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Print whole address block of some variable [closed]

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0

I have this code:

#include <stdio.h>

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, &a + 1, &a + 2, &a + 3);
 return 0;

Output:

First part of address block: 
0x7fffd029ecec 
All parts of address block: 
0x7fffd029ecec 
0x7fffd029ecf0 
0x7fffd029ecf4 
0x7fffd029ecf8 

In my opinion, address block of int a looks like this:

| 0x7fffd029ece**c** | 0x7fffd029ece**d** | 0x7fffd029ece**e** | 0x7fffd029ece**f** |
| 0000 | 0000 | 0011 | 1000 

c pointers hex memory-address

share|improve this question

edited Mar 26 at 14:50

abelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

asked Mar 26 at 14:28

John DoeJohn Doe

111 bronze badge

closed as unclear what you're asking by Lundin, Toby Speight, Abhishek Pandey, Tomtom, Mathieu Mar 27 at 7:47

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  int commonly has a size of 4, so that output looks right to me.
  
  – Blaze
  Mar 26 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You might want to read about "pointer arithmetics".
  
  – Gerhardh
  Mar 26 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You must cast to void * to get the proper type for %p. Also, print the size of a (printf("my int is %zu bytesn", sizeof a);) to make sure.
  
  – unwind
  Mar 26 at 14:31
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  What is the question?
  
  – Lundin
  Mar 26 at 14:52
  

  
  
  
  

add a comment | 

0

I have this code:

#include <stdio.h>

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, &a + 1, &a + 2, &a + 3);
 return 0;

Output:

First part of address block: 
0x7fffd029ecec 
All parts of address block: 
0x7fffd029ecec 
0x7fffd029ecf0 
0x7fffd029ecf4 
0x7fffd029ecf8 

In my opinion, address block of int a looks like this:

| 0x7fffd029ece**c** | 0x7fffd029ece**d** | 0x7fffd029ece**e** | 0x7fffd029ece**f** |
| 0000 | 0000 | 0011 | 1000 

c pointers hex memory-address

share|improve this question

edited Mar 26 at 14:50

abelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

asked Mar 26 at 14:28

John DoeJohn Doe

111 bronze badge

closed as unclear what you're asking by Lundin, Toby Speight, Abhishek Pandey, Tomtom, Mathieu Mar 27 at 7:47

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  int commonly has a size of 4, so that output looks right to me.
  
  – Blaze
  Mar 26 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You might want to read about "pointer arithmetics".
  
  – Gerhardh
  Mar 26 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You must cast to void * to get the proper type for %p. Also, print the size of a (printf("my int is %zu bytesn", sizeof a);) to make sure.
  
  – unwind
  Mar 26 at 14:31
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  What is the question?
  
  – Lundin
  Mar 26 at 14:52
  

  
  
  
  

add a comment | 

I have this code:

#include <stdio.h>

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, &a + 1, &a + 2, &a + 3);
 return 0;

Output:

First part of address block: 
0x7fffd029ecec 
All parts of address block: 
0x7fffd029ecec 
0x7fffd029ecf0 
0x7fffd029ecf4 
0x7fffd029ecf8 

In my opinion, address block of int a looks like this:

| 0x7fffd029ece**c** | 0x7fffd029ece**d** | 0x7fffd029ece**e** | 0x7fffd029ece**f** |
| 0000 | 0000 | 0011 | 1000 

c pointers hex memory-address

share|improve this question

edited Mar 26 at 14:50

abelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

asked Mar 26 at 14:28

John DoeJohn Doe

111 bronze badge

#include <stdio.h>

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, &a + 1, &a + 2, &a + 3);
 return 0;

First part of address block: 
0x7fffd029ecec 
All parts of address block: 
0x7fffd029ecec 
0x7fffd029ecf0 
0x7fffd029ecf4 
0x7fffd029ecf8 

| 0x7fffd029ece**c** | 0x7fffd029ece**d** | 0x7fffd029ece**e** | 0x7fffd029ece**f** |
| 0000 | 0000 | 0011 | 1000 

share|improve this question

edited Mar 26 at 14:50

abelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

asked Mar 26 at 14:28

John DoeJohn Doe

111 bronze badge

share|improve this question

edited Mar 26 at 14:50

abelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

asked Mar 26 at 14:28

John DoeJohn Doe

111 bronze badge

edited Mar 26 at 14:50

abelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

edited Mar 26 at 14:50

abelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

asked Mar 26 at 14:28

John DoeJohn Doe

111 bronze badge

asked Mar 26 at 14:28

John DoeJohn Doe

111 bronze badge

2 Answers
2

active

oldest

votes

3

C increment pointers not by bytes, but by size of type is pointed on.
That is, if you have pointer to int it increments by sizeof(int) - in your example is 4.

share|improve this answer

answered Mar 26 at 14:33

AlexAlex

39622 silver badges1313 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  When I have char a = 'A' and do the same thing as in my post: &a + 1, will it be +1? Am I right?
  
