Question

这是我到目前为止的代码：

#include <MsgBoxConstants.au3>
If FileExists(@HomeDrive & "Program Files (x86)") Then
   $arc = 64
Else
   $arc = 32
EndIf
MsgBox("Hello", "Bit", "The computer that the program is been run on is a " & $arc & "bit one.")

当它是64时，它会说我的电脑是32位。所以我查看了路径并制作了新代码：

#include <MsgBoxConstants.au3>
If FileExists(@HomeDrive & "Program Files (x86)") Then
   $arc = 64
Else
   $arc = 32
EndIf
$path = @HomeDrive & "\Program Files (x86)"
MsgBox("Hello", $path, "The computer that the program is been run on is a " & $arc & "bit one.")

这显示了路径，似乎没问题。我哪里错了？

Answer 1

您需要阅读@OSArch宏。

返回以下内容之一：＆＃34; X86＆＃34;，＆＃34; IA64＆＃34;，＆＃34; X64＆＃34; - 这是当前运行的操作系统的体系结构类型。

Answer 2

有没有人知道我哪里出错了。

根据Documentation - Macros：

@OSArch返回以下内容之一：＆＃34; X86＆＃34;，＆＃34; IA64＆＃34;，＆＃34; X64＆＃34; - 这是当前运行的操作系统的体系结构类型。

@CPUArch返回＆＃34; X86＆＃34;当CPU是32位CPU和＆＃34; X64＆＃34;当CPU为64位时。

示例：

Global Const $g_sTpl = 'OS  : %s\nCPU : %s\n'
Global Const $g_sMsg = StringFormat($g_sTpl, @OSArch, @CPUArch)

ConsoleWrite($g_sMsg)

Answer 3

试试这个

＆＃13;

#include <MsgBoxConstants.au3>
if FileExists (@HomeDrive&"\Program Files (x86)") then
   $arc=64
Else
   $arc=32
EndIf
MsgBox("Hello", "Bit", "The computer that the program is been run on is a "&$arc&"bit one.")

＆＃13;

或者

＆＃13;

#include <MsgBoxConstants.au3>
if FileExists (@HomeDrive&"\Program Files (x86)") then
   $arc=64
Else
   $arc=32
EndIf
$path=@HomeDrive&"\Program Files (x86)"
MsgBox("Hello", $path, "The computer that the program is been run on is a "&$arc&"bit one.")

＆＃13;

你的代码没有用，因为你现在没有反斜杠检查我编辑了你的代码

将Windows确定为32位或64位

3 个答案: