Bash 分解长字符串，然后将十六进制转换为十进制_Bash_Awk_Hex

Bash 分解长字符串，然后将十六进制转换为十进制

bash awk

Bash 分解长字符串，然后将十六进制转换为十进制,bash,awk,hex,Bash,Awk,Hex,从另一个输出中，我有一个长字符串： 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000033B2E3C9FD0803CE800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

从另一个输出中，我有一个长字符串：

00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000033B2E3C9FD0803CE800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

需要将其分解为64个字符的部分：

0000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000033B2E3C9FD0803CE8000000
000000000000000000000000000000000000000D610B7305BB52FC30A0000000
000000000000000000000000AF298D050E4395D69670B12B7F41000000000000
0000000000000000000000000000000000000000000000000000000000000000

然后，每一行都需要从十六进制转换为十进制，比如echo“obase=10；ibase=16；$hexNum”| bc

六边形部分很容易解决。我怀疑长字符串的“awk”操作可以将其分解为5个变量。。上面的bc命令可以转换为十进制。

这里有一种方法：

input='000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
部分=（）nums=（）
而[[-n$input]]；do parts+=（“${input:0:64}”）；input=${input:64}；完成
用于“${parts[@]}”中的h；do nums+=（$（echo“obase=10；ibase=16；$h”| bc））；完成
声明-p部件NUM

declare-a parts=（[0]=“0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000033B2E3C9FD0803CE8000000”[2]=“00000000000000000000000000000000000000000D610B7305BB52FC30A0000000”[3]=“00000000000000000000000000000000000AF298D050E4395D6970B12B7F41000000000”[4]="0000000000000000000000000000000000000000000000000000000000000000")
声明-a nums=（[0]=“0”[1]=“10000000000000000000000000”[2]=“1060000000000000000000000000000”[3]=“10000000000000000000000000000000000000000000000000000000000000000000000000000000000”[4]=“0”）

如果不需要保留64位十六进制数，一个循环就足够了：

nums=（）
对于（（i=0；i这里有一种方法：
input='000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
部分=（）nums=（）
而[[-n$input]]；do parts+=（“${input:0:64}”）；input=${input:64}；完成
对于“${parts[@]}”中的h；do nums+=（$（echo“obase=10；ibase=16；$h”| bc））；完成
声明-p部件NUM

declare-a parts=（[0]=“0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000033B2E3C9FD0803CE8000000”[2]=“00000000000000000000000000000000000000000D610B7305BB52FC30A0000000”[3]=“000000000000000000000000000000 AF298D050E4395D6970B12B7F41000000000”[4]=”0000000000000000000000000000000000000000000000000000000000000000")
声明-a nums=（[0]=“0”[1]=“10000000000000000000000000”[2]=“1060000000000000000000000000000”[3]=“10000000000000000000000000000000000000000000000000000000000000000000000000000000000”[4]=“0”）

如果不需要保留64位十六进制数，一个循环就足够了：
nums=（）
对于（（i=0；i另一种溶液
$ fold -w64 file | awk '{print strtonum("0x"$1)}'

0
1000000000000000013287555072
1060000000000000004189203726336
1000000000000000043845843045076197354634048000000
0

另一个解决方案
$ fold -w64 file | awk '{print strtonum("0x"$1)}'

0
1000000000000000013287555072
1060000000000000004189203726336
1000000000000000043845843045076197354634048000000
0

另一种方法是使用read-n64…
一次读取64个字符，例如
读取时-n64行&[“${line}”-eq 64]；执行
回音$线
echo“value$（bc另一种方法是使用read-n64…
一次读取64个字符，例如
读取时-n64行&[“${line}”-eq 64]；执行
回音$线
echo“value$（bc最后，我将字符串设置为变量（dc）
然后
最后，我将字符串设置为变量（dc）
然后
您可能需要split-b64
？您可能需要split-b64
？转换后的值似乎是四舍五入的。嗯……我没有意识到这一点。这似乎是obase/ibase
与bc
转换的结果，并与@glennjackman得到的结果相匹配。我必须看看那里发生了什么。考虑到值的数量级超过了64位数字，很明显，一个大数字的计算已经完成。我也不确定我的版本是否正确，但数字不可能是10的精确幂。我完全同意你关于：）
转换后的值似乎是四舍五入的。嗯……我没有意识到这一点。这似乎是obase/ibase
与bc
进行转换的结果，并与@glennjackman得到的结果相匹配。我必须看看那里发生了什么。考虑到这些值在数量级上超过了64位数字，这显然是一个大问题-数字计算已经完成。我也不确定我的版本是否正确，但数字不可能是10的精确幂。我完全同意你关于：）