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C: get the bits of an integer

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

-3

I have a program that will take two 4-byte integers as input and I need to store these into integer arrays like so...

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (int argc, char *argv[]) 
 int vals1[32], vals2[32];
 int num1 = atoi(argv[1]);
 int num2 = atoi(argv[2]);

 // so argv[1] might be 47 and I would want to set the values of the vals1 array to reflect that in binary form

Any suggestions?

c binary integer

share|improve this question

edited Mar 8 at 17:14

chqrlie

63.9k851108

asked Mar 8 at 16:00

Ben deVriesBen deVries

52

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Hints: for a positive number, '% 2' extracts the least significant binary digit, and / 2 removes the final binary digit.
  
  – Bathsheba
  Mar 8 at 16:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your question is unclear. So is your input a string? Do you want to just store that as an int into an int array? Do you want a binary representation of your int? What representation?
  
  – Superlokkus
  Mar 8 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What have you tried so far? Where specifically are you stuck?
  
  – Govind Parmar
  Mar 8 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Superlokkus, i am taking a cstring which i will cast to an integer using the atoi function. So, i start with "47" and turn it into the integer 47, but then want to extract the binary bits that represent 47 as an integer and store them in val1
  
  – Ben deVries
  Mar 8 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  int nthbit(unsigned long long x, int n) return (x >> n) & 1;
  
  – pmg
  Mar 8 at 17:24
  

  
  
  
  

add a comment | 

-3

I have a program that will take two 4-byte integers as input and I need to store these into integer arrays like so...

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (int argc, char *argv[]) 
 int vals1[32], vals2[32];
 int num1 = atoi(argv[1]);
 int num2 = atoi(argv[2]);

 // so argv[1] might be 47 and I would want to set the values of the vals1 array to reflect that in binary form

Any suggestions?

c binary integer

share|improve this question

edited Mar 8 at 17:14

chqrlie

63.9k851108

asked Mar 8 at 16:00

Ben deVriesBen deVries

52

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Hints: for a positive number, '% 2' extracts the least significant binary digit, and / 2 removes the final binary digit.
  
  – Bathsheba
  Mar 8 at 16:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your question is unclear. So is your input a string? Do you want to just store that as an int into an int array? Do you want a binary representation of your int? What representation?
  
  – Superlokkus
  Mar 8 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What have you tried so far? Where specifically are you stuck?
  
  – Govind Parmar
  Mar 8 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Superlokkus, i am taking a cstring which i will cast to an integer using the atoi function. So, i start with "47" and turn it into the integer 47, but then want to extract the binary bits that represent 47 as an integer and store them in val1
  
  – Ben deVries
  Mar 8 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  int nthbit(unsigned long long x, int n) return (x >> n) & 1;
  
  – pmg
  Mar 8 at 17:24
  

  
  
  
  

add a comment | 

I have a program that will take two 4-byte integers as input and I need to store these into integer arrays like so...

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (int argc, char *argv[]) 
 int vals1[32], vals2[32];
 int num1 = atoi(argv[1]);
 int num2 = atoi(argv[2]);

 // so argv[1] might be 47 and I would want to set the values of the vals1 array to reflect that in binary form

Any suggestions?

c binary integer

share|improve this question

edited Mar 8 at 17:14

chqrlie

63.9k851108

asked Mar 8 at 16:00

Ben deVriesBen deVries

52

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (int argc, char *argv[]) 
 int vals1[32], vals2[32];
 int num1 = atoi(argv[1]);
 int num2 = atoi(argv[2]);

