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What is “:-!!” in C code?memcpy - cast to pointer from integer of different sizeassignment makes pointer from integer without a cast [enabled by default] when assigning char pointer to character arrayCompiling an application for use in highly radioactive environmentsPop from Stack of arrays in C - assigning NULL to a int *C pointer difference Char Intwhy does it cause error when trying to give the original pointer value a new value?How to change content of char * const after it has been set (C)array name decaying to pointer in c - error compilingC the Heap using in Fuctions still not clear
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I have this code:
int *b;
b = 50;
printf("Pointer point to address: %p and also point to this value: %d", b, *b);
return 0
I got this error:
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
timeout: the monitored command dumped core
sh: line 1: 47524 Segmentation fault timeout 10s main
I wanna print value from fifty byte of memory.
Is my code right or will compiler do it work.
c pointers memory-management
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
|
show 1 more comment
I have this code:
int *b;
b = 50;
printf("Pointer point to address: %p and also point to this value: %d", b, *b);
return 0
I got this error:
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
timeout: the monitored command dumped core
sh: line 1: 47524 Segmentation fault timeout 10s main
I wanna print value from fifty byte of memory.
Is my code right or will compiler do it work.
c pointers memory-management
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
2
The compiler is always right. The compiler gave you a warning. You ignored it and look what happened. The warning is actually pretty obvious and something most C programmer will recognise...
– John3136
Mar 26 at 23:06
1
I strongly suggest you go thru any basic c tutorial about pointers and references.
– Marcin Orlowski
Mar 26 at 23:07
2
50is probably not a good (hard coded) pointer value. Please tryint a = 50; int *b = &a; ...
– Weather Vane
Mar 26 at 23:07
I think I get some value from memory, something like leak.
– John Doe
Mar 26 at 23:13
2
You have tried to read memory address50which probably does not belong to you, hence the segfault. The compiler objected because50is an integer value, not a pointer value, hence the warning.
– Weather Vane
Mar 26 at 23:15
|
show 1 more comment
I have this code:
int *b;
b = 50;
printf("Pointer point to address: %p and also point to this value: %d", b, *b);
return 0
I got this error:
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
timeout: the monitored command dumped core
sh: line 1: 47524 Segmentation fault timeout 10s main
I wanna print value from fifty byte of memory.
Is my code right or will compiler do it work.
c pointers memory-management
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
I have this code:
int *b;
b = 50;
printf("Pointer point to address: %p and also point to this value: %d", b, *b);
return 0
I got this error:
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
timeout: the monitored command dumped core
sh: line 1: 47524 Segmentation fault timeout 10s main
I wanna print value from fifty byte of memory.
Is my code right or will compiler do it work.
c pointers memory-management
c pointers memory-management
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
asked Mar 26 at 23:03
John DoeJohn Doe
11 bronze badge
11 bronze badge
2
The compiler is always right. The compiler gave you a warning. You ignored it and look what happened. The warning is actually pretty obvious and something most C programmer will recognise...
– John3136
Mar 26 at 23:06
1
I strongly suggest you go thru any basic c tutorial about pointers and references.
– Marcin Orlowski
Mar 26 at 23:07
2
50is probably not a good (hard coded) pointer value. Please tryint a = 50; int *b = &a; ...
– Weather Vane
Mar 26 at 23:07
I think I get some value from memory, something like leak.
– John Doe
Mar 26 at 23:13
2
You have tried to read memory address50which probably does not belong to you, hence the segfault. The compiler objected because50is an integer value, not a pointer value, hence the warning.
– Weather Vane
Mar 26 at 23:15
|
show 1 more comment
2
The compiler is always right. The compiler gave you a warning. You ignored it and look what happened. The warning is actually pretty obvious and something most C programmer will recognise...
– John3136
Mar 26 at 23:06
1
I strongly suggest you go thru any basic c tutorial about pointers and references.
– Marcin Orlowski
Mar 26 at 23:07
2
50is probably not a good (hard coded) pointer value. Please tryint a = 50; int *b = &a; ...
– Weather Vane
Mar 26 at 23:07
I think I get some value from memory, something like leak.
– John Doe
Mar 26 at 23:13
2
You have tried to read memory address50which probably does not belong to you, hence the segfault. The compiler objected because50is an integer value, not a pointer value, hence the warning.
– Weather Vane
Mar 26 at 23:15
2
2
The compiler is always right. The compiler gave you a warning. You ignored it and look what happened. The warning is actually pretty obvious and something most C programmer will recognise...
– John3136
Mar 26 at 23:06
The compiler is always right. The compiler gave you a warning. You ignored it and look what happened. The warning is actually pretty obvious and something most C programmer will recognise...
– John3136
Mar 26 at 23:06
1
1
I strongly suggest you go thru any basic c tutorial about pointers and references.
– Marcin Orlowski
Mar 26 at 23:07
I strongly suggest you go thru any basic c tutorial about pointers and references.
