需要awk变量赋值语句说明 [英] awk variable assignment statement explanation needed
问题描述
好吧，直截了当，这是代码，我对代码进行了一些格式化以使其易于阅读:
ok, straight to the point, here is the codes, I formatted the codes a little to make it easy to read:
awk '{
t=$0 ;
$0=t ; $0=// ; print "$0=// ; value of $0 is ",$0
$0=t ; $0=/./ ; print "$0=/./ ; value of $0 is ",$0
$0=t ; $0=/*/ ; print "$0=/*/ ; value of $0 is ",$0
$0=t ; $0=/**/ ; print "$0=/**/ ; value of $0 is ",$0
$0=t ; $0=/[0-9]/ ; print "$0=/[0-9]/ ; value of $0 is ",$0
$0=t ; $0=/[a-z]/ ; print "$0=/[a-z]/ ; value of $0 is ",$0
$0=t ; $0=/[0-9][a-z]/ ; print "$0=/[0-9][a-z]/ ; value of $0 is ",$0
$0=t ; $0=/5/ ; print "$0=/5/ ; value of $0 is ",$0
$0=t ; $0=/55/ ; print "$0=/55/ ; value of $0 is ",$0
$0=t ; $0=/x/ ; print "$0=/x/ ; value of $0 is ",$0
$0=t ; $0=/5x/ ; print "$0=/5x/ ; value of $0 is ",$0
$0=t ; $0=/x5/ ; print "$0=/x5/ ; value of $0 is ",$0
$0=t ; $0=/xoo/ ; print "$0=/xoo/ ; value of $0 is ",$0
$0=t ; $0=/500/ ; print "$0=/500/ ; value of $0 is ",$0
}'<<<"5x"
我得到了这个输出:
$0=// ; value of $0 is 1
$0=/./ ; value of $0 is 1
$0=/*/ ; value of $0 is 0
$0=/**/ ; value of $0 is 1
$0=/[0-9]/ ; value of $0 is 1
$0=/[a-z]/ ; value of $0 is 1
$0=/[0-9][a-z]/ ; value of $0 is 1
$0=/5/ ; value of $0 is 1
$0=/55/ ; value of $0 is 0
$0=/x/ ; value of $0 is 1
$0=/5x/ ; value of $0 is 1
$0=/x5/ ; value of $0 is 0
$0=/xoo/ ; value of $0 is 0
$0=/500/ ; value of $0 is 0
我不明白为什么得到此输出.我以为awk会抱怨赋值语句，但事实并非如此……当我执行$0=/xxx/时，awk似乎正在执行奇怪的正则表达式匹配检查?即$0=/pattern/是否与$0=$0~/pattern/相同?
I cannot understand why I got this output. I thought awk would complain about the assignment statement, but it didn't... It seems that awk is doing weird regex matching check when I did $0=/xxx/? that is, $0=/pattern/ is same as $0=$0~/pattern/?
然后我做了这个测试:
kent$ echo "xx"|awk '{y="777";y=/x/;print y}'
1
kent$ echo "xx"|awk '{y="777";y=/7/;print y}'
0
所以看起来foo=/pattern/与foo=$0~/pattern/
但是我不确定...在文档/手册页中找不到该信息.
but I am not sure... cannot find the info in document/man page.
当我在这里回答awk问题时，我发现了它.
I found it when I was answering an awk question here @ SO.
我用awk进行了测试:
I tested with my awk:
kent$ awk --version|head -1
GNU Awk 4.1.0, API: 1.0 (GNU MPFR 3.1.2, GNU MP 5.1.2)
我很高兴有人可以向我解释.预先感谢.
I appreciate if someone could explain me. Thanks in advance.
编辑
经过更多测试后，结果再次出现(不确定，因为我没有在文档中找到它)，在之内{...}，仅/pattern/是$0~/pattern/与{...}外部相同，例如/patter/{do something}.这样我们就可以做到:
after doing more tests, it turns out (again, not sure, since I didn't find it in document), within {...}, the /pattern/ alone is short form of $0~/pattern/ same as outside the {...}, like /patter/{do something} . so we could do:
kent$ echo "xx"|awk '{if(/x/)print "ok"}'
ok
我不知道它是否是awk脚本语法的标准功能.以及它是否适用于所有awk实施.我仍在搜索手册和手册页...
I don't know if it is a standard feature of awk script syntax. and if it works for all awk implementation. I am still searching manual and man page...
推荐答案
/regexp/只是$0 ~ /regexp/
var = ($0 ~ /regexp/)将var的比较结果分配给var，比较结果为1，否则为0.
var = ($0 ~ /regexp/) assigns var the result of the comparison, 1 for true, zero otherwise.
$0 = ($0 ~ /regexp/)为$ 0分配$ 0和/regexp/
$0 = ($0 ~ /regexp/) assigns $0 the result of the comparison of $0 and /regexp/
$0 = ($0 ~ //)向$ 0分配在$ 0中查找空字符串的结果，该字符串始终存在，因此条件为true，结果为1
$0 = ($0 ~ //) assigns $0 the result of looking for an empty string in $0, which there always is, so the condition is true so the result is 1
$0 = //相同
$0 = /*/正在$ 0中查找文字*，没有，因此结果为零.
$0 = /*/ is looking for a literal * in $0, which there isn't so the result is zero.
$0 = /**/正在寻找在$ 0中重复零次或多次的立即数*，它们为零，因此结果为1.
$0 = /**/ is looking for a literal * repeated ZERO or more times in $0, which there are zero of them so that result is 1.
没有神秘，没有奇怪...
No mystery, nothing weird...
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