String to binary converter

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1
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I did research about binary counting, and wanted to make an application in Java that converts a string to binary code:

 /**
 * This class converts strings by looping through the characters, gets the
 * decimal code for each, then divides the number by 2 untill the quotient is 0
 * 
 * @author Mr Pro Pop
 *
 */
public class Binary 

 /**
 * This function takes a string as a parameter and converts it to binary code
 * 
 * @param string The string to be converted to binary code
 * @return The binary code of the input string
 */

 public String getBinaryCode(String string) 

 if (string.length() == 0)
 return null;

 char characters = string.toCharArray();
 StringBuilder result = new StringBuilder();
 int number = 0;

 for (int i = 0; i < characters.length; i++) 
 number = getDecimal(characters[i]);
 for (int j = 0; j < 8; j++) 
 if (number % 2 == 0) 
 result.insert(i, 0);
 else 
 result.insert(i, 1);
 
 number = (int) (number / 2);
 
 result.append(" ");
 
 return result.toString();
 

 /**
 * This function gets the decimal number of a character
 * 
 * @param character The character that we want the decimal value of
 * @return The decimal of the character from the ascii table
 */
 public int getDecimal(char character) 
 for (int i = 0; i <= 255; i++) 
 if (character == (char) i) 
 return i;
 
 
 return -1;
 

 public static void main(String args) 
 String string = "a";
 Binary b = new Binary();
 System.out.println("Binary code for" + string + " is " + b.getBinaryCode(string));

Output:

Binary code for a is 01100001

Here it works well, however, spacing doesn't work. I do acknowledge that I could simply code this in a different easier method by Integer.toBinaryString(character[i]) but tried doing it my way to learn and especially with this method.

Any improvements I could do or anything to pay attention for?

edited 10 mins ago

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124k14144401

asked 2 hours ago

Mr Pro Pop

1204

add a commentÂ |Â

up vote
1
down vote

favorite

I did research about binary counting, and wanted to make an application in Java that converts a string to binary code:

 /**
 * This class converts strings by looping through the characters, gets the
 * decimal code for each, then divides the number by 2 untill the quotient is 0
 * 
 * @author Mr Pro Pop
 *
 */
public class Binary 

 /**
 * This function takes a string as a parameter and converts it to binary code
 * 
 * @param string The string to be converted to binary code
 * @return The binary code of the input string
 */

 public String getBinaryCode(String string) 

 if (string.length() == 0)
 return null;

 char characters = string.toCharArray();
 StringBuilder result = new StringBuilder();
 int number = 0;

 for (int i = 0; i < characters.length; i++) 
 number = getDecimal(characters[i]);
 for (int j = 0; j < 8; j++) 
 if (number % 2 == 0) 
 result.insert(i, 0);
 else 
 result.insert(i, 1);
 
 number = (int) (number / 2);
 
 result.append(" ");
 
 return result.toString();
 

 /**
 * This function gets the decimal number of a character
 * 
 * @param character The character that we want the decimal value of
 * @return The decimal of the character from the ascii table
 */
 public int getDecimal(char character) 
 for (int i = 0; i <= 255; i++) 
 if (character == (char) i) 
 return i;
 
 
 return -1;
 

 public static void main(String args) 
 String string = "a";
 Binary b = new Binary();
 System.out.println("Binary code for" + string + " is " + b.getBinaryCode(string));

Output:

Binary code for a is 01100001

Any improvements I could do or anything to pay attention for?

edited 10 mins ago

200_success

124k14144401

asked 2 hours ago

Mr Pro Pop

1204

add a commentÂ |Â

up vote
1
down vote

favorite

I did research about binary counting, and wanted to make an application in Java that converts a string to binary code:

 /**
 * This class converts strings by looping through the characters, gets the
 * decimal code for each, then divides the number by 2 untill the quotient is 0
 * 
 * @author Mr Pro Pop
 *
 */
public class Binary 

 /**
 * This function takes a string as a parameter and converts it to binary code
 * 
 * @param string The string to be converted to binary code
 * @return The binary code of the input string
 */

 public String getBinaryCode(String string) 

 if (string.length() == 0)
 return null;

 char characters = string.toCharArray();
 StringBuilder result = new StringBuilder();
 int number = 0;

 for (int i = 0; i < characters.length; i++) 
 number = getDecimal(characters[i]);
 for (int j = 0; j < 8; j++) 
 if (number % 2 == 0) 
 result.insert(i, 0);
 else 
 result.insert(i, 1);
 
 number = (int) (number / 2);
 
 result.append(" ");
 
 return result.toString();
 

