64位代码如何在OS-X 10.5上工作? [英] How does 64 bit code work on OS-X 10.5?
问题描述
我最初认为64位指令在OS-X 10.5上不起作用。我写了一个小测试程序并用编译了它, GCC -m64
。
对于我的64位整数,我使用了 long long
。
使用的汇编指令看起来像是64位。例如。 imultq
和 movq 8(%rbp),%rax
。
我只使用 printf
来显示64位值,使用%lld
。
- 这是预期行为吗?
- 是否有任何
gotcha的
会导致此失败? - 我允许在问题中提出多个问题吗?
- 这是否适用于其他操作系统?
为了使这一点完全清楚,下面是OS X上32位和64位可执行文件的情况:
-
32 - 和64位用户空间可执行文件可以在OS X 10.6的32位和64位内核上运行,无需仿真。在10.4和10.5上,32位和64位可执行文件都可以在32位内核上运行。 (在Windows上不是这样)
-
OS X和Linux为64位可执行文件使用LP64模型。指针和
long
是64位宽,int
仍然是32位,long long
仍然是64位。 (Windows使用LLP64模型,而不是 -long
在64位Windows中为32位宽)。
用户空间系统库和框架在10.5和10.6上构建为32/64位胖。无论您是在构建32位,64位还是两者,您都可以正常关联它们。一些库(基本上是POSIX层)在10.4上也被构建为32/64位胖,但其中很多不是。 在10.6上,构建工具默认生成64位可执行文件。在10.5和更早版本中,缺省值是32位。 在10.6上,默认情况下,构建fat的可执行文件将运行64位版本。在10.5和更早版本中,默认情况下会执行32位方面。 code> arch 命令。例如。 arch -arch i386 someCommandToRunThatIWantToRunIn32BitMode
。对于应用程序包,您可以从命令行启动它们,或者如果您在应用程序中获取信息,则可以选择首选项。
I initially thought that 64 bit instructions would not work on OS-X 10.5.
I wrote a little test program and compiled it with GCC -m64
.
I used long long
for my 64 bit integers.
The assembly instructions used look like they are 64 bit. eg. imultq
and movq 8(%rbp),%rax
.
I seems to work.
I am only using printf
to display the 64 bit values using %lld
.
- Is this the expected behaviour?
- Are there any
gotcha's
that would cause this to fail? - Am I allowed to ask multiple questions in a question?
- Does this work on other OS's?
Just to make this completely clear, here is the situation for 32- and 64-bit executables on OS X:
Both 32- and 64-bit user space executables can be run on both 32- and 64-bit kernels in OS X 10.6, without emulation. On 10.4 and 10.5, both 32- and 64-bit executables can run on the 32-bit kernel. (This is not true on Windows)
The user space system libraries and frameworks are built 32/64-bit fat on 10.5 and 10.6. You can link against them normally, whether you're building for 32-bit, 64-bit, or both. A few libraries (basically the POSIX layer) are also built 32/64-bit fat on 10.4, but many of them are not.
On 10.6, the build tools produce 64-bit executables by default. On 10.5 and earlier, the default is 32-bit.
On 10.6, executables that are built fat will run the 64-bit side by default. On 10.5 and earlier, the 32-bit side is executed by default.
You can always manually specify which slice of a fat executable to use by using the
arch
command. eg.arch -arch i386 someCommandToRunThatIWantToRunIn32BitMode
. For application bundles, you can either launch them from the command line, or there is a preference if you "get info" on the application.OS X and Linux use the LP64 model for 64-bit executables. Pointers and
long
are 64 bits wide,int
is still 32 bits, andlong long
is still 64 bits. (Windows uses the LLP64 model instead --long
is 32 bits wide in 64 bit Windows).
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