Calculating with dc

Introduction

dc(1) is a non standard, but commonly found, reverse-polish Desk Calculator. According to Ken Thompson, "dc is the oldest language on Unix; it was written on the PDP-7 and ported to the PDP-11 before Unix [itself] was ported".

Historically the standard bc(1) has been implemented as a front-end to dc.

Simple calculation

In brief, the reverse polish notation means that you first put the numbers (on a stack) and then apply an operation on these. Instead of writing 1+1, you write 1 1+.

By default dc, unlike bc, doesn't print anything, the result is just pushed on the stack. You have to use to use the command "p" to print the element at the top of the stack. dc. Thus a simple operation looks like:

$ dc <<< '1 1+pq'
2

I used here a "here string" present in bash 3.x, ksh93 and zsh. if your shell doesn't support this you can use echo '1 1+p' | dc or if you have GNU dc, you can use dc -e '1 1 +p'.

Of course, you can also just run dc and enter the commands.

The classic operations are present:

addition: +
subtraction: -
division: /
multiplication: *
remainder (modulo): %
exponentiation: ^
square root: v

GNU dc adds a couple more.

To input a negative number you need to use the _ (underscore) character:

$ dc <<< '1_1-p' 
2

You can use the digits 0 to 9 and the letters A to F for the numbers as well as a dot (.) to separate the decimals. The A to F must be capital letters in order not to be confused with the commands that are written with lower case characters. A number with a letter is considered as an hexadecimal number:

dc <<< 'Ap'
10

The output is converted in base 10 by default

Scale And Base

dc is a calulator with abitrary precision, by default this precision is 0. thus dc «< "5 4/p" prints "1".

We can increase the precision using the k command. It pops the value at the top of the stack and use it as the precision:

dc <<< '2k5 4/p' # prints 1.25
dc <<< '4k5 4/p' # prints 1.2500
dc <<< '100k 2vp' 
1.4142135623730950488016887242096980785696718753769480731766797379907\
324784621070388503875343276415727

You can use a very large precision.

You can change the base used to output (print) the numbers with o and the base used to input (type) the numbers with i:

dc << EOF
20 p# prints 20 output is in base 10
16o # the output is now done in bas2 16
20p # prints 14,  in hex
16i # the output is now done in hex
p   # prints 14 this doesn't modify the number present in the stack
10p # prints 10 the output is done in base 16
EOF

Take care that when the input value is modified the base is modified for all the commands, including i:

dc << EOF
16i 16o # base is 16 for input and output
10p # prints 10
10i # ! set the base to 10 in base 16 ie to 16 decimal
17p # prints 17 
EOF

This code prints 17 while we might think that 10i revert the base back to 10 and thus the number should be converted to hex and prints 11. The problem is that 10 was typed while the base for input was 16, thus the base was set to 10 in hexadecimal, i.e. to 16 decimal.

dc << EOF
16o16o10p #prints 10
Ai # set the base to A in hex ie 10
17p # prints 11 in base 16
EOF

Stack

There are two basic commands to manipulate the stack:

d duplicates the top of the stack
c clears the stack

$ dc << EOF
2   # put 2 on the stack
d   # duplicate ie put another 2 on the stack
*p  # multiply and print
c p # clear and print
EOF
4
dc: stack empty

c p results in an error, as we would expect, as c remove everything on the stack. Note how we can use # to put comments in the script.

If you are lost you can inspect (i.e. print) the stack using the command f. this command does not modify the stack:

dc <<< '1 2 d 4+f'
6
2
1

Note how the first element that will be poped from the stack is printed first, if you are used to HP calculator, it's the reverse.

Don't hesitate to put some f in the examples of this tutorial, it doesn't change the result and it's a good way to see what's going on.

Registers

The GNU dc manual says that dc has at least 256 registers depending on the range of unsigned char. I'm not sure how you are supposed to use the NUL byte... but anyway...

Using a register is easy:

dc <<EOF
12 # put 12 on the stack
sa # remove it from the stack and put (s) it in the register 'a'
10 # now put 10 on the stack
la # read (l) the value of the register 'a' and push it on the stack
+p # add the 2 values and print
EOF

The above snippet use newlines to embed comments, but it doesn't really matter, you can use echo '12sa10la+p'| dc, it works the same.

The register can contain more that just a value, each register is a stack on its own.

dc <<EOF
12sa #store 12 in 'a'
6Sa # with a capital S the 6 is removed
    # from the  main stack and pushed on the stack 'a'
lap # prints 6, the value at the top of the stack 'a'
lap # still prints 6
Lap # prints 6 also but with a capital L, it push the value in 'a'
    # on the main stack and pull it from the 'a' stack
lap # prints 12, which is now at the top of the stack
EOF

Macros

dc let's you push aribtray strings on the stack in when put inside []. You can print it with p: dc <<< '[Hello World!]p' and you can evalute it with x: dc <<< '[1 2+]xp'.

This is not that interesting until you combine this with registers. First let's say we want the to calculate the square of a number (don't forget to put some f in there if you are lost!):

dc << EOF
3 # push our number on the stack
d # duplicate it ie push 3 again
d**p # duplicate again and calculate the product and print
EOF

Now we have several cube to calculate, we could use dd** several times, or us a macro.

dc << EOF
[dd**] # push a string
sa # save it in the register a
3 # push 3 on the stack
lax # push the string "dd**" on the stack and execute it
p # print the result
4laxp # same thing fo 4, in one line
EOF

Conditionals and Loops

dc can execute a macro stored in a register using the lR x combo, but it can also execute macros conditionally. >a will execute the macro stored in the register a, if the top of the stack is greater than the second element of the stack. Take care, that the top of the stack is the number entered last so when written it's somehow the reverse of what we are used to read:

dc << EOF
[[Hello World]p] sR  # store in 'R' a macro that prints Hello World
2 1 >R   	     # do nothing 1 is at the top 2 is the second element
1 2 >R    	     # prints Hello World
EOF

Some dc have >R <R =R, GNU dc had some more, check your manual. Not that the test "consumes" its operands the 2 first elements are poped off the stack (you can verify that dc «< "[f]sR 2 1 >R 1 2 >R f" doesn't print anything)

Have you noticed how we can include a macro (string) in a macro? and as dc relies on a stack we can in fact use the macro recursively (have your favorite control-c key combo ready ;)) :

dc << EOF
[ [Hello World] p   # our macro starts by printing Hello World
  lRx            ]  # and then executes the macro in R
sR 		    # we store it in the register R
lRx                 # and finally executes it.
EOF

We have recursivity, we have test, we have loops:

dc << EOF
[ li       # put our index i on the stack 
  p 	   # print it, to see what's going on
  1 -      # we decrement the index by one
  si 	   # store decremented index (i=i-1)
 0 li >L   # if i > 0 the execute L
] sL       # store our macro with the name L
 
10 si      # let's give to our index the value 10
lLx        # and start our loop
EOF

Of course code written this way is far too easy to read! Make sure to remove all those extra spaces newlines and comments:

 
dc <<< '[lip1-si0li>L]sL10silLx' 
dc <<< '[p1-d0<L]sL10lLx' # use the stack instead of a register

I'll let you figure out the 2nd example, it's not hard, it uses the stack instead of a register for the index.