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Radar · Go · Archive · Week 38 · Sep 15 – 21, 2025

Expression to create pointer to simple types

Worth knowinglanguage

What changed
The proposal is to add support for new(expr) by amending the Allocation section to say that new(expr) allocates a variable of type T, initializes it to the value of expr, and returns its address.
Production impact
The source does not say.
Try it
Write p := new(123) and observe that *p is 123.
Source
github.com/golang/go/issues/45624

Understand it, then run it

Go lets you create a new variable and get a pointer to it with the built‑in new. Before the change you could only write new(T) where T is a type, and the new variable would be zero‑valued. The change adds a second form: new(expr). If expr is an expression of type T, new(expr) allocates a variable of type T, stores the value of expr in it, and returns a pointer to that variable. So new(123) now gives you a pointer to an int whose value is 123, just like new(int) gives you a pointer to an int whose value is 0.

Run it now

Todaygo
// This program demonstrates the new `new(expr)` feature.
// It prints the value of a pointer created with new(123).
package main

import "fmt"

func main() {
	// Allocate an int initialized to 123 and get its address.
	p := new(123)

	// Dereference the pointer to read the value.
	fmt.Println(*p) // Output: 123
}

What it printed when we ran it on Go 1.27.1

123

After the change ships

go · the proposal's code; it does not compile until the change ships

// This code does not compile until the change ships in Go 1.26.
package main

import "fmt"

func main() {
	p := new(123)
	fmt.Println(*p)
}

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Written by gpt-oss-20b from the linked source · claims checked against the sources · archive, not individually reviewed