string formatting in bash

Question

I have the following code that runs in a shell script

foo=`seq 1 1 100`
for i in $foo; do
  echo "input$i\_now"
done

Here's my question: Under some conditions, the output prints input1_now whereas othertimes it prints input1\_now. I'm sure something is different, but I can't figure out what makes it print one way or the other. If my code is

for i in $foo; do
   echo "input$i_now"
done

I will always get input with the rest of the line being omitted.

I know I can use input${i}_now instead and have it print correctly every time, but I'm mostly interested in understanding why the output is different under seemingly the same conditions.

UPDATE:

In the following example, the first part correctly formats the variables and text such that the \_ is replaced as _. However, the last part required me to place variables in curly brackets in order to have them formatted correctly.

echo "Enter Simulation #: "
read sim

g.mapset results

for i in `seq 1 1 100`; do

file=sim$sim\_run$i\_sum
g.copy $file\@expSim$sim\_$i,$file

file=sim$sim\_run$i\_average
g.copy $file\@expSim$sim\_$i,$file

for year in `seq 2004 1 2006`; do   

    file=sim$sim\_$year\_run$i\_sum
    g.copy $file\@expSim$sim\_$i,$file

    file=sim$sim\_$year\_run$i\_average
    g.copy $file\@expSim$sim\_$i,$file

done

years="2004 2005 2006"
times=`seq -w 1 16 365`
runs=`seq 1 1 100`

for year in $years; do
for ptime in $times; do
    for i in $runs; do
        if [ $i -eq 1 ]; then
            g.copy  vect=sim${sim}_pts_${year}_${ptime}_run${i}@expSim${sim}_${i},sim${sim}_pts_${year}_${ptime}
        fi
        if [ $i -gt  1 ]; then  
        v.patch input=sim${sim}_pts_${year}_${ptime}_run${i}@expSim${sim}_${i} output=sim${sim}_pts_${year}_${ptime} -e -a --o
        fi
    done
done
done

Answer 1

Is _ supposed to be a placeholder that is sometimes a different character?

In bash, "input$i\_now" with an actual _ will always produce input1\_now. Inside double-quotes, bash only removes a \ when it is followed by a $, a `, a ", a \, or a newline. See “Double Quotes” in the Bash Reference Manual. This is the POSIX standard behavior; see “Double-Quotes” in Shell Command Language.

UPDATE

If you write "input$i_now", bash will just print input. It will not print input1 or input1_now. It does this because _ is a valid parameter name character, so bash thinks you are asking for the value of the i_now parameter. Unless you have set i_now to a non-empty string, bash will expand $i_now to the empty string, thus turning "input$i_now" into input.

UPDATE 2

Now that you have posted real code, we can see what's going on.

First of all, in the real code you posted, you never used double-quotes around a parameter expansion. This makes a difference.

Outside of double-quotes, a \ is always removed. See “Quote Removal” in the Bash Reference Manual. Hence input$i\_now (with no surrounding double-quotes) expands to input1_now.

However, as I explained in my first update, _ is a parameter name character. See “Name” in Shell Command Language. So when bash sees input$i_now, it takes i_now as the parameter name.

Whether or not you're using double-quotes, you must separate the parameter name from the following character, if bash would otherwise treat the following character as part of the parameter name. You can do this by putting \ after the parameter name, or you can do it by putting the parameter name in {...}.

It is safer to always use {...}, because (as you have discovered?) \ is handled differently depending on whether it's inside double-quotes. If you go back and add double-quotes later, and you have used \, you will need to change the \ to {...} anyway.

UPDATE 3

Here is a demonstration of the effects of \, {...}, and double-quoting. First, we set up some variables:

$ year=2004 ptime=1 i=1 sim=123

Here's what happens with no quoting whatsoever:

$ echo vect=sim$sim_pts_$year_$ptime_run$i@expSim$sim_$i,sim$sim_pts_$year_$ptime
vect=sim1@expSim1,sim1

Here's what happens if we just use {...} without double-quotes:

$ echo vect=sim${sim}_pts_${year}_${ptime}_run${i}@expSim${sim}_${i},sim${sim}_pts_${year}_${ptime}
vect=sim123_pts_2004_1_run1@expSim123_1,sim123_pts_2004_1

If we add double-quotes, they have no effect:

$ echo "vect=sim${sim}_pts_${year}_${ptime}_run${i}@expSim${sim}_${i},sim${sim}_pts_${year}_${ptime}"
vect=sim123_pts_2004_1_run1@expSim123_1,sim123_pts_2004_1

Here's what happens if we just use \:

$ echo vect=sim$sim\_pts\_$year\_$ptime\_run$i@expSim$sim\_$i,sim$sim\_pts\_$year\_$ptime
vect=sim123_pts_2004_1_run1@expSim123_1,sim123_pts_2004_1

Notice that each \ was removed. The shell removes a \ if it's not quoted.

If we add double-quotes, they prevent the shell from removing each \:

$ echo "vect=sim$sim\_pts\_$year\_$ptime\_run$i@expSim$sim\_$i,sim$sim\_pts\_$year\_$ptime"
vect=sim123\_pts\_2004\_1\_run1@expSim123\_1,sim123\_pts\_2004\_1