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if (counter & (1

What are bitwise shift (bit-shift) operators and how do they work?What does Str1[j] = (portData >> j) & 0x1; mean in C#What does 0 mean when initializing an object?What does “static” mean in C?How do function pointers in C work?What does the question mark and the colon (?: ternary operator) mean in objective-c?What does “dereferencing” a pointer mean?What does “CONST func(arg);” mean in C language?What does the C ??!??! operator do?Why does ENOENT mean “No such file or directory”?What does (x ^ 0x1) != 0 mean?LNK2091 error for OCIObjectGetAttr and OCIObjectSetAttr

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

5

3

This question already has an answer here:

• 
  What are bitwise shift (bit-shift) operators and how do they work?
  
  8 answers
  
  

i was working on an algorithm of sub sequences.
please tell me the meaning of this statement.
program is as below.

void printSubsequences(int arr[], int n)

 unsigned int opsize = pow(2, n);

 for (int counter = 1; counter < opsize; counter++)
 
 for (int j = 0; j < n; j++)
 
 if (counter & (1<<j))
 cout << arr[j] << " ";
 
 cout << endl;
 

c

share|improve this question

edited Jan 13 '17 at 7:43

Codor

15.7k82648

asked Jan 13 '17 at 7:40

Sarathi ShahSarathi Shah

3215

marked as duplicate by Mitch Wheat, Lundin c
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Jan 13 '17 at 9:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  << means that the left-hand operand is multiplied by 2, for as many times as the number in the right-hand operand. E.g. 1 << 3 means 1*2*2*2.
  
  – M.M
  Jan 13 '17 at 7:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The counter & (stuff) results in a bitwise AND of counter and stuff
  
  – David C. Rankin
  Jan 13 '17 at 7:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What are bitwise shift (bit-shift) operators and how do they work?
  
  – phuclv
  Jan 13 '17 at 7:47
  

  
  
  
  

add a comment | 

5

3

This question already has an answer here:

• 
  What are bitwise shift (bit-shift) operators and how do they work?
  
  8 answers
  
  

i was working on an algorithm of sub sequences.
please tell me the meaning of this statement.
program is as below.

void printSubsequences(int arr[], int n)

 unsigned int opsize = pow(2, n);

 for (int counter = 1; counter < opsize; counter++)
 
 for (int j = 0; j < n; j++)
 
 if (counter & (1<<j))
 cout << arr[j] << " ";
 
 cout << endl;
 

c

share|improve this question

edited Jan 13 '17 at 7:43

Codor

15.7k82648

asked Jan 13 '17 at 7:40

Sarathi ShahSarathi Shah

3215

marked as duplicate by Mitch Wheat, Lundin c
Users with the  c badge can single-handedly close c questions as duplicates and reopen them as needed.

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Jan 13 '17 at 9:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  << means that the left-hand operand is multiplied by 2, for as many times as the number in the right-hand operand. E.g. 1 << 3 means 1*2*2*2.
  
  – M.M
  Jan 13 '17 at 7:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The counter & (stuff) results in a bitwise AND of counter and stuff
  
  – David C. Rankin
  Jan 13 '17 at 7:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What are bitwise shift (bit-shift) operators and how do they work?
  
  – phuclv
  Jan 13 '17 at 7:47
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  What are bitwise shift (bit-shift) operators and how do they work?
  
  8 answers
  
  

i was working on an algorithm of sub sequences.
please tell me the meaning of this statement.
program is as below.

void printSubsequences(int arr[], int n)

 unsigned int opsize = pow(2, n);

 for (int counter = 1; counter < opsize; counter++)
 
 for (int j = 0; j < n; j++)
 
 if (counter & (1<<j))
 cout << arr[j] << " ";
 
 cout << endl;
 

c

share|improve this question

edited Jan 13 '17 at 7:43

Codor

15.7k82648

asked Jan 13 '17 at 7:40

Sarathi ShahSarathi Shah

3215

void printSubsequences(int arr[], int n)

 unsigned int opsize = pow(2, n);

 for (int counter = 1; counter < opsize; counter++)
 
 for (int j = 0; j < n; j++)
 
 if (counter & (1<<j))
 cout << arr[j] << " ";
 
 cout << endl;
 

share|improve this question

edited Jan 13 '17 at 7:43

Codor

15.7k82648

asked Jan 13 '17 at 7:40

Sarathi ShahSarathi Shah

3215

share|improve this question

edited Jan 13 '17 at 7:43

Codor

15.7k82648

asked Jan 13 '17 at 7:40

Sarathi ShahSarathi Shah

3215

edited Jan 13 '17 at 7:43

Codor

15.7k82648

edited Jan 13 '17 at 7:43

Codor

15.7k82648

asked Jan 13 '17 at 7:40

Sarathi ShahSarathi Shah

3215

asked Jan 13 '17 at 7:40

Sarathi ShahSarathi Shah

3215

1 Answer
1

active

oldest

votes

12

The statement:

if (counter & (1<<j))

checks if the j-th bit of counter is set. In more detail, 1 << j uses shifting of 1 to generate a bit mask in which only the j-th bit is set. The & operator then masks out the j-bit of counter; if the result is not zero (which means that the j-th bit of counter was set), the condition is satisfied.

