BEGIN块中的十六进制字符

Question

I can print a hex character in the process block

$ awk '{printf "%c", $0}' <<< 0x21
!

However the same character will not print in the BEGIN block

$ awk 'BEGIN {printf "%c", 0x21}'
0

How can I print a hex character in the BEGIN block?

Answer 1

GNU awk supports hex notation but traditional awk does not . The POSIX standard for awk is here and it states:

An integer constant cannot begin with 0x or include the hexadecimal digits 'a', 'b', 'c', 'd', 'e', 'f', 'A', 'B', 'C', 'D', 'E', or 'F'.

Here is one method that uses bash to supply the constant string that you want to a POSIX awk:

awk -v p=$'\x21' 'BEGIN {printf "%c", p}'

References

• -vp=string

  The -v option allows awk variables to be defined via the command line. This is documented in the POSIX spec under "options" here .

• $'\\x21'

  The $'...' construct allows many special characters to be added to bash strings. Here, we add a hexadecimal. This is documented in the bash manual here .

Answer 2

When reading a string awk will recognize hex numbers.

When reading a number awk will not recognize hex numbers, it will toss out everything starting with the first non number.

So with the first example $0 is a string and all is well. With the second example x21 is tossed out and 0 is left. A workaround is to pass 0x21 as a string

$ awk 'BEGIN {printf "%c", +"0x21"}'
!