 // so argv[1] might be 47 and I would want to set the values of the vals1 array to reflect that in binary form

share|improve this question

edited Mar 8 at 17:14

chqrlie

63.9k851108

asked Mar 8 at 16:00

Ben deVriesBen deVries

52

share|improve this question

edited Mar 8 at 17:14

chqrlie

63.9k851108

asked Mar 8 at 16:00

Ben deVriesBen deVries

52

edited Mar 8 at 17:14

chqrlie

63.9k851108

edited Mar 8 at 17:14

chqrlie

63.9k851108

asked Mar 8 at 16:00

Ben deVriesBen deVries

52

asked Mar 8 at 16:00

Ben deVriesBen deVries

52

1 Answer
1

active

oldest

votes

1

First task would be to convert a char * to an int, which you said you can. So here comes the next part i.e. getting the binary representation. For getting the binary representation of any data type, usage of Shift Operator is one of the best ways. And you can get it by performing shift on the data type and then performing Bitwise AND i.e. & with 1. For example, if n is an integer

int n = 47;
for (int i = 0; i < 32; i++)

 /* It's going to print the values of bits starting from least significant. */
 printf("Bit%d = %drn", i, (unsigned int)((n >> i) & 1));

So, using shift operator, solution to your problem would be something like

void fun(int n1, int n2)

 int i, argv1[32], argv2[32];

 for (i = 0; i < 32; i++)
 
 argv1[i] = ((unsigned int)n1 >> i) & 1;
 argv2[i] = ((unsigned int)n2 >> i) & 1;
 

You have to be careful about the bit order i.e. which bit are being stored at which of the array index.

share|improve this answer

edited Mar 8 at 18:00

chqrlie

63.9k851108

answered Mar 8 at 17:01

MazharMazhar

564520

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  n1 >> i has implementation defined behavior if n1 is negative. You should use (unsigned int)n1 >> i
  
  – chqrlie
  Mar 8 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Like what exactly what happen if n1 is negative for example, please elaborate a bit more. Thanks a lot.
  
  – Mazhar
  Mar 8 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Mazhar This explains it.
  
  – Michail
  Mar 8 at 17:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Another point: you make the silent assumption that sizeof(int) == 4 because you use for (i = 0; i < (8 * sizeof(int)); i++) and the sizes have a fixed size of 32. Better write for (i = 0; i < 32; i++)
  
  – chqrlie
  Mar 8 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I took the liberty to fix the parentheses in the answer's code: n1 must be cast before the >> operation.
  
  – chqrlie
  Mar 8 at 18:00
  

  
  
  
  

 | 
show 2 more comments

Your Answer

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1

First task would be to convert a char * to an int, which you said you can. So here comes the next part i.e. getting the binary representation. For getting the binary representation of any data type, usage of Shift Operator is one of the best ways. And you can get it by performing shift on the data type and then performing Bitwise AND i.e. & with 1. For example, if n is an integer

int n = 47;
for (int i = 0; i < 32; i++)

 /* It's going to print the values of bits starting from least significant. */
 printf("Bit%d = %drn", i, (unsigned int)((n >> i) & 1));

So, using shift operator, solution to your problem would be something like

void fun(int n1, int n2)

 int i, argv1[32], argv2[32];

 for (i = 0; i < 32; i++)
 
 argv1[i] = ((unsigned int)n1 >> i) & 1;
 argv2[i] = ((unsigned int)n2 >> i) & 1;
 

You have to be careful about the bit order i.e. which bit are being stored at which of the array index.

share|improve this answer

edited Mar 8 at 18:00

chqrlie

63.9k851108

answered Mar 8 at 17:01

MazharMazhar

564520

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  n1 >> i has implementation defined behavior if n1 is negative. You should use (unsigned int)n1 >> i
  
  – chqrlie
  Mar 8 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Like what exactly what happen if n1 is negative for example, please elaborate a bit more. Thanks a lot.
  
  – Mazhar
  Mar 8 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Mazhar This explains it.
  
  – Michail
  Mar 8 at 17:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Another point: you make the silent assumption that sizeof(int) == 4 because you use for (i = 0; i < (8 * sizeof(int)); i++) and the sizes have a fixed size of 32. Better write for (i = 0; i < 32; i++)
  
  – chqrlie
  Mar 8 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I took the liberty to fix the parentheses in the answer's code: n1 must be cast before the >> operation.
  
