Bash BC problem

mcfc1900 · 04-05-2011, 01:24 PM

My line of code which isn't working is the following:

variable=$(ibase=16; $hex | bc)

I have also tried:

variable=$(echo "ibase=16; $hex" | bc)

Neither will work and gives me the error "(Standard_in) 1: parse error".

$hex has been defined two lines above this piece of code...

Can anyone see where I am going wrong please?

Thanks

Kenhelm · 04-05-2011, 02:38 PM

The error message could be caused by the letters in $hex not being in upper case.
Lower case letters are used for variable names in bc.

Code:

hex=45de
echo "ibase=16; $hex" | bc
(standard_in) 1: parse error

hex=45DE
echo "ibase=16; $hex" | bc
17886

Telengard · 04-05-2011, 02:49 PM

Quote:

Originally Posted by mcfc1900

variable=$(ibase=16; $hex | bc)

I can't see how you expect that to work. Assuming $hex holds a valid hexadecimal value, ibase=16; $hex needs to be supplied to bc on its standard input. The | (pipe) implies that some other program will be emitting the program from its standard output.

This works fine for me.

Code:

foo$ hex=10
foo$ variable=$(echo "ibase=16; $hex" | bc)
foo$ echo $variable
16
foo$

I think we will need to see more of your program to determine the cause of your error.

Edit

Quote:

Originally Posted by Kenhelm

The error message could be caused by the letters in $hex not being in upper case.
Lower case letters are used for variable names in bc.

Nice catch!

mcfc1900 · 04-05-2011, 03:52 PM

Thank you both

The problem was that the hex value was lower case, soon as I changed that it started working.

Cheers

Nominal Animal · 04-05-2011, 03:57 PM

Quote:

Originally Posted by mcfc1900

variable=$(echo "ibase=16; $hex" | bc)

Why not use

Code:

variable=$[0x$hex]