9

localtime在尝试公开Perl 6 中的功能时，我似乎做错了什么：

use NativeCall;
my class TimeStruct is repr<CStruct> {
    has int32 $!tm_sec;
    has int32 $!tm_min;
    has int32 $!tm_hour;
    has int32 $!tm_mday;
    has int32 $!tm_mon;
    has int32 $!tm_year;
    has int32 $!tm_wday;
    has int32 $!tm_yday;
    has int32 $!tm_isdst;
    has Str   $!tm_zone;
    has long  $!tm_gmtoff;
}

sub localtime(uint32 $epoch --> TimeStruct) is native {*}
dd localtime(time);  # segfault

在 下运行perl6-lldb-m，我得到：

Process 82758 stopped
* thread #1, queue = 'com.apple.main-thread', stop reason = EXC_BAD_ACCESS (code=1, address=0x5ae5dda1)
    frame #0: 0x00007fffe852efb4 libsystem_c.dylib`_st_localsub + 13
libsystem_c.dylib`_st_localsub:
->  0x7fffe852efb4 <+13>: movq   (%rdi), %rax
    0x7fffe852efb7 <+16>: movq   %rax, -0x20(%rbp)
    0x7fffe852efbb <+20>: movq   0x8e71d3e(%rip), %rbx     ; lclptr
    0x7fffe852efc2 <+27>: testq  %rbx, %rbx
Target 0: (moar) stopped.

我在这里做错了什么明显的事情吗？

更新：最终的工作解决方案：

class TimeStruct is repr<CStruct> {
    has int32 $.tm_sec;    # *must* be public attributes
    has int32 $.tm_min;
    has int32 $.tm_hour;
    has int32 $.tm_mday;
    has int32 $.tm_mon;
    has int32 $.tm_year;
    has int32 $.tm_wday;
    has int32 $.tm_yday;
    has int32 $.tm_isdst;
    has long  $.tm_gmtoff; # these two were
    has Str   $.time_zone; # in the wrong order
}

sub localtime(int64 $epoch is rw --> TimeStruct) is native {*}

my int64 $epoch = time;  # needs a separate definition somehow
dd localtime($epoch);
4

1 回答 1

8

localtime()需要一个类型的指针time_t*作为参数。假设time_t并且uint32_t是您特定平台上的兼容类型，

sub localtime(uint32 $epoch is rw --> TimeStruct) is native {*}
my uint32 $t = time;
dd localtime($t);

应该这样做（尽管除非您将属性公开，否则您将看不到任何东西）。

我有点惊讶您time_t不是 64 位类型，并且刚刚 googleapple time.h过，我还怀疑您的 struct 声明中的最后两个属性的顺序错误......

于 2018-04-29T15:23:09.877 回答