Psuedocode for converting binary to decimal and decimal to binary.

Question

I am studying python and my question is how to write a pseudocode that explains the steps in converting 1) binary to decimal 2) decimal to binary. (If example needed then binary 1010 to decimal 10)

Answer 1

The answer depends on how big the numbers are.

If they are very big, then the problem is difficult.

Imagine a number containing millions of digits.

After seeing only a small number of digits in one radix, say radix 2,

we need to produce the leading digit in the other radix, say radix 10.

I do not know how to solve that.

So I will just show you a pseudocode for the simple problem, where

the numbers involved fit into a machine representable integer.

So below are two pieces of pseudocode:

One converts a string of digits of any given radix (2 or 10) into an integer k.

And the other converts an integer k into a string of digits in any given radix.

In both cases all integers are non-negative.

Negative integers are yet another story, called "two's complement".

PSEUDOCODE 1:

# Convert an array A of length n with digits in radix r into an integer k.

# A[0] contains the least significant digit

# A[n-1] contains the most significant digit

k = 0 # the result k will keep accumulating the final value

i = 0 # start iterating from the least significant digit

v = 1 # always v= rⁱ is the contribution of a digit in position i

loop for i = 0 ... n-1

k += A[i]*v # k gets the contribution of digit i

v *= r # v becomes the value of the next more significant position

PSEUDOCODE 2:

# Convert a given non-negative integer k into an array A with digits in radix r.

# We assume that A has been allocated big enough.

# The pseudocode will compute

# n -- the number of digits in radix r needed to represent k

# A -- the digits of the representation

# A[0] will contains the least significant digit

# A[n-1] will contains the most significant digit

n = 0 # start iterating from the least significant digit

loop until break

A[n] = k % r # each digit is the remainder of k if divided by r

n += 1 # prepare for next digit

k /= r # integer division discarding remainder

if (k == 0) # the whole number k converted

then break from the loop