bc and its ibase/obase options:

问题:

I stumbled over a curious bug, I think:

I tried to read "512" as a number to base 6, and output it as base 16:

echo "ibase=6;obase=16;512" | bc
161

As you can see, the output is 161, but it should be bc(sic!). I tried with base 10:

echo "ibase=6;obase=10;512" | bc
512

The value is unchanged. Curious! Default obase is 10. If I omit it:

echo "ibase=6;512" | bc
188

Well, that seems right. In a two step process, it works:

echo "obase=16;"$(echo "ibase=6;512" | bc) | bc
BC

So I made a script for different bases, but it keeps me puzzled:

for ib in {6,8,10,16}; do echo $ib; for ob in {10,16}; do echo -en $ib $ob"     \t => " ; echo "ibase=$ib;obase=$ob;333" | bc ; done; done; 
6
6 10         => 333
6 16         => 108
8
8 10         => 333
8 16         => 119
10
10 10        => 333
10 16        => 14D
16
16 10        => 333
16 16        =>  01 15 05

Shall I file a bugreport, or do I miss the obvious? I can't really believe that such a basic tool is broken.

回答1:

Not a bug.

As soon as ibase=6 is interpreted, numbers are read in base 6. So ibase=6;obase=16 makes obase's value to be 16_{base 6} which is invalid, and is interpreted as 11_decimal.

From the man page:

For multi-digit numbers, bc changes all input digits greater or equal to ibase to the value of ibase-1.

So 16 is interpreted as 15_{base 6} which is 11_decimal. And the conversion is correct.

Set obase before ibase, or make sure to specify your obase in base ibase.

$ echo "obase=16;ibase=6;512" | bc
BC

回答2:

See http://docstore.mik.ua/orelly/unix/upt/ch49_03.htm

When you set ibase or obase, it's expressed with the current base of ibase. So set obase before you set ibase if you want to express obase in decimal.

See also http://www.gnu.org/software/bc/manual/html_mono/bc.html#SEC9

Input numbers may contain the characters 0-9 and A-F. (Note: They must be capitals. Lower case letters are variable names.) Single digit numbers always have the value of the digit regardless of the value of ibase. (i.e. A = 10.) For multi-digit numbers, bc changes all input digits greater or equal to ibase to the value of ibase-1. This makes the number FFF always be the largest 3 digit number of the input base.

So for obase=16 in ibase=6, the 6 becomes a 5, and it is equivalent to an output base of decimal 6 * 1 + 1 * 5 == 11, thus:

$ echo "obase=11;ibase=6;512" | bc
161

回答3:

Slightly modify the code (two ways) and your expected results appear:

for ib in {6,8,10,16}; do 
    echo $ib; for ob in {10,16}; do 
        echo -en $ib $ob"     \t => " ; 
        ob=`echo "obase=$ib;$ob" | bc`
        echo "ibase=$ib;obase=$ob;333" | bc ; 
    done; 
done;

for ib in {6,8,10,16}; do 
    echo $ib; for ob in {10,16}; do 
        echo -en $ib $ob"     \t => " ; 
        echo "obase=$ob;ibase=$ib;333" | bc ; 
    done; 
done;

execute in bash sandbox

results for both ways are:

6
6 10         => 129
6 16         => 81
8
8 10         => 219
8 16         => DB
10
10 10        => 333
10 16        => 14D
16
16 10        => 819
16 16        => 333