  – chqrlie
  Mar 8 at 18:00
  

  
  
  
  

 | 
show 2 more comments

1

First task would be to convert a char * to an int, which you said you can. So here comes the next part i.e. getting the binary representation. For getting the binary representation of any data type, usage of Shift Operator is one of the best ways. And you can get it by performing shift on the data type and then performing Bitwise AND i.e. & with 1. For example, if n is an integer

int n = 47;
for (int i = 0; i < 32; i++)

 /* It's going to print the values of bits starting from least significant. */
 printf("Bit%d = %drn", i, (unsigned int)((n >> i) & 1));

So, using shift operator, solution to your problem would be something like

void fun(int n1, int n2)

 int i, argv1[32], argv2[32];

 for (i = 0; i < 32; i++)
 
 argv1[i] = ((unsigned int)n1 >> i) & 1;
 argv2[i] = ((unsigned int)n2 >> i) & 1;
 

You have to be careful about the bit order i.e. which bit are being stored at which of the array index.

share|improve this answer

edited Mar 8 at 18:00

chqrlie

63.9k851108

answered Mar 8 at 17:01

MazharMazhar

564520

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  n1 >> i has implementation defined behavior if n1 is negative. You should use (unsigned int)n1 >> i
  
  – chqrlie
  Mar 8 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Like what exactly what happen if n1 is negative for example, please elaborate a bit more. Thanks a lot.
  
  – Mazhar
  Mar 8 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Mazhar This explains it.
  
  – Michail
  Mar 8 at 17:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Another point: you make the silent assumption that sizeof(int) == 4 because you use for (i = 0; i < (8 * sizeof(int)); i++) and the sizes have a fixed size of 32. Better write for (i = 0; i < 32; i++)
  
  – chqrlie
  Mar 8 at 17:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I took the liberty to fix the parentheses in the answer's code: n1 must be cast before the >> operation.
  
  – chqrlie
  Mar 8 at 18:00
  

  
  
  
  

 | 
show 2 more comments

First task would be to convert a char * to an int, which you said you can. So here comes the next part i.e. getting the binary representation. For getting the binary representation of any data type, usage of Shift Operator is one of the best ways. And you can get it by performing shift on the data type and then performing Bitwise AND i.e. & with 1. For example, if n is an integer

int n = 47;
for (int i = 0; i < 32; i++)

 /* It's going to print the values of bits starting from least significant. */
 printf("Bit%d = %drn", i, (unsigned int)((n >> i) & 1));

So, using shift operator, solution to your problem would be something like

void fun(int n1, int n2)

 int i, argv1[32], argv2[32];

 for (i = 0; i < 32; i++)
 
 argv1[i] = ((unsigned int)n1 >> i) & 1;
 argv2[i] = ((unsigned int)n2 >> i) & 1;
 

You have to be careful about the bit order i.e. which bit are being stored at which of the array index.

share|improve this answer

edited Mar 8 at 18:00

chqrlie

63.9k851108

answered Mar 8 at 17:01

MazharMazhar

564520

int n = 47;
for (int i = 0; i < 32; i++)

 /* It's going to print the values of bits starting from least significant. */
 printf("Bit%d = %drn", i, (unsigned int)((n >> i) & 1));

void fun(int n1, int n2)

 int i, argv1[32], argv2[32];

 for (i = 0; i < 32; i++)
 
 argv1[i] = ((unsigned int)n1 >> i) & 1;
 argv2[i] = ((unsigned int)n2 >> i) & 1;
 

share|improve this answer

edited Mar 8 at 18:00

chqrlie

63.9k851108

answered Mar 8 at 17:01

MazharMazhar

564520

edited Mar 8 at 18:00

chqrlie

63.9k851108

edited Mar 8 at 18:00

chqrlie

63.9k851108

answered Mar 8 at 17:01

MazharMazhar

564520

answered Mar 8 at 17:01

MazharMazhar

564520