Creating a variable creates a copy, dereferencing does not. Why?What are the differences between a pointer variable and a reference variable in C++?With arrays, why is it the case that a[5] == 5[a]?What does “dereferencing” a pointer mean?Why does the arrow (->) operator in C exist?Why should I use a pointer rather than the object itself?How does pointer dereferencing work in golang?Is it safe to use the memory address of descoped variables?If you malloc a struct* does it create local variables?Is it possible to initialize a C pointer to NULL?Why is the golang range operator implemented with declaration of a local variable?
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Creating a variable creates a copy, dereferencing does not. Why?
What are the differences between a pointer variable and a reference variable in C++?With arrays, why is it the case that a[5] == 5[a]?What does “dereferencing” a pointer mean?Why does the arrow (->) operator in C exist?Why should I use a pointer rather than the object itself?How does pointer dereferencing work in golang?Is it safe to use the memory address of descoped variables?If you malloc a struct* does it create local variables?Is it possible to initialize a C pointer to NULL?Why is the golang range operator implemented with declaration of a local variable?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
Why does putting a value into a variable create a copy but de-referencing does not?
Is it a simple optimization of the compiler that knows that it can just use the address of the original structure, while creating a variable always allocates new memory?
Example 1:
x1 := &struct x int x: 0
y1 := *x1
z1 := &y1
z1.x++
fmt.Printf("--- 1:n%#vn%#vn", x1, z1)
Example 2:
x2 := &struct x int x: 0
z2 := &*x2
z2.x++
fmt.Printf("--- 2:n%#vn%#vn", x2, z2)
Run here: https://play.golang.org/p/myugNmjrQFj
Is there a a part of the go documentation that describes this behavior?
pointers go
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
add a comment |
Why does putting a value into a variable create a copy but de-referencing does not?
Is it a simple optimization of the compiler that knows that it can just use the address of the original structure, while creating a variable always allocates new memory?
Example 1:
x1 := &struct x int x: 0
y1 := *x1
z1 := &y1
z1.x++
fmt.Printf("--- 1:n%#vn%#vn", x1, z1)
Example 2:
x2 := &struct x int x: 0
z2 := &*x2
z2.x++
fmt.Printf("--- 2:n%#vn%#vn", x2, z2)
Run here: https://play.golang.org/p/myugNmjrQFj
Is there a a part of the go documentation that describes this behavior?
pointers go
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
4
Neither creating a variable, nor dereferencing create copies. Copy happens on assignment.= &*x2copies (assigns) the pointer value to your new variable.
– JimB
Mar 26 at 22:47
So basically, by creating a variable I allocate memory and assigning then fills that memory by copying the data. In my second example the data is never copied because there is no memory allocated where the data could go in between de-referencing and getting the pointer. Did I understand that correctly?
– sirion
Mar 26 at 22:54
2
I guess you could put it that way, but I don't think you need to complicate it so. The simple concept "assignment == copy" works for everything. The odd expression&*x2doesn't assign anything, therefor doesn't copy anything.
– JimB
Mar 26 at 22:59
add a comment |
Why does putting a value into a variable create a copy but de-referencing does not?
Is it a simple optimization of the compiler that knows that it can just use the address of the original structure, while creating a variable always allocates new memory?
Example 1:
x1 := &struct x int x: 0
y1 := *x1
z1 := &y1
z1.x++
fmt.Printf("--- 1:n%#vn%#vn", x1, z1)
Example 2:
x2 := &struct x int x: 0
z2 := &*x2
z2.x++
fmt.Printf("--- 2:n%#vn%#vn", x2, z2)
Run here: https://play.golang.org/p/myugNmjrQFj
Is there a a part of the go documentation that describes this behavior?
pointers go
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
Why does putting a value into a variable create a copy but de-referencing does not?
Is it a simple optimization of the compiler that knows that it can just use the address of the original structure, while creating a variable always allocates new memory?
Example 1:
x1 := &struct x int x: 0
y1 := *x1
z1 := &y1
z1.x++
fmt.Printf("--- 1:n%#vn%#vn", x1, z1)
Example 2:
x2 := &struct x int x: 0
z2 := &*x2
z2.x++
fmt.Printf("--- 2:n%#vn%#vn", x2, z2)
Run here: https://play.golang.org/p/myugNmjrQFj
Is there a a part of the go documentation that describes this behavior?
pointers go
pointers go
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
asked Mar 26 at 22:41
sirionsirion
9335 silver badges12 bronze badges
9335 silver badges12 bronze badges
4
Neither creating a variable, nor dereferencing create copies. Copy happens on assignment.= &*x2copies (assigns) the pointer value to your new variable.
– JimB
Mar 26 at 22:47
So basically, by creating a variable I allocate memory and assigning then fills that memory by copying the data. In my second example the data is never copied because there is no memory allocated where the data could go in between de-referencing and getting the pointer. Did I understand that correctly?
– sirion
Mar 26 at 22:54
2
I guess you could put it that way, but I don't think you need to complicate it so. The simple concept "assignment == copy" works for everything. The odd expression&*x2doesn't assign anything, therefor doesn't copy anything.
– JimB
Mar 26 at 22:59
add a comment |
4
Neither creating a variable, nor dereferencing create copies. Copy happens on assignment.= &*x2copies (assigns) the pointer value to your new variable.
