简单的方法来分配int指针值？ [英] Easy way to assign int pointer values?

问题描述

给定一个 struct ，看起来像

  type foo struct { 
i * int 
} 
  

如果我想设置 i 为1，我必须

  throwAway：= 1 
 instance：= foo {i：& throwAway} 
  

有没有办法在一行中做到这一点，而不必给我新的 i 赋值它自己的名字（在这种情况下 throwaway ）？

解决方案

正如邮件列表，你可以这样做：

  func intPtr（int）* int {
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
  instance：= foo {i：intPtr（1）} 
  
如果你必须经常这样做。  intPtr 获取内联（见 go build -gcflags'-m'输出），所以它应该接近没有性能罚款。
 
Given a struct that looks like
type foo struct {
 i *int
}
if I want to set i to 1, I must
throwAway := 1
instance := foo { i: &throwAway }
Is there any way to do this in a single line without having to give my new i value it's own name (in this case throwaway)?
解决方案
As pointed in the mailing list, you can just do this:
func intPtr(i int) *int {
    return &i
}
and then
instance := foo { i: intPtr(1) }
if you have to do it often. intPtr gets inlined (see go build -gcflags '-m' output), so it should have next to no performance penalty.

                        
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