我想使用AWK将文件中的十进制数字列表转换为二进制,但似乎没有内置的方法。示例文件如下:
134218506
134218250
134217984
1610612736
16384
33554432
以下是一种awk的方法,已经转化为函数方便使用:
awk '
function d2b(d, b) {
while(d) {
b=d%2b
d=int(d/2)
}
return(b)
}
{
print d2b($0)
}' file
前三个记录的输出:
1000000000000000001100001010
1000000000000000001000001010
1000000000000000000100000000
以下是一种方法,首先将十进制转换为十六进制,然后将每个十六进制字符转换为其二进制等效物:
$ cat dec2bin.awk
BEGIN {
h2b["0"] = "0000"; h2b["8"] = "1000"
h2b["1"] = "0001"; h2b["9"] = "1001"
h2b["2"] = "0010"; h2b["a"] = "1010"
h2b["3"] = "0011"; h2b["b"] = "1011"
h2b["4"] = "0100"; h2b["c"] = "1100"
h2b["5"] = "0101"; h2b["d"] = "1101"
h2b["6"] = "0110"; h2b["e"] = "1110"
h2b["7"] = "0111"; h2b["f"] = "1111"
}
{ print dec2bin($0) }
function hex2bin(hex, n,i,bin) {
n = length(hex)
for (i=1; i<=n; i++) {
bin = bin h2b[substr(hex,i,1)]
}
sub(/^0+/,"",bin)
return bin
}
function dec2bin(dec, hex, bin) {
hex = sprintf("%x\n", dec)
bin = hex2bin(hex)
return bin
}
$ awk -f dec2bin.awk file
1000000000000000001100001010
1000000000000000001000001010
1000000000000000000100000000
1100000000000000000000000000000
100000000000000
10000000000000000000000000
你可以尝试使用Perl的一行命令
$ cat hamdani.txt
134218506
134218250
134217984
134217984
1610612736
16384
33554432
$ perl -nle ' printf("%b\n",$_) ' hamdani.txt
1000000000000000001100001010
1000000000000000001000001010
1000000000000000000100000000
1000000000000000000100000000
1100000000000000000000000000000
100000000000000
10000000000000000000000000
$
# -f infile : Use infile for data
# after -e , it is there are the dc command
dc -f infile -e '
z # number of values
sa # keep in register a
2
o # set the output radix to 2 : binary
[
Sb # keep all the value of infile in the register b
# ( b is use here as a stack)
z
0 <M # until there is no more value
] sM # define macro M in [ and ]
lMx # execute macro M to populate stack b
[
Lb # get all values one at a time from stack b
p # print this value in binary
la # get the number of value
1
- # decremente it
d # duplicate
sa # keep one in register a
0<N # the other is use here
]sN # define macro N
lNx' # execute macro N to print each values in binary
# gawk binary number functions
# RPC 09OCT2022
# convert an 8 bit binary number to an integer
function bin_to_n(i)
{
n = 0;
#printf(">> %s:", i);
for (k = 1; k < 9; k++) {
n = n * 2;
b = substr(i, k, 1);
if (b == "1") {
n = n + 1;
}
}
return (n);
}
# convert a number to a binary number
function dectobin(n)
{
printf("dectobin: n in %d ",n);
binstring = "0b"; # some c compilers allow 0bXXXXXXXX format numbers
bn = 128;
for(k=0;k<8;k++) {
if (n >= bn) {
binstring = binstring "1";
n = n - bn;
} else {
binstring = binstring "0"
}
printf(" bn %d",bn);
bn = bn / 2;
}
return binstring;
}
BEGIN {
FS = " ";