 /**
 * This function gets the decimal number of a character
 * 
 * @param character The character that we want the decimal value of
 * @return The decimal of the character from the ascii table
 */
 public int getDecimal(char character) 
 for (int i = 0; i <= 255; i++) 
 if (character == (char) i) 
 return i;
 
 
 return -1;
 

 public static void main(String args) 
 String string = "a";
 Binary b = new Binary();
 System.out.println("Binary code for" + string + " is " + b.getBinaryCode(string));

Output:

Binary code for a is 01100001

Any improvements I could do or anything to pay attention for?

edited 10 mins ago

200_success

124k14144401

asked 2 hours ago

Mr Pro Pop

1204

I did research about binary counting, and wanted to make an application in Java that converts a string to binary code:

 /**
 * This class converts strings by looping through the characters, gets the
 * decimal code for each, then divides the number by 2 untill the quotient is 0
 * 
 * @author Mr Pro Pop
 *
 */
public class Binary 

 /**
 * This function takes a string as a parameter and converts it to binary code
 * 
 * @param string The string to be converted to binary code
 * @return The binary code of the input string
 */

 public String getBinaryCode(String string) 

 if (string.length() == 0)
 return null;

 char characters = string.toCharArray();
 StringBuilder result = new StringBuilder();
 int number = 0;

 for (int i = 0; i < characters.length; i++) 
 number = getDecimal(characters[i]);
 for (int j = 0; j < 8; j++) 
 if (number % 2 == 0) 
 result.insert(i, 0);
 else 
 result.insert(i, 1);
 
 number = (int) (number / 2);
 
 result.append(" ");
 
 return result.toString();
 

 /**
 * This function gets the decimal number of a character
 * 
 * @param character The character that we want the decimal value of
 * @return The decimal of the character from the ascii table
 */
 public int getDecimal(char character) 
 for (int i = 0; i <= 255; i++) 
 if (character == (char) i) 
 return i;
 
 
 return -1;
 

 public static void main(String args) 
 String string = "a";
 Binary b = new Binary();
 System.out.println("Binary code for" + string + " is " + b.getBinaryCode(string));

Output:

Binary code for a is 01100001

Any improvements I could do or anything to pay attention for?

java reinventing-the-wheel number-systems

edited 10 mins ago

200_success

124k14144401

asked 2 hours ago

Mr Pro Pop

1204

edited 10 mins ago

200_success

124k14144401

asked 2 hours ago

Mr Pro Pop

1204

edited 10 mins ago

200_success

124k14144401

edited 10 mins ago

200_success

124k14144401

edited 10 mins ago

200_success

124k14144401

asked 2 hours ago

Mr Pro Pop

1204

asked 2 hours ago

Mr Pro Pop

1204

asked 2 hours ago

Mr Pro Pop

1204

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
3
down vote

accepted

I would change

public int getDecimal(char character)

 for (int i = 0; i <= 255; i++)
 
 if (character == (char) i)
 
 return i;
 
 
 return -1;

To just

public int getDecimal(char character)

 if(character<0

, As char is a number, too.

Also try using >> operator to get nth binary digit, which means division by 2.

number = (int) (number / 2);

number = number >>1;

I don't know if it would be faster in java, but it works faster in old computer's assembly program(Division was expensive than just shifting)

If you use assembly you can access carry flags and optimize it.

For more information about it, refer here.

Provided that al holds the target number and bx holds the target string address, and cx holds 8,

L1: ; This is the loop
mov dl, '0' ; Ascii character zero
shl al, 1 ; Upper bit now in carry flag
adc dl, 0 ; Adds carry flag - 0 or 1
mov [bx], dl ; Save digit to current position
inc bx ; Next position
loop L1 ; Counts down cx
mov [bx], 0 ; Zero terminate (might need to use register)

edited 1 hour ago

answered 1 hour ago

KYHSGeekCode

1464

New contributor

Oh thanks!!! Changed the getDecimal method. One thing I didn't quite get is the >> operator, what do you mean? And also one more question please, why is my append(" ") putting space after the result? So say I pass "sup" it would give the binary code of each character not separated and at the end put 3 spaces.
â€“Â Mr Pro Pop
1 hour ago

@MrProPop Maybe because of the insert method?
â€“Â KYHSGeekCode
1 hour ago

Aha, thanks for the tip. Btw, yes your right, the problem is because of insert, but if I don't use insert, then I would get the binary code the other way round, like flipped. Do I run a for loop to order them back? Just thought of it would consume more resources considering a large string looping through each character.
â€“Â Mr Pro Pop
1 hour ago

1

@MrProPop Maybe you can do that. Or if you use assembly language you can access carry flag which is set when it is shifted in certain conditions like the ooerand is even or its highst bit is 1, etc. Then you can naturally find bits from its highest bits.
â€“Â KYHSGeekCode
1 hour ago

add a commentÂ |Â

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
3
down vote

accepted

I would change

public int getDecimal(char character)

 for (int i = 0; i <= 255; i++)
 
 if (character == (char) i)
 
 return i;
 
 
 return -1;

To just

public int getDecimal(char character)

 if(character<0

, As char is a number, too.

