我有2个测试文件.在一个文件中,我想使用状态变量作为开关来提取中间部分,而在另一个文件中,我想使用状态变量来保存所看到的数字之和.
文件一:
section 0; state 0; not needed
= start section 1 =
state 1; needed
= end section 1 =
section 2; state 2; not needed
文件二:
1
2
3
4
5
处理文件一的代码:
cat file1 | perl6 -ne 'state $x = 0; say " x is ", $x; if $_ ~~ m/ start / { $x = 1; }; .say if $x == 1; if $_ ~~ m/ end / { $x = 2; }'
结果有错误:
x is (Any)
Use of uninitialized value of type Any in numeric context
in block at -e line 1
x is (Any)
= start section 1 =
x is 1
state 1; needed
x is 1
= end section 1 =
x is 2
x is 2
处理文件二的代码是
cat file2 | perl6 -ne 'state $x=0; if $_ ~~ m/ \d+ / { $x += $/.Str; } ; say $x; '
结果如预期:
1
3
6
10
15
是什么让状态变量在第一个代码中初始化失败,但在第二个代码中没问题?
我发现在第一个代码中,如果我让状态变量做一些事情,比如添加,那么它就可以了.为什么这样?
cat file1 | perl6 -ne 'state $x += 0; say " x is ", $x; if $_ ~~ m/ start / { $x = 1; }; .say if $x == 1; if $_ ~~ m/ end / { $x = 2; }'
# here, $x += 0 instead of $x = 0; and the results have no errors:
x is 0
x is 0
= start section 1 =
x is 1
state 1; needed
x is 1
= end section 1 =
x is 2
x is 2
谢谢你的帮助.
最佳答案 这在smls的评论中得到了回答:
Looks like a Rakudo bug. Simpler test-case:
echo Hello | perl6 -ne 'state $x = 42; dd $x'
.
It seems that top-level state variables are
not initialized when the-n
or-p
switch is used. As a work-around, you can manually initialize the variable in a separate statement, using the//=
(assign if undefined) operator:
state $x; $x //= 42;