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-3

0

I'm fairly new to C in general but have a basic code knowlege. i'm having a tough time getting pointers into my head and was wondering if anybody has any good way of thinking of them (like getting '*' and '&' the right way around). I'm trying to make a header file for console inputs and in doing so am returning a pointer to an array of answers. I want to make the header file as easy to use for the end user.

This is what I have and it works but is there a way of eliminating the pointer (*) in printf("%sn", list[*i]); by modifying the for loop attributes?

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int *i = pointer; *i >= 0; i++)
 
 printf("%sn", list[*i]);
 

Final Solution:

#include "inquire.h"

int main(int argc, char *argv[])

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int i = 0; pointer[i] >= 0; i++)
 
 printf("%sn", list[pointer[i]]);
 

c pointers

share|improve this question

edited Mar 27 at 10:09

hamish sams

asked Mar 24 at 16:17

hamish samshamish sams

606

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  what do you think you need the pointer for? what does multiChoice do?
  
  – Antti Haapala
  Mar 24 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @AnttiHaapala MultiChoice returns a pointer to an array of answers
  
  – hamish sams
  Mar 24 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  But the choices are list[0], list[1], list[2], list[3] and list[4], right, like count from 0 to 4, or count from 0 until you list[i] is NULL. I ask again: what does multiChoice do, please provide the code!
  
  – Antti Haapala
  Mar 24 at 16:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I find this question hard to answer, because I think pointers are rather beautiful :)
  
  – Josh Greifer
  Mar 24 at 16:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I suppose you can use array-like syntax instead of pointer arithmetic and dereferencing (i.e. make i an integer starting at 0, and use pointer[i] instead of *i). But the pointer syntax is idiomatic C.
  
  – interjay
  Mar 24 at 16:37
  
  

  
  
  
  

 | 
show 3 more comments

-3

0

I'm fairly new to C in general but have a basic code knowlege. i'm having a tough time getting pointers into my head and was wondering if anybody has any good way of thinking of them (like getting '*' and '&' the right way around). I'm trying to make a header file for console inputs and in doing so am returning a pointer to an array of answers. I want to make the header file as easy to use for the end user.

This is what I have and it works but is there a way of eliminating the pointer (*) in printf("%sn", list[*i]); by modifying the for loop attributes?

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int *i = pointer; *i >= 0; i++)
 
 printf("%sn", list[*i]);
 

Final Solution:

#include "inquire.h"

int main(int argc, char *argv[])

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int i = 0; pointer[i] >= 0; i++)
 
 printf("%sn", list[pointer[i]]);
 

c pointers

share|improve this question

edited Mar 27 at 10:09

hamish sams

asked Mar 24 at 16:17

hamish samshamish sams

606

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  what do you think you need the pointer for? what does multiChoice do?
  
  – Antti Haapala
  Mar 24 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @AnttiHaapala MultiChoice returns a pointer to an array of answers
  
  – hamish sams
  Mar 24 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  But the choices are list[0], list[1], list[2], list[3] and list[4], right, like count from 0 to 4, or count from 0 until you list[i] is NULL. I ask again: what does multiChoice do, please provide the code!
  
  – Antti Haapala
  Mar 24 at 16:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I find this question hard to answer, because I think pointers are rather beautiful :)
  
  – Josh Greifer
  Mar 24 at 16:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I suppose you can use array-like syntax instead of pointer arithmetic and dereferencing (i.e. make i an integer starting at 0, and use pointer[i] instead of *i). But the pointer syntax is idiomatic C.
  
  – interjay
  Mar 24 at 16:37
  
  

  
  
  
  

 | 
show 3 more comments

I'm fairly new to C in general but have a basic code knowlege. i'm having a tough time getting pointers into my head and was wondering if anybody has any good way of thinking of them (like getting '*' and '&' the right way around). I'm trying to make a header file for console inputs and in doing so am returning a pointer to an array of answers. I want to make the header file as easy to use for the end user.

This is what I have and it works but is there a way of eliminating the pointer (*) in printf("%sn", list[*i]); by modifying the for loop attributes?

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int *i = pointer; *i >= 0; i++)
 
 printf("%sn", list[*i]);
 

Final Solution:

#include "inquire.h"

int main(int argc, char *argv[])

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int i = 0; pointer[i] >= 0; i++)
 
 printf("%sn", list[pointer[i]]);
 

c pointers

share|improve this question

edited Mar 27 at 10:09

hamish sams

asked Mar 24 at 16:17

hamish samshamish sams

606

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int *i = pointer; *i >= 0; i++)
 
 printf("%sn", list[*i]);
 

#include "inquire.h"

int main(int argc, char *argv[])

 int *pointer;
 char *list[] = "Hello1", "Hello2", "Hello3", "Hello4", "Hello5", NULL;

 pointer = multiChoice(list);

 printf("Choices:n");
 for (int i = 0; pointer[i] >= 0; i++)
 
 printf("%sn", list[pointer[i]]);
 

share|improve this question

edited Mar 27 at 10:09

hamish sams

asked Mar 24 at 16:17

hamish samshamish sams

606

share|improve this question

edited Mar 27 at 10:09

hamish sams

asked Mar 24 at 16:17

hamish samshamish sams

606

asked Mar 24 at 16:17

hamish samshamish sams

606

asked Mar 24 at 16:17

hamish samshamish sams

606

1 Answer
1

active

oldest

votes

-5

As you might know, pointers store addresses of, for example, variables and functions. The '*' operator is used for two different operations in this case. One in declaring a pointer, as you did with pointer. The other operation is dereferencing, which returns the value stored at the address the pointer points to.