– JimB
Mar 26 at 22:47
So basically, by creating a variable I allocate memory and assigning then fills that memory by copying the data. In my second example the data is never copied because there is no memory allocated where the data could go in between de-referencing and getting the pointer. Did I understand that correctly?
– sirion
Mar 26 at 22:54
2
I guess you could put it that way, but I don't think you need to complicate it so. The simple concept "assignment == copy" works for everything. The odd expression&*x2doesn't assign anything, therefor doesn't copy anything.
– JimB
Mar 26 at 22:59
4
4
Neither creating a variable, nor dereferencing create copies. Copy happens on assignment.
= &*x2 copies (assigns) the pointer value to your new variable.– JimB
Mar 26 at 22:47
Neither creating a variable, nor dereferencing create copies. Copy happens on assignment.
= &*x2 copies (assigns) the pointer value to your new variable.– JimB
Mar 26 at 22:47
So basically, by creating a variable I allocate memory and assigning then fills that memory by copying the data. In my second example the data is never copied because there is no memory allocated where the data could go in between de-referencing and getting the pointer. Did I understand that correctly?
– sirion
Mar 26 at 22:54
So basically, by creating a variable I allocate memory and assigning then fills that memory by copying the data. In my second example the data is never copied because there is no memory allocated where the data could go in between de-referencing and getting the pointer. Did I understand that correctly?
– sirion
Mar 26 at 22:54
2
2
I guess you could put it that way, but I don't think you need to complicate it so. The simple concept "assignment == copy" works for everything. The odd expression
&*x2 doesn't assign anything, therefor doesn't copy anything.– JimB
Mar 26 at 22:59
I guess you could put it that way, but I don't think you need to complicate it so. The simple concept "assignment == copy" works for everything. The odd expression
&*x2 doesn't assign anything, therefor doesn't copy anything.– JimB
Mar 26 at 22:59
add a comment |
1 Answer
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Is there a a part of the go documentation that describes this behavior?
Yes, the language spec. See https://golang.org/ref/spec
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
Gesturing vaguely at a 100-page document covering every aspect of the language doesn't answer the question. Directing them to the relevant headings would offer more value to the asker and to future developers landing on this question from searches. Also quoting the relevant sections would make this an actual answer; a link with no details does not.
– Adrian
Mar 27 at 14:09
add a comment |
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Is there a a part of the go documentation that describes this behavior?
Yes, the language spec. See https://golang.org/ref/spec
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
Gesturing vaguely at a 100-page document covering every aspect of the language doesn't answer the question. Directing them to the relevant headings would offer more value to the asker and to future developers landing on this question from searches. Also quoting the relevant sections would make this an actual answer; a link with no details does not.
– Adrian
Mar 27 at 14:09
add a comment |
Is there a a part of the go documentation that describes this behavior?
Yes, the language spec. See https://golang.org/ref/spec
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
Gesturing vaguely at a 100-page document covering every aspect of the language doesn't answer the question. Directing them to the relevant headings would offer more value to the asker and to future developers landing on this question from searches. Also quoting the relevant sections would make this an actual answer; a link with no details does not.
– Adrian
Mar 27 at 14:09
add a comment |
Is there a a part of the go documentation that describes this behavior?
Yes, the language spec. See https://golang.org/ref/spec
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
Is there a a part of the go documentation that describes this behavior?
Yes, the language spec. See https://golang.org/ref/spec
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
answered Mar 27 at 4:28
VolkerVolker
22.5k3 gold badges52 silver badges60 bronze badges
22.5k3 gold badges52 silver badges60 bronze badges
Gesturing vaguely at a 100-page document covering every aspect of the language doesn't answer the question. Directing them to the relevant headings would offer more value to the asker and to future developers landing on this question from searches. Also quoting the relevant sections would make this an actual answer; a link with no details does not.
– Adrian
Mar 27 at 14:09
add a comment |
Gesturing vaguely at a 100-page document covering every aspect of the language doesn't answer the question. Directing them to the relevant headings would offer more value to the asker and to future developers landing on this question from searches. Also quoting the relevant sections would make this an actual answer; a link with no details does not.
– Adrian
Mar 27 at 14:09
Gesturing vaguely at a 100-page document covering every aspect of the language doesn't answer the question. Directing them to the relevant headings would offer more value to the asker and to future developers landing on this question from searches. Also quoting the relevant sections would make this an actual answer; a link with no details does not.
– Adrian
Mar 27 at 14:09
Gesturing vaguely at a 100-page document covering every aspect of the language doesn't answer the question. Directing them to the relevant headings would offer more value to the asker and to future developers landing on this question from searches. Also quoting the relevant sections would make this an actual answer; a link with no details does not.
– Adrian
Mar 27 at 14:09
add a comment |
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4
Neither creating a variable, nor dereferencing create copies. Copy happens on assignment.
= &*x2copies (assigns) the pointer value to your new variable.– JimB
Mar 26 at 22:47
So basically, by creating a variable I allocate memory and assigning then fills that memory by copying the data. In my second example the data is never copied because there is no memory allocated where the data could go in between de-referencing and getting the pointer. Did I understand that correctly?
– sirion
Mar 26 at 22:54
2
I guess you could put it that way, but I don't think you need to complicate it so. The simple concept "assignment == copy" works for everything. The odd expression
&*x2doesn't assign anything, therefor doesn't copy anything.– JimB
Mar 26 at 22:59