问题
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"
I got this output:
$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
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=$0~/pattern/?
then I did this test:
kent$ echo "xx"|awk '{y="777";y=/x/;print y}'
1
kent$ echo "xx"|awk '{y="777";y=/7/;print y}'
0
so it seems, foo=/pattern/ is same as foo=$0~/pattern/
but I am not sure... cannot find the info in document/man page.
I found it when I was answering an awk question here @ SO.
I tested with my awk:
kent$ awk --version|head -1
GNU Awk 4.1.0, API: 1.0 (GNU MPFR 3.1.2, GNU MP 5.1.2)
I appreciate if someone could explain me. Thanks in advance.
EDIT
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
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...
回答1:
/regexp/ is just shorthand for $0 ~ /regexp/
var = ($0 ~ /regexp/) assigns var the result of the comparison, 1 for true, zero otherwise.
$0 = ($0 ~ /regexp/) assigns $0 the result of the comparison of $0 and /regexp/
$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 = // does the same
$0 = /*/ is looking for a literal * in $0, which there isn't so the result is zero.
$0 = /**/ is looking for a literal * repeated ZERO or more times in $0, which there are zero of them so that result is 1.
No mystery, nothing weird...
回答2:
There is not mystery here you are just setting the value of $0 to the value of the evaluated regular expression match. The right hand side is evaluated first, if the line matches the regular expression match the results is True i.e. 1, else the results is False i.e. 0.
来源：https://stackoverflow.com/questions/18795408/awk-variable-assignment-statement-explanation-needed