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3

$begingroup$

This is the code I came up with.

I added comments to make the solution more verbose.

int findComplement(int num) 
 // b is the answer which will be returned
 int b = 0;

 // One bit will be taken at a time from num, will be inverted and stored in n for adding to result
 int n = 0;

 // k will be used to shift bit to be inserted in correct position
 int k = 0;

 while(num)
 // Invert bit of current number
 n = !(num & 1);

 // Shift the given number one bit right to accesss next bit in next iteration
 num = num >>1 ;

 // Add the inverted bit after shifting
 b = b + (n<<k);

 // Increment the number by which to shift next bit
 k++;
 
 return b;

Is there any redundant statment in my code which can be removed? Or any other better logic to invert bits of a given integer

c++ mathematics bitwise

share|improve this question

asked 8 hours ago

KshitijV97KshitijV97

8066 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Are you re-inventing the binary not operator (~)?
  $endgroup$
  – Martin R
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't want to sound dumb, But honestly, I did not know that ~ operator existed which inverts all bits of a given integer.
  $endgroup$
  – KshitijV97
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Many easy ways. ~num or -1 - num, or 0xFFFFFFFF - num, or 0xFFFFFFFF ^ num or (-1) ^ num. Doing it one bit at a time is most definitely the hard way.
  $endgroup$
  – AJNeufeld
  6 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

This is the code I came up with.

I added comments to make the solution more verbose.

int findComplement(int num) 
 // b is the answer which will be returned
 int b = 0;

 // One bit will be taken at a time from num, will be inverted and stored in n for adding to result
 int n = 0;

 // k will be used to shift bit to be inserted in correct position
 int k = 0;

 while(num)
 // Invert bit of current number
 n = !(num & 1);

 // Shift the given number one bit right to accesss next bit in next iteration
 num = num >>1 ;

 // Add the inverted bit after shifting
 b = b + (n<<k);

 // Increment the number by which to shift next bit
 k++;
 
 return b;

Is there any redundant statment in my code which can be removed? Or any other better logic to invert bits of a given integer

c++ mathematics bitwise

share|improve this question

asked 8 hours ago

KshitijV97KshitijV97

8066 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Are you re-inventing the binary not operator (~)?
  $endgroup$
  – Martin R
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't want to sound dumb, But honestly, I did not know that ~ operator existed which inverts all bits of a given integer.
  $endgroup$
  – KshitijV97
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Many easy ways. ~num or -1 - num, or 0xFFFFFFFF - num, or 0xFFFFFFFF ^ num or (-1) ^ num. Doing it one bit at a time is most definitely the hard way.
  $endgroup$
  – AJNeufeld
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is the code I came up with.

I added comments to make the solution more verbose.

int findComplement(int num) 
 // b is the answer which will be returned
 int b = 0;

 // One bit will be taken at a time from num, will be inverted and stored in n for adding to result
 int n = 0;

 // k will be used to shift bit to be inserted in correct position
 int k = 0;

 while(num)
 // Invert bit of current number
 n = !(num & 1);

 // Shift the given number one bit right to accesss next bit in next iteration
 num = num >>1 ;

 // Add the inverted bit after shifting
 b = b + (n<<k);

 // Increment the number by which to shift next bit
 k++;
 
 return b;

Is there any redundant statment in my code which can be removed? Or any other better logic to invert bits of a given integer

c++ mathematics bitwise

share|improve this question

asked 8 hours ago

KshitijV97KshitijV97

8066 bronze badges

$endgroup$

int findComplement(int num) 
 // b is the answer which will be returned
 int b = 0;

 // One bit will be taken at a time from num, will be inverted and stored in n for adding to result
 int n = 0;

 // k will be used to shift bit to be inserted in correct position
 int k = 0;

 while(num)
 // Invert bit of current number
 n = !(num & 1);

 // Shift the given number one bit right to accesss next bit in next iteration
 num = num >>1 ;

 // Add the inverted bit after shifting
 b = b + (n<<k);

 // Increment the number by which to shift next bit
 k++;
 
 return b;

share|improve this question

asked 8 hours ago

KshitijV97KshitijV97

8066 bronze badges

share|improve this question

asked 8 hours ago

KshitijV97KshitijV97

8066 bronze badges

asked 8 hours ago

KshitijV97KshitijV97

8066 bronze badges

asked 8 hours ago

KshitijV97KshitijV97

8066 bronze badges

1 Answer
1

active

oldest

votes

2

$begingroup$

int n = 0; This initialization is not used. It could simply be int n;, or could be int n = !(num & 1); inside the loop, to restrict the scope of n.