I don't know what multiChoice() does, so I can not eliminate the pointer pointer variable. But I would, in fact, rewrite in initialization if i, because of simplicity. It is shorter and more readable to dereference pointer once, instead of dereferencing i three times.

for (int i = *pointer; i >= 0; i++)

 printf("%sn", list[i]);

share|improve this answer

edited Mar 24 at 16:30

answered Mar 24 at 16:28

Martin PekárMartin Pekár

274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your version can only access consecutive elements of list, but OP's version can access any element (skipping over some). Also, your loop will never terminate.
  
  – interjay
  Mar 24 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you force the chosen indexes to be consecutive
  
  – bruno
  Mar 24 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's on purpose. I didn't want to fix the errors, just decreasing the amount of pointers.
  
  – Martin Pekár
  Mar 24 at 16:31
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Seriously? Obviously the point was to remove pointers without changing the functionality of the code or introducing bugs. Your answer is about as useless as telling him to delete all his code, as that will also get rid of the pointers.
  
  – interjay
  Mar 24 at 17:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Not semantically identical to the original.
  
  – Clifford
  Mar 24 at 18:23
  
  

  
  
  
  

 | 
show 7 more comments

Your Answer

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-5

As you might know, pointers store addresses of, for example, variables and functions. The '*' operator is used for two different operations in this case. One in declaring a pointer, as you did with pointer. The other operation is dereferencing, which returns the value stored at the address the pointer points to.

I don't know what multiChoice() does, so I can not eliminate the pointer pointer variable. But I would, in fact, rewrite in initialization if i, because of simplicity. It is shorter and more readable to dereference pointer once, instead of dereferencing i three times.

for (int i = *pointer; i >= 0; i++)

 printf("%sn", list[i]);

share|improve this answer

edited Mar 24 at 16:30

answered Mar 24 at 16:28

Martin PekárMartin Pekár

274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your version can only access consecutive elements of list, but OP's version can access any element (skipping over some). Also, your loop will never terminate.
  
  – interjay
  Mar 24 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you force the chosen indexes to be consecutive
  
  – bruno
  Mar 24 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's on purpose. I didn't want to fix the errors, just decreasing the amount of pointers.
  
  – Martin Pekár
  Mar 24 at 16:31
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Seriously? Obviously the point was to remove pointers without changing the functionality of the code or introducing bugs. Your answer is about as useless as telling him to delete all his code, as that will also get rid of the pointers.
  
  – interjay
  Mar 24 at 17:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Not semantically identical to the original.
  
  – Clifford
  Mar 24 at 18:23
  
  

  
  
  
  

 | 
show 7 more comments

-5

As you might know, pointers store addresses of, for example, variables and functions. The '*' operator is used for two different operations in this case. One in declaring a pointer, as you did with pointer. The other operation is dereferencing, which returns the value stored at the address the pointer points to.

I don't know what multiChoice() does, so I can not eliminate the pointer pointer variable. But I would, in fact, rewrite in initialization if i, because of simplicity. It is shorter and more readable to dereference pointer once, instead of dereferencing i three times.

for (int i = *pointer; i >= 0; i++)

 printf("%sn", list[i]);

share|improve this answer

edited Mar 24 at 16:30

answered Mar 24 at 16:28

Martin PekárMartin Pekár

274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your version can only access consecutive elements of list, but OP's version can access any element (skipping over some). Also, your loop will never terminate.
  
  – interjay
  Mar 24 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you force the chosen indexes to be consecutive
  
  – bruno
  Mar 24 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's on purpose. I didn't want to fix the errors, just decreasing the amount of pointers.
  
  – Martin Pekár
  Mar 24 at 16:31
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Seriously? Obviously the point was to remove pointers without changing the functionality of the code or introducing bugs. Your answer is about as useless as telling him to delete all his code, as that will also get rid of the pointers.
  
  – interjay
  Mar 24 at 17:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Not semantically identical to the original.
  
  – Clifford
  Mar 24 at 18:23
  
  

  
  
  
  

 | 
show 7 more comments

As you might know, pointers store addresses of, for example, variables and functions. The '*' operator is used for two different operations in this case. One in declaring a pointer, as you did with pointer. The other operation is dereferencing, which returns the value stored at the address the pointer points to.

I don't know what multiChoice() does, so I can not eliminate the pointer pointer variable. But I would, in fact, rewrite in initialization if i, because of simplicity. It is shorter and more readable to dereference pointer once, instead of dereferencing i three times.

for (int i = *pointer; i >= 0; i++)

 printf("%sn", list[i]);

share|improve this answer

edited Mar 24 at 16:30

answered Mar 24 at 16:28

Martin PekárMartin Pekár

274

for (int i = *pointer; i >= 0; i++)

 printf("%sn", list[i]);

share|improve this answer

edited Mar 24 at 16:30

answered Mar 24 at 16:28

Martin PekárMartin Pekár

274

answered Mar 24 at 16:28

Martin PekárMartin Pekár

274

answered Mar 24 at 16:28

Martin PekárMartin Pekár

274