# binary to decimal

Paul Salerno

Ranch Hand

Posts: 172

posted 14 years ago

I'm getting two conflicting answers concerning the most significant bit:

The Math Doctor Says:

Check that against the decimal equivalent of 10110111 binary:

1 0 1 1 0 1 1 1

^ ^ ^ ^ ^ ^ ^ ^

| | | | | | | |_________> 1 x 2^0 = 1

| | | | | | |___________> 1 x 2^1 = 2

| | | | | |_____________> 1 x 2^2 = 4

| | | | |_______________> 0 x 2^3 = 8

| | | |_________________> 1 x 2^4 = 16

| | |___________________> 1 x 2^5 = 32

| |_____________________> 0 x 2^6 = 64

|_______________________> 1 x 2^7 = 128

183 decimal

However a JavaRanch user states:

the most significant byte being 1 or (2 ^ 7)=-128

therefore shouldnt the answer be = -73

I dont see the math doctor taking the most significant digit into accout. Whos right?

TIA

The Math Doctor Says:

Check that against the decimal equivalent of 10110111 binary:

1 0 1 1 0 1 1 1

^ ^ ^ ^ ^ ^ ^ ^

| | | | | | | |_________> 1 x 2^0 = 1

| | | | | | |___________> 1 x 2^1 = 2

| | | | | |_____________> 1 x 2^2 = 4

| | | | |_______________> 0 x 2^3 = 8

| | | |_________________> 1 x 2^4 = 16

| | |___________________> 1 x 2^5 = 32

| |_____________________> 0 x 2^6 = 64

|_______________________> 1 x 2^7 = 128

183 decimal

However a JavaRanch user states:

the most significant byte being 1 or (2 ^ 7)=-128

therefore shouldnt the answer be = -73

I dont see the math doctor taking the most significant digit into accout. Whos right?

TIA

Terence Doyle

Ranch Hand

Posts: 328

posted 14 years ago

Hi,

The difference between the two ( perfectly possible ) interpretations lie in if the number is signed or positive.

If it's signed then the most significant bit is considered to be negative and with 8 data bits (one byte )we would get a range from - 128 to + 127.

If the number is unsigned then the the range is from 0 to 255 inclusive.

Hope that helps,

The difference between the two ( perfectly possible ) interpretations lie in if the number is signed or positive.

If it's signed then the most significant bit is considered to be negative and with 8 data bits (one byte )we would get a range from - 128 to + 127.

If the number is unsigned then the the range is from 0 to 255 inclusive.

Hope that helps,

Raising Flares debut album 'Ignition' out now

http://www.raisingflares.com

Terry Doyle <br />SCPJ 1.4 , SCWCD , SCMAD(Beta)

Valentin Crettaz

Gold Digger

Sheriff

Sheriff

Posts: 7610

posted 14 years ago

from JLS 4.2 Primitive Types and Values

Thus when dealing with byte, short, int, and long the number should be treated as signed which means the high-level bit is always the sign bit.

When dealing with char the high-level bit is nothing, just another bit.

HIH

...

The integral types are byte, short, int, and long, whose values are 8-bit, 16-bit, 32-bit and 64-bitsignedtwo's-complement integers, respectively, and char, whose values are

16-bitunsignedintegers representing Unicode characters.

...

Thus when dealing with byte, short, int, and long the number should be treated as signed which means the high-level bit is always the sign bit.

When dealing with char the high-level bit is nothing, just another bit.

HIH

SCJP 5, SCJD, SCBCD, SCWCD, SCDJWS, IBM XML

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Erik Dark

Ranch Hand

Posts: 107

posted 14 years ago

I agree with the JavaRancher. The way I treat this:

1) a negative number is represented by inverting each bit of the corresponding

positive number and then adding 1 (twos complement form)

2) consider the 7 right bits as the number and the 1 most left as the sign

10110111 - negative (remember!) so go back trough the statements above:

*0110111 - just consider the 7 right bits

*0110110 - minus 1 (the reversed way)

*1001001 - invert and start counting: 64+0+0+8+0+0+1=73, get the sign back: -73

Erik Dark

1) a negative number is represented by inverting each bit of the corresponding

positive number and then adding 1 (twos complement form)

2) consider the 7 right bits as the number and the 1 most left as the sign

10110111 - negative (remember!) so go back trough the statements above:

*0110111 - just consider the 7 right bits

*0110110 - minus 1 (the reversed way)

*1001001 - invert and start counting: 64+0+0+8+0+0+1=73, get the sign back: -73

Erik Dark