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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

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

Quote:

Originally Posted by mcfc1900 (Post 4314843)

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 (Post 4314920)

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! :)

Thank you both :)

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

Cheers

Quote:

Originally Posted by mcfc1900 (Post 4314843)

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

Why not use

Code:

variable=$[0x$hex]