双击打开避免一闪而逝，命令行自动忽略

废话不多说直接上代码，很多人在写程序时双击打开都会一闪而逝，因此都会在程序执行最后加上获取输入的代码。但是命令行时又不想再敲一次回车。下面代码就能解决你的烦恼，原理就是判断父进程是否为cmd.exe，如果不是则说明不是命令行打开，则加上获取输入回车。

package main

import (
	"fmt"
	"syscall"
	"unsafe"
)

func main() {
	if name, err := getParentProcessName(); err == nil && name != "cmd.exe" {
		defer fmt.Scanln() // 不是命令行时避免一闪而逝
	}
	fmt.Println("hello word!")
}

func getParentProcessName() (string, error) {
	snapshot, err := syscall.CreateToolhelp32Snapshot(syscall.TH32CS_SNAPPROCESS, 0)
	if err != nil {
		return "", err
	}
	defer syscall.CloseHandle(snapshot)
	var procEntry syscall.ProcessEntry32
	procEntry.Size = uint32(unsafe.Sizeof(procEntry))
	if err = syscall.Process32First(snapshot, &procEntry); err != nil {
		return "", err
	}
	var (
		pid      = uint32(syscall.Getpid())
		pName    = make(map[uint32]string, 32)
		parentId = uint32(1<<32 - 1)
	)
	for {
		pName[procEntry.ProcessID] = syscall.UTF16ToString(procEntry.ExeFile[:])
		if procEntry.ProcessID == pid {
			parentId = procEntry.ParentProcessID
		}
		if s, ok := pName[parentId]; ok {
			return s, nil
		}
		err = syscall.Process32Next(snapshot, &procEntry)
		if err != nil {
			return "", err
		}
	}
}