publish pointers
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@ -14,7 +14,7 @@ Multiple Return Values
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Variadic Functions
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Closures
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Recursion
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# Pointers
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Pointers
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# New
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# Structs
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# Methods
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@ -1,14 +1,39 @@
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// Go supports <em><a href="http://en.wikipedia.org/wiki/Pointer_(computer_programming)">pointers</a></em>,
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// allowing you to pass references to values and records
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// within your program.
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package main
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import "fmt"
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func zero(xPtr *int) {
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*xPtr = 0
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// We'll show how pointers work in contrast to values with
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// 2 functions: `zeroval` and `zeroptr`. `zeroval` has an
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// `int` parameter, so arguments will be passed to it by
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// value. `zeroval` will get a copy of `ival` distinct
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// from the one in the calling function.
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func zeroval(ival int) {
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ival = 0
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}
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// `zeroptr` in contrast has an `*int` parameter, meaning
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// that it takes an `int` pointer. The `*iptr` code in the
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// function body then _dereferences_ the pointer from its
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// memory address to the current value at that address.
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// Assigning a value to a dereferenced pointer changes the
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// value at the referenced address.
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func zeroptr(iptr *int) {
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*iptr = 0
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}
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func main() {
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x := 5
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fmt.Println(x)
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zero(&x)
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fmt.Println(x)
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i := 1
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fmt.Println("initial:", i)
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zeroval(i)
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fmt.Println("zeroval:", i)
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// The `&i` syntax gives the memory address of `i`,
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// i.e. a pointer to `i`.
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zeroptr(&i)
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fmt.Println("zeroptr:", i)
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}
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7
examples/pointers/pointers.sh
Normal file
7
examples/pointers/pointers.sh
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@ -0,0 +1,7 @@
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# `zeroval` doesn't change the `i` in `main`, but
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# `zeroptr` does because it has a reference to
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# the memory address for that variable.
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$ go run pointers.go
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initial: 1
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zeroval: 1
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zeroptr: 0
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