# gawk (I think) has no atoi() funciton or equiv. So a table of all
# chars (well 256 ascii) can be used with the index function to get
# round this
for (i = 0; i < 255; i++) {
table = sprintf("%s%c", table, i);
}
}
{
# assume on stdin a buffer of 8 bit binary numbers "01000001 01000010" is AB etc
for (i = 1; i <= NF; i++)
printf("bin-num#%d: %x --> %c\n", i, bin_to_n($i), bin_to_n($i));
s = "ABC123string to test";
for (i = 0; i < length(s); i++) {
nn = index(table, substr(s,i+1,1))-1;
printf("substr :%s:%x:",ss,nn);
printf(" :%d: %s\n", i, dectobin(nn));
}
}
除了别人已经提到的内容,这个函数还有一个快捷方式,可以用于非负整数幂次为2的情况。
—-(因为它们总是有二进制模式 / ^ [1] [0] * $ /
)
版本1:以3位一组的方式处理,而不是逐位处理:
{m,g}awk '
BEGIN {
1 CONVFMT="%.250g"
1 _^=OFMT="%.25g"
}
($++NF=________v1($_))^!_
function ________v1(__,___,_,____,_____)
{
6 if (+__==(_+=_^=____="")^(___=log(__)/log(_))) { # 2
2 return \
___<=_^_^_ \
? (_+_*_*_)^___ \
: sprintf("%.f%0*.f",--_,___,--_)
}
4 ___=(!_!_!_!!_) (_^((_____=_*_*_)+_)-_^_^_+(++_))
4 gsub("..", "&0&1", ___)
41 while(__) {
41 ____ = substr(___,
__%_____*_+(__=int(__/_____))^!_,_)____
}
4 return substr(__=____, index(__, _^(! _)))
}'
sprintf()
将数字转换为八进制,然后再映射到二进制。 function ________v2(__,___,_,____,_____)
{
6 if (+__==(_+=_^=____="")^(___=log(__)/log(_))) { # 2
2 return \
___<=_^_^_ \
? (_+_*_*_)^___ \
: sprintf("%.f%0*.f",--_,___,--_)
}
4 ___=(!_!_!_!!_) (_^((_____=_*_*_)+_)-_^_^_+(++_))
4 gsub("..", "&0&1", ___)
4 _____=___
4 __=sprintf("%o%.*o", int(__/(___=++_^(_*--_+_))),
_*_+!!_, __%___)
4 sub("^[0]+", "", __)
41 for (___=length(__); ___; ___--) {
41 ____ = substr(_____, substr(__,
___,!!_)*_ + !!_,_)____
}
4 return substr(____, index(____,!!_))
}
|
134218506 1000000000000000001100001010
134218250 1000000000000000001000001010
134217984 1000000000000000000100000000
1610612736 1100000000000000000000000000000
16384 100000000000000
33554432 10000000000000000000000000
29.5 MB/s
吞吐量)。
.
{m,g,n}awk '
1 function ________(_______,_, __,____,______)
{
1 split(_=__=____=______="", _______, _)
2 for (_^=_<_; -_<=+_; _--) {
4 for (__^=_<_; -__<=+__; __--) {
8 for (____^=_<_; -____<=+____; ____--) {
16 for (______^=_<_; -______<=+______; ______--) {
16 _______[_+_+_+_+_+_+_+_+__+__+\
__+__+____+____+______]=\
(_)__ (____)______
}
}
}
}
1 return _^(_<_)
}
BEGIN {
1 CONVFMT = "%." ((_+=(_^=_<_)+(_+=_))*_)(!_)"g"
1 OFMT = "%." (_*_) "g"
1 _ = ________(_____)
}
($++NF=___($_))^!_
function ___(__,____,_,______)
{
6 if ((__=int(__))<(______=\
(_*=_+=_+=_^=____="")*_)) {
return _____[int(__/_)]_____[__%_]
}
16 do { ____=_____[int(__/_)%_]_____[__%_]____
} while (______<=(__=int(__/______)))
6 return int(_____[int(__/_)%_]\
_____[ (__) %_])____
}
你不应该使用awk
,而是用bc
:
$ bc <<EOF
ibase=10
obase=2
$(cat file)
EOF
或者
bc <<< $(awk 'BEGIN{ print "ibase=10; obase=2"}1' file)