在试图在Perl 6中公开localtime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perl6-lldb-m下运行,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.我在这里做错什么了吗?
更新:最后工作解决方案:
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);发布于 2018-04-29 15:23:09
localtime()需要一个类型为time_t*的指针作为参数。假设time_t和uint32_t在特定平台上是兼容的类型,
sub localtime(uint32 $epoch is rw --> TimeStruct) is native {*}
my uint32 $t = time;
dd localtime($t);应该这样做(除非公开属性,否则什么也看不见)。
我有点惊讶于您的time_t不是64位类型,而且我刚刚搜索了apple time.h,我还怀疑您的struct声明中的最后两个属性的顺序是错误的.
https://stackoverflow.com/questions/50087913
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