Automatic variable, pointer after function still gets it
0
compiling under Windows with gcc, from this code:
#include <stdio.h>
int f(int ** iptr)
{
printf("f initial value: %in",**iptr);
int a =10;
*iptr = &a;
printf("f changed value: %in",**iptr);
return 0;
}
int main()
{
int * p = 0;
int i =7;
p = &i;
printf("main initial value: %in",*p);
f(&p);
printf("main after f value under p: %in",*p);
printf("again value under p: %in",*p);
printf("value of i: %in",i);
return 0;
}
I get this result:
main initial value: 7
f initial value: 7
f changed value: 10
main after f value under p: 10
again value under p: 1963422240
value of i: 7
Any idea, why this happens so?
Thanks!
P.S. And formatting of this website complains, that I did not write enough conversation here... So, being social... :)
c pointers variables
asked Nov 24 '18 at 21:04
MarcinMarcin
11
add a comment |
0
compiling under Windows with gcc, from this code:
#include <stdio.h>
int f(int ** iptr)
{
printf("f initial value: %in",**iptr);
int a =10;
*iptr = &a;
printf("f changed value: %in",**iptr);
return 0;
}
int main()
{
int * p = 0;
int i =7;
p = &i;
printf("main initial value: %in",*p);
f(&p);
printf("main after f value under p: %in",*p);
printf("again value under p: %in",*p);
printf("value of i: %in",i);
return 0;
}
I get this result:
main initial value: 7
f initial value: 7
f changed value: 10
main after f value under p: 10
again value under p: 1963422240
value of i: 7
Any idea, why this happens so?
Thanks!
P.S. And formatting of this website complains, that I did not write enough conversation here... So, being social... :)
c pointers variables
asked Nov 24 '18 at 21:04
MarcinMarcin
11
1
That's undefined behavior in a nutshell. Besides that, the memory the pointer is pointing to doesn't physically disappear, it's just invalid to dereference it in any way.
– Some programmer dude
Nov 24 '18 at 21:07
Yes, but the first dereference still gets value 10, like it seems to survive the function call. I will have to check, if this is a matter of time. Like after longer time it will not show the correct value.
– Marcin
Nov 24 '18 at 21:14
And yes, I placed a delay(100); function and this first dereference no longer showed 10. It is just about short timing.
– Marcin
Nov 24 '18 at 21:22
Or, it is actually about stack being used by delay() function.
– Marcin
Nov 24 '18 at 21:25
It's not a "matter of time", it's just plain wrong. Don't dereference pointers that are invalid.
– Some programmer dude
Nov 25 '18 at 9:01
add a comment |
0
0
0
compiling under Windows with gcc, from this code:
#include <stdio.h>
int f(int ** iptr)
{
printf("f initial value: %in",**iptr);
int a =10;
*iptr = &a;
printf("f changed value: %in",**iptr);
return 0;
}
int main()
{
int * p = 0;
int i =7;
p = &i;
printf("main initial value: %in",*p);
f(&p);
printf("main after f value under p: %in",*p);
printf("again value under p: %in",*p);
printf("value of i: %in",i);
return 0;
}
I get this result:
main initial value: 7
f initial value: 7
f changed value: 10
main after f value under p: 10
again value under p: 1963422240
value of i: 7
Any idea, why this happens so?
Thanks!
P.S. And formatting of this website complains, that I did not write enough conversation here... So, being social... :)
c pointers variables
asked Nov 24 '18 at 21:04
MarcinMarcin
11
compiling under Windows with gcc, from this code:
#include <stdio.h>
int f(int ** iptr)
{
printf("f initial value: %in",**iptr);
int a =10;
*iptr = &a;
printf("f changed value: %in",**iptr);
return 0;
}
int main()
{
int * p = 0;
int i =7;
p = &i;
printf("main initial value: %in",*p);
f(&p);
printf("main after f value under p: %in",*p);
printf("again value under p: %in",*p);
printf("value of i: %in",i);
return 0;
}
I get this result:
main initial value: 7
f initial value: 7
f changed value: 10
main after f value under p: 10
again value under p: 1963422240
value of i: 7
Any idea, why this happens so?
Thanks!
