Print whole address block of some variable [closed]What are the differences between a pointer variable and a reference variable in C++?Improve INSERT-per-second performance of SQLite?Correct format specifier to print pointer or address?When defining a variable in C for example, where is the memory address for that variable stored?Storage of variables and dereferencing themVariable addresses are not aligned when they are not printedPrinting pointer's address in CHow I can remove all odds bits in C?Return binary results with newline incrementsPrinting type-casted types in C
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Print whole address block of some variable [closed]
What are the differences between a pointer variable and a reference variable in C++?Improve INSERT-per-second performance of SQLite?Correct format specifier to print pointer or address?When defining a variable in C for example, where is the memory address for that variable stored?Storage of variables and dereferencing themVariable addresses are not aligned when they are not printedPrinting pointer's address in CHow I can remove all odds bits in C?Return binary results with newline incrementsPrinting type-casted types in C
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom: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
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
asked Mar 26 at 14:28
John DoeJohn Doe
11 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.
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
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
asked Mar 26 at 14:28
John DoeJohn Doe
11 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.
intcommonly has a size of4, 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 tovoid *to get the proper type for%p. Also, print the size ofa(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
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
asked Mar 26 at 14:28
John DoeJohn Doe
11 bronze badge
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
c pointers hex memory-address
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
asked Mar 26 at 14:28
John DoeJohn Doe
11 bronze badge
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
asked Mar 26 at 14:28
John DoeJohn Doe
11 bronze badge
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
edited Mar 26 at 14:50
abelenky
48.8k21 gold badges87 silver badges137 bronze badges
48.8k21 gold badges87 silver badges137 bronze badges
asked Mar 26 at 14:28
John DoeJohn Doe
11 bronze badge
asked Mar 26 at 14:28
John DoeJohn Doe
11 bronze badge
asked Mar 26 at 14:28
John DoeJohn Doe
11 bronze badge
11 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.
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.
intcommonly has a size of4, 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 tovoid *to get the proper type for%p. Also, print the size ofa(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 |
intcommonly has a size of4, 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 tovoid *to get the proper type for%p. Also, print the size ofa(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
int commonly has a size of 4, so that output looks right to me.– Blaze
Mar 26 at 14:30
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 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
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
3
What is the question?
– Lundin
Mar 26 at 14:52
What is the question?
– Lundin
Mar 26 at 14:52
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 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, sincesizeof(char)is always 1
– Spikatrix
Mar 26 at 14:48
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
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 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 triedprintf("Value: %d", *((char*)&a)), but it does not work.
– John Doe
Mar 26 at 15:02
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 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, sincesizeof(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.
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 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, sincesizeof(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.
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 bronze badges
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.
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 bronze badges
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 bronze badges
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 bronze badges
answered Mar 26 at 14:33
AlexAlex
3962 silver badges13 bronze badges
3962 silver badges13 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, sincesizeof(char)is always 1
– Spikatrix
Mar 26 at 14:48
add a comment |
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, sincesizeof(char)is always 1
– Spikatrix
Mar 26 at 14:48
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
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
@JohnDoe Yes, since
sizeof(char) is always 1– Spikatrix
Mar 26 at 14:48
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
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 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 triedprintf("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
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 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 triedprintf("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
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 bronze badges
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
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 bronze badges
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 bronze badges
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 bronze badges
answered Mar 26 at 14:42
abelenkyabelenky
48.8k21 gold badges87 silver badges137 bronze badges
48.8k21 gold badges87 silver badges137 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 triedprintf("Value: %d", *((char*)&a)), but it does not work.
– John Doe
Mar 26 at 15:02
add a comment |
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 triedprintf("Value: %d", *((char*)&a)), but it does not work.
– John Doe
Mar 26 at 15:02
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
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 |
intcommonly has a size of4, 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 ofa(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