Re: [Help-bash] something about bc

[lina]

On 29 Dec, 2011, at 0:18, Greg Wooledge <address@hidden> wrote:

> On Wed, Dec 28, 2011 at 09:56:00AM -0600, address@hidden wrote:
>> On Wed, 28 Dec 2011, lina wrote:
>>> some wrong try:
>>> 
>>> $ a=1.5 ; a+=1 ; echo $a
>>> 1.51
> 
> You are doing string concatenation here.  The two strings "1.5" and "1"
> are concatenated to make "1.51".
> 
>>> $ a=1.5 ; a+=1 | bc; echo $a
>>> 1.5
> 
> Here, you ran three separate commands:
> 
> a=1.5        # variable assignment
> a+=1 | bc    # variable assignment piped to bc
> echo $a      # expand a variable, perform word splitting and globbing, write
> 
> Since the second variable assignment does not write anything to standard
> output, the pipe to bc has no input, and therefore bc does nothing.
> 
> Since the second variable assignment was performed inside a pipeline, it
> takes place in a subshell (child process), and no change to the variable
> takes place in the main shell.
> 
>>> $ a=1.5 ; let "a+=1" | bc; echo $a
>>> bash: let: 1.5: syntax error: invalid arithmetic operator (error token is 
>>> ".5")
>>> 1.5
> 
> Bash cannot perform floating-point arithmetic.  It only has integer
> arithmetic.
> 
> imadev:~$ let a=1.5+1
> bash: let: a=1.5+1: syntax error: invalid arithmetic operator (error token is 
> ".5+1")
> 
> If you want to perform floating-point math, you have to use bc *instead of*
> let, not *in addition to* let.
> 
>> echo "a=1.5; a+=1; print a;" | bc
> 
> This must be GNU bc.  It doesn't work with standard bc:
> 
> imadev:~$ echo "a=1.5; a+=1; print a;" | bc
> syntax error on line 1, 
> 
> Here is how it works with standard bc:
> 
> imadev:~$ echo "a=1.5; a=a+1; a" | bc
> 2.5
> 
> Or more simply:
> 
> imadev:~$ echo "1.5+1" | bc
> 2.5
> 
> And, taking a wild-ass guess, here is what I think lina wanted to do:
> 
> imadev:~$ a=1.5
> imadev:~$ a=$(echo "$a + 1" | bc)
> imadev:~$ echo "$a"
> 2.5
> 

Cool. Thanks for both of you.  I should try more times.

Best wishes,