit is stored in it's own 4 bytes, and then they are pushed together. What you are describing is Binary Coded Decimal (and i have a clock on my desk that tells time that way).

in Java (and other languages), the number is stored in base 2, or binary. each position, as we move from the least significant up, represents an increasingly larger power of 2. so, the least sig. digit represents 2^0 (that's 2 raised to the 0th power), the next over is 2^1, then 2^2... or in other words 1, 2, 4... etc.

there are many ways to convert from 14 base 10 into base 2. if the numbers are small enough, i just start figuring out which powers of 2 i need to add up to the value i want.

for 14, i need 8 + 4 + 2. those are 2^3, 2^2, and 2^1, so i would have

00001110.

for 23, i need 16 + 4 + 2 + 1, or 2^4 + 2^2 + 2^1 + 2^0. that give us

00010111

in Java (and other languages), the number is stored in base 2, or binary. each position, as we move from the least significant up, represents an increasingly larger power of 2. so, the least sig. digit represents 2^0 (that's 2 raised to the 0th power), the next over is 2^1, then 2^2... or in other words 1, 2, 4... etc.

there are many ways to convert from 14 base 10 into base 2. if the numbers are small enough, i just start figuring out which powers of 2 i need to add up to the value i want.

for 14, i need 8 + 4 + 2. those are 2^3, 2^2, and 2^1, so i would have

00001110.

for 23, i need 16 + 4 + 2 + 1, or 2^4 + 2^2 + 2^1 + 2^0. that give us

00010111

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Corey McGlone

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Posts: 3271

Yogesh Chhawasaria

Ranch Hand

Posts: 53

posted 12 years ago
When you have eliminated all which is impossible, then whatever remains, however improbable, must be the truth.

Thanks that was great explanation

So what i understand is this

17 is (16 + 1) i.e. 00010001

12 is (8+4) i.e 00001100

and 17 ^ 12 is 00010001

00001100

------------------------------

00011101 (1+4+8+16) = 29

Gottit

Thanks a lot.

So what i understand is this

17 is (16 + 1) i.e. 00010001

12 is (8+4) i.e 00001100

and 17 ^ 12 is 00010001

00001100

------------------------------

00011101 (1+4+8+16) = 29

Gottit

Thanks a lot.