Consider the following example. If counter is 320, its binary representation is 101000000, which means that the 6th bit (the one corresponding to the value of 64) is set; let's test for that bit. The bit mask is generated by shifting 1, which has the binary representation 000000001, 6 places to the right, which results in the binary value 001000000. The value of counter, namely:

101000000

is combined with &, which is the bitwise and-operator, with the bit mask as follows:

 101000000
& 001000000
 ---------
 001000000

The value 001000000 again corresponds to the value of 64; however this is not important here, it only matters that it is not zero (as it has a nonzero bit, namely the bit for which we intended to check). In total, the condition

if ( 64 )

is satisfied. In the semantics of C (which does not feature a native Boolean data type) any nonzero value is treated as true when checked with if.

share|improve this answer

edited Jun 20 '18 at 4:32

Azeem

3,10941224

answered Jan 13 '17 at 7:46

CodorCodor

15.7k82648

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can you please give me step by step execution of it?
  
  – Sarathi Shah
  Jan 13 '17 at 8:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The first loop generates all possible combinations and the second loop print according to those combinations. @SarathiShah
  
  – shinzou
  Mar 2 '18 at 21:50
  
  

  
  
  
  

add a comment | 

12

The statement:

if (counter & (1<<j))

checks if the j-th bit of counter is set. In more detail, 1 << j uses shifting of 1 to generate a bit mask in which only the j-th bit is set. The & operator then masks out the j-bit of counter; if the result is not zero (which means that the j-th bit of counter was set), the condition is satisfied.

Consider the following example. If counter is 320, its binary representation is 101000000, which means that the 6th bit (the one corresponding to the value of 64) is set; let's test for that bit. The bit mask is generated by shifting 1, which has the binary representation 000000001, 6 places to the right, which results in the binary value 001000000. The value of counter, namely:

101000000

is combined with &, which is the bitwise and-operator, with the bit mask as follows:

 101000000
& 001000000
 ---------
 001000000

The value 001000000 again corresponds to the value of 64; however this is not important here, it only matters that it is not zero (as it has a nonzero bit, namely the bit for which we intended to check). In total, the condition

if ( 64 )

is satisfied. In the semantics of C (which does not feature a native Boolean data type) any nonzero value is treated as true when checked with if.

share|improve this answer

edited Jun 20 '18 at 4:32

Azeem

3,10941224

answered Jan 13 '17 at 7:46

CodorCodor

15.7k82648

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can you please give me step by step execution of it?
  
  – Sarathi Shah
  Jan 13 '17 at 8:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The first loop generates all possible combinations and the second loop print according to those combinations. @SarathiShah
  
  – shinzou
  Mar 2 '18 at 21:50
  
  

  
  
  
  

add a comment | 

12

The statement:

if (counter & (1<<j))

checks if the j-th bit of counter is set. In more detail, 1 << j uses shifting of 1 to generate a bit mask in which only the j-th bit is set. The & operator then masks out the j-bit of counter; if the result is not zero (which means that the j-th bit of counter was set), the condition is satisfied.

Consider the following example. If counter is 320, its binary representation is 101000000, which means that the 6th bit (the one corresponding to the value of 64) is set; let's test for that bit. The bit mask is generated by shifting 1, which has the binary representation 000000001, 6 places to the right, which results in the binary value 001000000. The value of counter, namely:

101000000

is combined with &, which is the bitwise and-operator, with the bit mask as follows:

 101000000
& 001000000
 ---------
 001000000

The value 001000000 again corresponds to the value of 64; however this is not important here, it only matters that it is not zero (as it has a nonzero bit, namely the bit for which we intended to check). In total, the condition

if ( 64 )

is satisfied. In the semantics of C (which does not feature a native Boolean data type) any nonzero value is treated as true when checked with if.

share|improve this answer

edited Jun 20 '18 at 4:32

Azeem

3,10941224

answered Jan 13 '17 at 7:46

CodorCodor

15.7k82648

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can you please give me step by step execution of it?
  
  – Sarathi Shah
  Jan 13 '17 at 8:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The first loop generates all possible combinations and the second loop print according to those combinations. @SarathiShah
  
  – shinzou
  Mar 2 '18 at 21:50
  
  

  
  
  
  

add a comment | 

The statement:

if (counter & (1<<j))

checks if the j-th bit of counter is set. In more detail, 1 << j uses shifting of 1 to generate a bit mask in which only the j-th bit is set. The & operator then masks out the j-bit of counter; if the result is not zero (which means that the j-th bit of counter was set), the condition is satisfied.

Consider the following example. If counter is 320, its binary representation is 101000000, which means that the 6th bit (the one corresponding to the value of 64) is set; let's test for that bit. The bit mask is generated by shifting 1, which has the binary representation 000000001, 6 places to the right, which results in the binary value 001000000. The value of counter, namely:

101000000

is combined with &, which is the bitwise and-operator, with the bit mask as follows:

 101000000
& 001000000
 ---------
 001000000

The value 001000000 again corresponds to the value of 64; however this is not important here, it only matters that it is not zero (as it has a nonzero bit, namely the bit for which we intended to check). In total, the condition

if ( 64 )

is satisfied. In the semantics of C (which does not feature a native Boolean data type) any nonzero value is treated as true when checked with if.

share|improve this answer

edited Jun 20 '18 at 4:32

Azeem

3,10941224

answered Jan 13 '17 at 7:46

CodorCodor

15.7k82648

if (counter & (1<<j))

101000000

 101000000
& 001000000
 ---------
 001000000

if ( 64 )

share|improve this answer

edited Jun 20 '18 at 4:32

Azeem

3,10941224

answered Jan 13 '17 at 7:46

CodorCodor

15.7k82648

edited Jun 20 '18 at 4:32

Azeem

3,10941224

edited Jun 20 '18 at 4:32

Azeem

3,10941224

answered Jan 13 '17 at 7:46

CodorCodor

15.7k82648

answered Jan 13 '17 at 7:46

CodorCodor

15.7k82648