  – John Doe
  Mar 26 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JohnDoe Yes, since sizeof(char) is always 1
  
  – Spikatrix
  Mar 26 at 14:48
  

  
  
  
  

add a comment | 

0

As Alex said, pointer math increments by the size of the type, not the byte.

You can make it meet your expectation by casting &a to a char* (aka. byte-pointer):

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, ((char*)&a) + 1, ((char*)&a) + 2, ((char*)&a) + 3);
 return 0;

Here is how I'd write it to make it handle the platform size:

#include <stdio.h>

int main(void) 
 int a = 56;
 char* p = (char*)&a;
 printf("First part of address block:n");
 printf("%pn", &a);
 printf("All parts of address block:n");
 for(int i=0;i<sizeof(a); ++i) printf("%p : [%d] : 0x%02xn", p+i, i, *(p+i));
 return 0;

Output

Success #stdin #stdout 0s 9424KB
First part of address block:
0x7ffff2e39c5c
All parts of address block:
0x7ffff2e39c5c : [0] : 0x38
0x7ffff2e39c5d : [1] : 0x00
0x7ffff2e39c5e : [2] : 0x00
0x7ffff2e39c5f : [3] : 0x00

share|improve this answer

answered Mar 26 at 14:42

abelenkyabelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you give me code how to print also value? If first part is 0000, value will be 0 ... last block is 1000, value will be 8. I tried printf("Value: %d", *((char*)&a)), but it does not work.
  
  – John Doe
  Mar 26 at 15:02
  
  

  
  
  
  

add a comment | 

3

C increment pointers not by bytes, but by size of type is pointed on.
That is, if you have pointer to int it increments by sizeof(int) - in your example is 4.

share|improve this answer

answered Mar 26 at 14:33

AlexAlex

39622 silver badges1313 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  When I have char a = 'A' and do the same thing as in my post: &a + 1, will it be +1? Am I right?
  
  – John Doe
  Mar 26 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JohnDoe Yes, since sizeof(char) is always 1
  
  – Spikatrix
  Mar 26 at 14:48
  

  
  
  
  

add a comment | 

3

C increment pointers not by bytes, but by size of type is pointed on.
That is, if you have pointer to int it increments by sizeof(int) - in your example is 4.

share|improve this answer

answered Mar 26 at 14:33

AlexAlex

39622 silver badges1313 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  When I have char a = 'A' and do the same thing as in my post: &a + 1, will it be +1? Am I right?
  
  – John Doe
  Mar 26 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JohnDoe Yes, since sizeof(char) is always 1
  
  – Spikatrix
  Mar 26 at 14:48
  

  
  
  
  

add a comment | 

C increment pointers not by bytes, but by size of type is pointed on.
That is, if you have pointer to int it increments by sizeof(int) - in your example is 4.

share|improve this answer

answered Mar 26 at 14:33

AlexAlex

39622 silver badges1313 bronze badges

share|improve this answer

answered Mar 26 at 14:33

AlexAlex

39622 silver badges1313 bronze badges

answered Mar 26 at 14:33

AlexAlex

39622 silver badges1313 bronze badges

answered Mar 26 at 14:33

AlexAlex

39622 silver badges1313 bronze badges

0

As Alex said, pointer math increments by the size of the type, not the byte.

