ring.Link() function in Golang is used to connect two ring r and ring s ring.next() function connect last node of ring r to first node of ring s .so that for r.next() must not be empty. Otherwise, it will throw an error. It works like a circular linked list.
Syntax:
func (r *Ring) Link(s *Ring) *Ring
It does not return anything.
Example 1:
// Golang program to illustrate // the ring.Link() function package main import ( "container/ring"
"fmt"
) // Main function func main() { // Create two rings, a and b, of size 2
a := ring.New(4)
b := ring.New(4)
// Get the length of the ring
m := a.Len()
n := b.Len()
// Initialize a with 0s
for j := 0; j < m; j++ {
a.Value = 0
a = a.Next()
}
// Initialize b with 1s
for i := 0; i < n; i++ {
b.Value = 1
b = b.Next()
}
// Link ring a and ring b
ab := a.Link(b)
ab.Do(func(p interface{}) {
fmt.Println(p.( int ))
})
} |
Output:
0 0 0 0 1 1 1 1
Example 2:
// Golang program to illustrate // the ring.Link() function package main import ( "container/ring"
"fmt"
) // Main function func main() { // Create two rings, r and s, of size 2
r := ring.New(2)
s := ring.New(2)
// Get the length of the ring
lr := r.Len()
ls := s.Len()
// Initialize r with "GFG"
for i := 0; i < lr; i++ {
r.Value = "GFG"
r = r.Next()
}
// Initialize s with "COURSE"
for j := 0; j < ls; j++ {
s.Value = "COURSE"
s = s.Next()
}
// Link ring r and ring s
rs := r.Link(s)
// Iterate through the combined
// ring and print its contents
rs.Do(func(p interface{}) {
fmt.Println(p.(string))
})
} |
Output:
GFG GFG COURSE COURSE
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