Also try using >> operator to get nth binary digit, which means division by 2.

number = (int) (number / 2);

number = number >>1;

I don't know if it would be faster in java, but it works faster in old computer's assembly program(Division was expensive than just shifting)

If you use assembly you can access carry flags and optimize it.

For more information about it, refer here.

Provided that al holds the target number and bx holds the target string address, and cx holds 8,

L1: ; This is the loop
mov dl, '0' ; Ascii character zero
shl al, 1 ; Upper bit now in carry flag
adc dl, 0 ; Adds carry flag - 0 or 1
mov [bx], dl ; Save digit to current position
inc bx ; Next position
loop L1 ; Counts down cx
mov [bx], 0 ; Zero terminate (might need to use register)

edited 1 hour ago

answered 1 hour ago

KYHSGeekCode

1464

New contributor

Oh thanks!!! Changed the getDecimal method. One thing I didn't quite get is the >> operator, what do you mean? And also one more question please, why is my append(" ") putting space after the result? So say I pass "sup" it would give the binary code of each character not separated and at the end put 3 spaces.
â€“Â Mr Pro Pop
1 hour ago

@MrProPop Maybe because of the insert method?
â€“Â KYHSGeekCode
1 hour ago

Aha, thanks for the tip. Btw, yes your right, the problem is because of insert, but if I don't use insert, then I would get the binary code the other way round, like flipped. Do I run a for loop to order them back? Just thought of it would consume more resources considering a large string looping through each character.
â€“Â Mr Pro Pop
1 hour ago

1

@MrProPop Maybe you can do that. Or if you use assembly language you can access carry flag which is set when it is shifted in certain conditions like the ooerand is even or its highst bit is 1, etc. Then you can naturally find bits from its highest bits.
â€“Â KYHSGeekCode
1 hour ago

add a commentÂ |Â

up vote
3
down vote

accepted

I would change

public int getDecimal(char character)

 for (int i = 0; i <= 255; i++)
 
 if (character == (char) i)
 
 return i;
 
 
 return -1;

To just

public int getDecimal(char character)

 if(character<0

, As char is a number, too.

Also try using >> operator to get nth binary digit, which means division by 2.

number = (int) (number / 2);

number = number >>1;

I don't know if it would be faster in java, but it works faster in old computer's assembly program(Division was expensive than just shifting)

If you use assembly you can access carry flags and optimize it.

For more information about it, refer here.

Provided that al holds the target number and bx holds the target string address, and cx holds 8,

L1: ; This is the loop
mov dl, '0' ; Ascii character zero
shl al, 1 ; Upper bit now in carry flag
adc dl, 0 ; Adds carry flag - 0 or 1
mov [bx], dl ; Save digit to current position
inc bx ; Next position
loop L1 ; Counts down cx
mov [bx], 0 ; Zero terminate (might need to use register)

edited 1 hour ago

answered 1 hour ago

KYHSGeekCode

1464

New contributor

Oh thanks!!! Changed the getDecimal method. One thing I didn't quite get is the >> operator, what do you mean? And also one more question please, why is my append(" ") putting space after the result? So say I pass "sup" it would give the binary code of each character not separated and at the end put 3 spaces.
â€“Â Mr Pro Pop
1 hour ago

@MrProPop Maybe because of the insert method?
â€“Â KYHSGeekCode
1 hour ago

Aha, thanks for the tip. Btw, yes your right, the problem is because of insert, but if I don't use insert, then I would get the binary code the other way round, like flipped. Do I run a for loop to order them back? Just thought of it would consume more resources considering a large string looping through each character.
â€“Â Mr Pro Pop
1 hour ago

1

@MrProPop Maybe you can do that. Or if you use assembly language you can access carry flag which is set when it is shifted in certain conditions like the ooerand is even or its highst bit is 1, etc. Then you can naturally find bits from its highest bits.
â€“Â KYHSGeekCode
1 hour ago

add a commentÂ |Â

up vote
3
down vote

accepted

I would change

public int getDecimal(char character)

 for (int i = 0; i <= 255; i++)
 
 if (character == (char) i)
 
 return i;
 
 
 return -1;

To just

public int getDecimal(char character)

 if(character<0

, As char is a number, too.

Also try using >> operator to get nth binary digit, which means division by 2.

number = (int) (number / 2);

number = number >>1;

I don't know if it would be faster in java, but it works faster in old computer's assembly program(Division was expensive than just shifting)

If you use assembly you can access carry flags and optimize it.

For more information about it, refer here.