– Marcin Orlowski
Mar 26 at 23:07
2
2
50 is probably not a good (hard coded) pointer value. Please try int a = 50; int *b = &a; ...– Weather Vane
Mar 26 at 23:07
50 is probably not a good (hard coded) pointer value. Please try int a = 50; int *b = &a; ...– Weather Vane
Mar 26 at 23:07
I think I get some value from memory, something like leak.
– John Doe
Mar 26 at 23:13
I think I get some value from memory, something like leak.
– John Doe
Mar 26 at 23:13
2
2
You have tried to read memory address
50 which probably does not belong to you, hence the segfault. The compiler objected because 50 is an integer value, not a pointer value, hence the warning.– Weather Vane
Mar 26 at 23:15
You have tried to read memory address
50 which probably does not belong to you, hence the segfault. The compiler objected because 50 is an integer value, not a pointer value, hence the warning.– Weather Vane
Mar 26 at 23:15
|
show 1 more comment
1 Answer
1
active
oldest
votes
Assuming you are running the program on a recent OS like Linux, Mac or Windows, 50 will no be the bytes located at the address 50 in your physical memory ; that's an address in a virtual space.
Then your process (program) has only access to a very limited range in that virtual space, which 50 is unlikely to be from ; in that case the OS protects the illegal access and stops your process (segfault) ; anyway you could even get a result that may or may not be the correct one, this is called undefined behavior, and you better not rely in this case on a apparently working executable.
To access directly the physical memory, you either need to build a kernel module, or boot from a DOS-like OS, for instance.
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
This being said, you need to cast 50 as a int * to clear the warning.
b = (int *)50;
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assuming you are running the program on a recent OS like Linux, Mac or Windows, 50 will no be the bytes located at the address 50 in your physical memory ; that's an address in a virtual space.
Then your process (program) has only access to a very limited range in that virtual space, which 50 is unlikely to be from ; in that case the OS protects the illegal access and stops your process (segfault) ; anyway you could even get a result that may or may not be the correct one, this is called undefined behavior, and you better not rely in this case on a apparently working executable.
To access directly the physical memory, you either need to build a kernel module, or boot from a DOS-like OS, for instance.
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
This being said, you need to cast 50 as a int * to clear the warning.
b = (int *)50;
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
add a comment |
Assuming you are running the program on a recent OS like Linux, Mac or Windows, 50 will no be the bytes located at the address 50 in your physical memory ; that's an address in a virtual space.
Then your process (program) has only access to a very limited range in that virtual space, which 50 is unlikely to be from ; in that case the OS protects the illegal access and stops your process (segfault) ; anyway you could even get a result that may or may not be the correct one, this is called undefined behavior, and you better not rely in this case on a apparently working executable.
To access directly the physical memory, you either need to build a kernel module, or boot from a DOS-like OS, for instance.
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
This being said, you need to cast 50 as a int * to clear the warning.
b = (int *)50;
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
add a comment |
Assuming you are running the program on a recent OS like Linux, Mac or Windows, 50 will no be the bytes located at the address 50 in your physical memory ; that's an address in a virtual space.
Then your process (program) has only access to a very limited range in that virtual space, which 50 is unlikely to be from ; in that case the OS protects the illegal access and stops your process (segfault) ; anyway you could even get a result that may or may not be the correct one, this is called undefined behavior, and you better not rely in this case on a apparently working executable.
To access directly the physical memory, you either need to build a kernel module, or boot from a DOS-like OS, for instance.
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
This being said, you need to cast 50 as a int * to clear the warning.
b = (int *)50;
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
Assuming you are running the program on a recent OS like Linux, Mac or Windows, 50 will no be the bytes located at the address 50 in your physical memory ; that's an address in a virtual space.
Then your process (program) has only access to a very limited range in that virtual space, which 50 is unlikely to be from ; in that case the OS protects the illegal access and stops your process (segfault) ; anyway you could even get a result that may or may not be the correct one, this is called undefined behavior, and you better not rely in this case on a apparently working executable.
To access directly the physical memory, you either need to build a kernel module, or boot from a DOS-like OS, for instance.
main.c:6:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
This being said, you need to cast 50 as a int * to clear the warning.
b = (int *)50;
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
edited Mar 27 at 0:01
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
answered Mar 26 at 23:43
Ring ØRing Ø
22.5k4 gold badges59 silver badges86 bronze badges
22.5k4 gold badges59 silver badges86 bronze badges
add a comment |
add a comment |
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2
The compiler is always right. The compiler gave you a warning. You ignored it and look what happened. The warning is actually pretty obvious and something most C programmer will recognise...
– John3136
Mar 26 at 23:06
1
I strongly suggest you go thru any basic c tutorial about pointers and references.
– Marcin Orlowski
Mar 26 at 23:07
2
50is probably not a good (hard coded) pointer value. Please tryint a = 50; int *b = &a; ...– Weather Vane
Mar 26 at 23:07
I think I get some value from memory, something like leak.
– John Doe
Mar 26 at 23:13
2
You have tried to read memory address
50which probably does not belong to you, hence the segfault. The compiler objected because50is an integer value, not a pointer value, hence the warning.– Weather Vane
Mar 26 at 23:15