P.S. And formatting of this website complains, that I did not write enough conversation here... So, being social... :)
c pointers variables
c pointers variables
asked Nov 24 '18 at 21:04
MarcinMarcin
11
asked Nov 24 '18 at 21:04
MarcinMarcin
11
edited Nov 24 '18 at 21:07
Marcin
asked Nov 24 '18 at 21:04
MarcinMarcin
11
asked Nov 24 '18 at 21:04
MarcinMarcin
11
asked Nov 24 '18 at 21:04
MarcinMarcin
11
11
1
That's undefined behavior in a nutshell. Besides that, the memory the pointer is pointing to doesn't physically disappear, it's just invalid to dereference it in any way.
– Some programmer dude
Nov 24 '18 at 21:07
Yes, but the first dereference still gets value 10, like it seems to survive the function call. I will have to check, if this is a matter of time. Like after longer time it will not show the correct value.
– Marcin
Nov 24 '18 at 21:14
And yes, I placed a delay(100); function and this first dereference no longer showed 10. It is just about short timing.
– Marcin
Nov 24 '18 at 21:22
Or, it is actually about stack being used by delay() function.
– Marcin
Nov 24 '18 at 21:25
It's not a "matter of time", it's just plain wrong. Don't dereference pointers that are invalid.
– Some programmer dude
Nov 25 '18 at 9:01
add a comment |
1
That's undefined behavior in a nutshell. Besides that, the memory the pointer is pointing to doesn't physically disappear, it's just invalid to dereference it in any way.
– Some programmer dude
Nov 24 '18 at 21:07
Yes, but the first dereference still gets value 10, like it seems to survive the function call. I will have to check, if this is a matter of time. Like after longer time it will not show the correct value.
– Marcin
Nov 24 '18 at 21:14
And yes, I placed a delay(100); function and this first dereference no longer showed 10. It is just about short timing.
– Marcin
Nov 24 '18 at 21:22
Or, it is actually about stack being used by delay() function.
– Marcin
Nov 24 '18 at 21:25
It's not a "matter of time", it's just plain wrong. Don't dereference pointers that are invalid.
– Some programmer dude
Nov 25 '18 at 9:01
1
1
That's undefined behavior in a nutshell. Besides that, the memory the pointer is pointing to doesn't physically disappear, it's just invalid to dereference it in any way.
– Some programmer dude
Nov 24 '18 at 21:07
That's undefined behavior in a nutshell. Besides that, the memory the pointer is pointing to doesn't physically disappear, it's just invalid to dereference it in any way.
– Some programmer dude
Nov 24 '18 at 21:07
Yes, but the first dereference still gets value 10, like it seems to survive the function call. I will have to check, if this is a matter of time. Like after longer time it will not show the correct value.
– Marcin
Nov 24 '18 at 21:14
Yes, but the first dereference still gets value 10, like it seems to survive the function call. I will have to check, if this is a matter of time. Like after longer time it will not show the correct value.
– Marcin
Nov 24 '18 at 21:14
And yes, I placed a delay(100); function and this first dereference no longer showed 10. It is just about short timing.
– Marcin
Nov 24 '18 at 21:22
And yes, I placed a delay(100); function and this first dereference no longer showed 10. It is just about short timing.
– Marcin
Nov 24 '18 at 21:22
Or, it is actually about stack being used by delay() function.
– Marcin
Nov 24 '18 at 21:25
Or, it is actually about stack being used by delay() function.
– Marcin
Nov 24 '18 at 21:25
It's not a "matter of time", it's just plain wrong. Don't dereference pointers that are invalid.
– Some programmer dude
Nov 25 '18 at 9:01
It's not a "matter of time", it's just plain wrong. Don't dereference pointers that are invalid.
– Some programmer dude
Nov 25 '18 at 9:01
add a comment |
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1
That's undefined behavior in a nutshell. Besides that, the memory the pointer is pointing to doesn't physically disappear, it's just invalid to dereference it in any way.
– Some programmer dude
Nov 24 '18 at 21:07
Yes, but the first dereference still gets value 10, like it seems to survive the function call. I will have to check, if this is a matter of time. Like after longer time it will not show the correct value.
– Marcin
Nov 24 '18 at 21:14
And yes, I placed a delay(100); function and this first dereference no longer showed 10. It is just about short timing.
– Marcin
Nov 24 '18 at 21:22
Or, it is actually about stack being used by delay() function.
– Marcin
Nov 24 '18 at 21:25
It's not a "matter of time", it's just plain wrong. Don't dereference pointers that are invalid.
– Some programmer dude
Nov 25 '18 at 9:01