snippetgoCritical
How do I print the pointer value of a Go object? What does the pointer value mean?
Viewed 0 times
objecthowpointertheprintdoesmeanvaluewhat
Problem
I am just playing around with Go and do not yet have a good mental model of when structs are passed by value or by reference.
This may be a very dumb question but I just want to experiment a bit and see if I am still working on the same object or I have made a copy of it (passed it by value).
Is there a way to print the pointer (or internal id if pointer value is changed by gc) of an object?
gives on my windows (8g compiled version):
Why does the pointer value from within the go routine show a different value? The quantity on the original object did get changed so it was working with the same object. Is there a way to see an object id that is persistent?
This may be a very dumb question but I just want to experiment a bit and see if I am still working on the same object or I have made a copy of it (passed it by value).
Is there a way to print the pointer (or internal id if pointer value is changed by gc) of an object?
package main
import ( "runtime" )
type Something struct {
number int
queue chan int
}
func gotest( s *Something, done chan bool ) {
println( "from gotest:")
println( &s )
for num := range s.queue {
println( num )
s.number = num
}
done <- true
}
func main() {
runtime.GOMAXPROCS(4)
s := new(Something)
println(&s)
s.queue = make(chan int)
done := make(chan bool)
go gotest(s, done)
s.queue <- 42
close(s.queue)
<- done
println(&s)
println(s.number)
}gives on my windows (8g compiled version):
0x4930d4
from gotest:
0x4974d8
42
0x4930d4
42Why does the pointer value from within the go routine show a different value? The quantity on the original object did get changed so it was working with the same object. Is there a way to see an object id that is persistent?
Solution
Go function arguments are passed by value.
First, let's discard the irrelevant parts of your example, so that we can easily see that you are merely passing an argument by value. For example,
Output:
In function
In function
In function
In function
In function
In function
In your example, the function
In your example, (
You should take mkb's sage advice and stop using
Pointers have the same value when they point to the same memory location; pointers have different values when they point to different memory locations.
First, let's discard the irrelevant parts of your example, so that we can easily see that you are merely passing an argument by value. For example,
package main
import "fmt"
func byval(q *int) {
fmt.Printf("3. byval -- q %T: &q=%p q=&i=%p *q=i=%v\n", q, &q, q, *q)
*q = 4143
fmt.Printf("4. byval -- q %T: &q=%p q=&i=%p *q=i=%v\n", q, &q, q, *q)
q = nil
}
func main() {
i := int(42)
fmt.Printf("1. main -- i %T: &i=%p i=%v\n", i, &i, i)
p := &i
fmt.Printf("2. main -- p %T: &p=%p p=&i=%p *p=i=%v\n", p, &p, p, *p)
byval(p)
fmt.Printf("5. main -- p %T: &p=%p p=&i=%p *p=i=%v\n", p, &p, p, *p)
fmt.Printf("6. main -- i %T: &i=%p i=%v\n", i, &i, i)
}Output:
1. main -- i int: &i=0xf840000040 i=42
2. main -- p *int: &p=0xf8400000f0 p=&i=0xf840000040 *p=i=42
3. byval -- q *int: &q=0xf8400000d8 q=&i=0xf840000040 *q=i=42
4. byval -- q *int: &q=0xf8400000d8 q=&i=0xf840000040 *q=i=4143
5. main -- p *int: &p=0xf8400000f0 p=&i=0xf840000040 *p=i=4143
6. main -- i int: &i=0xf840000040 i=4143In function
main, i is an int variable at memory location (&i) 0xf800000040 with an initial value (i) 42.In function
main, p is a pointer to an int variable at memory location (&p) 0xf8000000f0 with a value (p=&i) 0xf800000040 which points to an int value (*p=i) 42.In function
main, byval(p) is a function call which assigns the value (p=&i) 0xf800000040 of the argument at memory location (&p) 0xf8000000f0 to the function byval parameter q at memory location (&q) 0xf8000000d8. In other words, memory is allocated for the byval parameter q and the value of the main byval argument p is assigned to it; the values of p and q are initially the same, but the variables p and q are distinct.In function
byval, using pointer q (int), which is a copy of pointer p (int), integer *q (i) is set to a new int value 4143. At the end before returning. the pointer q is set to nil (zero value), which has no effect on p since q is a copy.In function
main, p is a pointer to an int variable at memory location (&p) 0xf8000000f0 with a value (p=&i) 0xf800000040 which points to a new int value (*p=i) 4143.In function
main, i is an int variable at memory location (&i) 0xf800000040 with a final value (i) 4143.In your example, the function
main variable s used as an argument to the function gotest call is not the same as the function gotest parameter s. They have the same name, but are different variables with different scopes and memory locations. The function parameter s hides the function call argument s. That's why in my example, I named the argument and parameter variables p and q respectively to emphasize the difference.In your example, (
&s) 0x4930d4 is the address of the memory location for the variable s in function main that is used as an argument to the function call gotest(s, done), and 0x4974d8 is the address of the memory location for the function gotest parameter s. If you set parameter s = nil at the end of function gotest, it has no effect on variable s in main; s in main and s in gotest are distinct memory locations. In terms of types, &s is **Something, s is Something, and s is Something. &s is a pointer to (address of memory location) s, which is a pointer to (address of memory location) an anonymous variable of type Something. In terms of values, main.&s != gotest.&s, main.s == gotest.s, main.s == gotest.s, and main.s.number == gotest.s.number.You should take mkb's sage advice and stop using
println(&s). Use the fmt package, for example,fmt.Printf("%v %p %v\n", &s, s, *s)Pointers have the same value when they point to the same memory location; pointers have different values when they point to different memory locations.
Code Snippets
package main
import "fmt"
func byval(q *int) {
fmt.Printf("3. byval -- q %T: &q=%p q=&i=%p *q=i=%v\n", q, &q, q, *q)
*q = 4143
fmt.Printf("4. byval -- q %T: &q=%p q=&i=%p *q=i=%v\n", q, &q, q, *q)
q = nil
}
func main() {
i := int(42)
fmt.Printf("1. main -- i %T: &i=%p i=%v\n", i, &i, i)
p := &i
fmt.Printf("2. main -- p %T: &p=%p p=&i=%p *p=i=%v\n", p, &p, p, *p)
byval(p)
fmt.Printf("5. main -- p %T: &p=%p p=&i=%p *p=i=%v\n", p, &p, p, *p)
fmt.Printf("6. main -- i %T: &i=%p i=%v\n", i, &i, i)
}1. main -- i int: &i=0xf840000040 i=42
2. main -- p *int: &p=0xf8400000f0 p=&i=0xf840000040 *p=i=42
3. byval -- q *int: &q=0xf8400000d8 q=&i=0xf840000040 *q=i=42
4. byval -- q *int: &q=0xf8400000d8 q=&i=0xf840000040 *q=i=4143
5. main -- p *int: &p=0xf8400000f0 p=&i=0xf840000040 *p=i=4143
6. main -- i int: &i=0xf840000040 i=4143fmt.Printf("%v %p %v\n", &s, s, *s)Context
Stack Overflow Q#4938612, score: 151
Revisions (0)
No revisions yet.