You can make it meet your expectation by casting &a to a char* (aka. byte-pointer):

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, ((char*)&a) + 1, ((char*)&a) + 2, ((char*)&a) + 3);
 return 0;

Here is how I'd write it to make it handle the platform size:

#include <stdio.h>

int main(void) 
 int a = 56;
 char* p = (char*)&a;
 printf("First part of address block:n");
 printf("%pn", &a);
 printf("All parts of address block:n");
 for(int i=0;i<sizeof(a); ++i) printf("%p : [%d] : 0x%02xn", p+i, i, *(p+i));
 return 0;

Output

Success #stdin #stdout 0s 9424KB
First part of address block:
0x7ffff2e39c5c
All parts of address block:
0x7ffff2e39c5c : [0] : 0x38
0x7ffff2e39c5d : [1] : 0x00
0x7ffff2e39c5e : [2] : 0x00
0x7ffff2e39c5f : [3] : 0x00

share|improve this answer

answered Mar 26 at 14:42

abelenkyabelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you give me code how to print also value? If first part is 0000, value will be 0 ... last block is 1000, value will be 8. I tried printf("Value: %d", *((char*)&a)), but it does not work.
  
  – John Doe
  Mar 26 at 15:02
  
  

  
  
  
  

add a comment | 

0

As Alex said, pointer math increments by the size of the type, not the byte.

You can make it meet your expectation by casting &a to a char* (aka. byte-pointer):

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, ((char*)&a) + 1, ((char*)&a) + 2, ((char*)&a) + 3);
 return 0;

Here is how I'd write it to make it handle the platform size:

#include <stdio.h>

int main(void) 
 int a = 56;
 char* p = (char*)&a;
 printf("First part of address block:n");
 printf("%pn", &a);
 printf("All parts of address block:n");
 for(int i=0;i<sizeof(a); ++i) printf("%p : [%d] : 0x%02xn", p+i, i, *(p+i));
 return 0;

Output

Success #stdin #stdout 0s 9424KB
First part of address block:
0x7ffff2e39c5c
All parts of address block:
0x7ffff2e39c5c : [0] : 0x38
0x7ffff2e39c5d : [1] : 0x00
0x7ffff2e39c5e : [2] : 0x00
0x7ffff2e39c5f : [3] : 0x00

share|improve this answer

answered Mar 26 at 14:42

abelenkyabelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you give me code how to print also value? If first part is 0000, value will be 0 ... last block is 1000, value will be 8. I tried printf("Value: %d", *((char*)&a)), but it does not work.
  
  – John Doe
  Mar 26 at 15:02
  
  

  
  
  
  

add a comment | 

As Alex said, pointer math increments by the size of the type, not the byte.

You can make it meet your expectation by casting &a to a char* (aka. byte-pointer):

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, ((char*)&a) + 1, ((char*)&a) + 2, ((char*)&a) + 3);
 return 0;

Here is how I'd write it to make it handle the platform size:

#include <stdio.h>

int main(void) 
 int a = 56;
 char* p = (char*)&a;
 printf("First part of address block:n");
 printf("%pn", &a);
 printf("All parts of address block:n");
 for(int i=0;i<sizeof(a); ++i) printf("%p : [%d] : 0x%02xn", p+i, i, *(p+i));
 return 0;

Output

Success #stdin #stdout 0s 9424KB
First part of address block:
0x7ffff2e39c5c
All parts of address block:
0x7ffff2e39c5c : [0] : 0x38
0x7ffff2e39c5d : [1] : 0x00
0x7ffff2e39c5e : [2] : 0x00
0x7ffff2e39c5f : [3] : 0x00

share|improve this answer

answered Mar 26 at 14:42

abelenkyabelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

int main(void) 
 int a = 56;
 printf("First part of address block:n");
 printf("%p", &a);
 printf("All parts of address block:n");
 printf("%pn%pn%pn%p", &a, ((char*)&a) + 1, ((char*)&a) + 2, ((char*)&a) + 3);
 return 0;

#include <stdio.h>

int main(void) 
 int a = 56;
 char* p = (char*)&a;
 printf("First part of address block:n");
 printf("%pn", &a);
 printf("All parts of address block:n");
 for(int i=0;i<sizeof(a); ++i) printf("%p : [%d] : 0x%02xn", p+i, i, *(p+i));
 return 0;

Success #stdin #stdout 0s 9424KB
First part of address block:
0x7ffff2e39c5c
All parts of address block:
0x7ffff2e39c5c : [0] : 0x38
0x7ffff2e39c5d : [1] : 0x00
0x7ffff2e39c5e : [2] : 0x00
0x7ffff2e39c5f : [3] : 0x00

share|improve this answer

answered Mar 26 at 14:42

abelenkyabelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

answered Mar 26 at 14:42

abelenkyabelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges

answered Mar 26 at 14:42

abelenkyabelenky

48.8k2121 gold badges8787 silver badges137137 bronze badges