---

This loop:

int k = 0;
while (num) 
 ...
 k++;

could be written as:

for(int k = 0; num; k++) 
 ...

---

Since you are doing bit manipulation, instead of using addition, you should probably use a “binary or” operation to merge the bit into your accumulator:

 b = b | (n << k);

or simply:

 b |= n << k;

---

Bug

You are not inverting the most significant zero bits. Assuming an 8-bit word size, the binary compliment of 9 (0b00001001) should be 0b11110110, not 0b00000110. And the compliment of that should return to the original number (0b00001001), but instead yields 0b00000001.

---

And, as mentioned by @Martin R, you could simply return ~num;

share|improve this answer

answered 5 hours ago

AJNeufeldAJNeufeld

10.9k11 gold badge88 silver badges3333 bronze badges

$endgroup$

add a comment | 

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2

$begingroup$

int n = 0; This initialization is not used. It could simply be int n;, or could be int n = !(num & 1); inside the loop, to restrict the scope of n.

---

This loop:

int k = 0;
while (num) 
 ...
 k++;

could be written as:

for(int k = 0; num; k++) 
 ...

---

Since you are doing bit manipulation, instead of using addition, you should probably use a “binary or” operation to merge the bit into your accumulator:

 b = b | (n << k);

or simply:

 b |= n << k;

---

Bug

You are not inverting the most significant zero bits. Assuming an 8-bit word size, the binary compliment of 9 (0b00001001) should be 0b11110110, not 0b00000110. And the compliment of that should return to the original number (0b00001001), but instead yields 0b00000001.

---

And, as mentioned by @Martin R, you could simply return ~num;

share|improve this answer

answered 5 hours ago

AJNeufeldAJNeufeld

10.9k11 gold badge88 silver badges3333 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

int n = 0; This initialization is not used. It could simply be int n;, or could be int n = !(num & 1); inside the loop, to restrict the scope of n.

---

This loop:

int k = 0;
while (num) 
 ...
 k++;

could be written as:

for(int k = 0; num; k++) 
 ...

---

Since you are doing bit manipulation, instead of using addition, you should probably use a “binary or” operation to merge the bit into your accumulator:

 b = b | (n << k);

or simply:

 b |= n << k;

---

Bug

You are not inverting the most significant zero bits. Assuming an 8-bit word size, the binary compliment of 9 (0b00001001) should be 0b11110110, not 0b00000110. And the compliment of that should return to the original number (0b00001001), but instead yields 0b00000001.

---

And, as mentioned by @Martin R, you could simply return ~num;

share|improve this answer

answered 5 hours ago

AJNeufeldAJNeufeld

10.9k11 gold badge88 silver badges3333 bronze badges

$endgroup$

add a comment | 

$begingroup$

int n = 0; This initialization is not used. It could simply be int n;, or could be int n = !(num & 1); inside the loop, to restrict the scope of n.

---

This loop:

int k = 0;
while (num) 
 ...
 k++;

could be written as:

for(int k = 0; num; k++) 
 ...

---

Since you are doing bit manipulation, instead of using addition, you should probably use a “binary or” operation to merge the bit into your accumulator:

 b = b | (n << k);

or simply:

 b |= n << k;

---

Bug

You are not inverting the most significant zero bits. Assuming an 8-bit word size, the binary compliment of 9 (0b00001001) should be 0b11110110, not 0b00000110. And the compliment of that should return to the original number (0b00001001), but instead yields 0b00000001.

---

And, as mentioned by @Martin R, you could simply return ~num;

share|improve this answer

answered 5 hours ago

AJNeufeldAJNeufeld

10.9k11 gold badge88 silver badges3333 bronze badges

$endgroup$

int k = 0;
while (num) 
 ...
 k++;

for(int k = 0; num; k++) 
 ...

 b = b | (n << k);

 b |= n << k;

share|improve this answer

answered 5 hours ago

AJNeufeldAJNeufeld

10.9k11 gold badge88 silver badges3333 bronze badges

answered 5 hours ago

AJNeufeldAJNeufeld

10.9k11 gold badge88 silver badges3333 bronze badges

answered 5 hours ago

AJNeufeldAJNeufeld

10.9k11 gold badge88 silver badges3333 bronze badges