Provided that al holds the target number and bx holds the target string address, and cx holds 8,

L1: ; This is the loop
mov dl, '0' ; Ascii character zero
shl al, 1 ; Upper bit now in carry flag
adc dl, 0 ; Adds carry flag - 0 or 1
mov [bx], dl ; Save digit to current position
inc bx ; Next position
loop L1 ; Counts down cx
mov [bx], 0 ; Zero terminate (might need to use register)

edited 1 hour ago

answered 1 hour ago

KYHSGeekCode

1464

New contributor

I would change

public int getDecimal(char character)

 for (int i = 0; i <= 255; i++)
 
 if (character == (char) i)
 
 return i;
 
 
 return -1;

To just

public int getDecimal(char character)

 if(character<0

, As char is a number, too.

Also try using >> operator to get nth binary digit, which means division by 2.

number = (int) (number / 2);

number = number >>1;

I don't know if it would be faster in java, but it works faster in old computer's assembly program(Division was expensive than just shifting)

If you use assembly you can access carry flags and optimize it.

For more information about it, refer here.

Provided that al holds the target number and bx holds the target string address, and cx holds 8,

L1: ; This is the loop
mov dl, '0' ; Ascii character zero
shl al, 1 ; Upper bit now in carry flag
adc dl, 0 ; Adds carry flag - 0 or 1
mov [bx], dl ; Save digit to current position
inc bx ; Next position
loop L1 ; Counts down cx
mov [bx], 0 ; Zero terminate (might need to use register)

edited 1 hour ago

answered 1 hour ago

KYHSGeekCode

1464

New contributor

edited 1 hour ago

answered 1 hour ago

KYHSGeekCode

1464

New contributor

answered 1 hour ago

KYHSGeekCode

1464

answered 1 hour ago

KYHSGeekCode

1464

New contributor

KYHSGeekCode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Oh thanks!!! Changed the getDecimal method. One thing I didn't quite get is the >> operator, what do you mean? And also one more question please, why is my append(" ") putting space after the result? So say I pass "sup" it would give the binary code of each character not separated and at the end put 3 spaces.
â€“Â Mr Pro Pop
1 hour ago

@MrProPop Maybe because of the insert method?
â€“Â KYHSGeekCode
1 hour ago

Aha, thanks for the tip. Btw, yes your right, the problem is because of insert, but if I don't use insert, then I would get the binary code the other way round, like flipped. Do I run a for loop to order them back? Just thought of it would consume more resources considering a large string looping through each character.
â€“Â Mr Pro Pop
1 hour ago

1

@MrProPop Maybe you can do that. Or if you use assembly language you can access carry flag which is set when it is shifted in certain conditions like the ooerand is even or its highst bit is 1, etc. Then you can naturally find bits from its highest bits.
â€“Â KYHSGeekCode
1 hour ago

add a commentÂ |Â

Oh thanks!!! Changed the getDecimal method. One thing I didn't quite get is the >> operator, what do you mean? And also one more question please, why is my append(" ") putting space after the result? So say I pass "sup" it would give the binary code of each character not separated and at the end put 3 spaces.
â€“Â Mr Pro Pop
1 hour ago

@MrProPop Maybe because of the insert method?
â€“Â KYHSGeekCode
1 hour ago

Aha, thanks for the tip. Btw, yes your right, the problem is because of insert, but if I don't use insert, then I would get the binary code the other way round, like flipped. Do I run a for loop to order them back? Just thought of it would consume more resources considering a large string looping through each character.
â€“Â Mr Pro Pop
1 hour ago

1

@MrProPop Maybe you can do that. Or if you use assembly language you can access carry flag which is set when it is shifted in certain conditions like the ooerand is even or its highst bit is 1, etc. Then you can naturally find bits from its highest bits.
â€“Â KYHSGeekCode
1 hour ago

Oh thanks!!! Changed the getDecimal method. One thing I didn't quite get is the >> operator, what do you mean? And also one more question please, why is my append(" ") putting space after the result? So say I pass "sup" it would give the binary code of each character not separated and at the end put 3 spaces.
â€“Â Mr Pro Pop
1 hour ago

@MrProPop Maybe because of the insert method?
â€“Â KYHSGeekCode
1 hour ago

Aha, thanks for the tip. Btw, yes your right, the problem is because of insert, but if I don't use insert, then I would get the binary code the other way round, like flipped. Do I run a for loop to order them back? Just thought of it would consume more resources considering a large string looping through each character.
â€“Â Mr Pro Pop
1 hour ago

@MrProPop Maybe you can do that. Or if you use assembly language you can access carry flag which is set when it is shifted in certain conditions like the ooerand is even or its highst bit is 1, etc. Then you can naturally find bits from its highest bits.
â€“Â KYHSGeekCode
1 hour ago

add a commentÂ |Â

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