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
        }
    }
}

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