Number bases

In base b, each digit has a value from 0 to b - 1. Positions from right to left have weights 1, b, b * b, and so on. Decimal uses 0..9; binary uses only 0 and 1.

Decode a representation#

Binary 1101 means 1 * 8 + 1 * 4 + 0 * 2 + 1 = 13. Distinguish the string "1101", the decimal number 1101, and the value 13 represented by those binary digits.

To process digits from left to right, multiply the current value by the base, then add the next digit.

1text = "1101"2value = 03FOR i FROM 0 TO LEN(text) - 14  digit = ORD(CHAR(text, i)) - ORD("0")5  value = value * 2 + digit6END7PRINT value

The result is 13. Intermediate values are 1, 3, 6, 13. The example assumes a non-empty string containing binary digits only; validate characters before processing external input.

Decimal to binary#

Repeated division by two produces digits in reverse order. Prepend each remainder to the result string.

1n = 132x = n3binary = ""4IF x == 0 THEN5  binary = "0"6END7WHILE x > 08  binary = STR(x MOD 2) + binary9  x = x DIV 210END11PRINT binary

The remainders are 1, 0, 1, 1; the final string is 1101. Zero needs a separate case or the result would be empty. This algorithm requires a non-negative integer.

For bases 2..10, replace divisor 2 with the chosen base. Higher bases need extra digit symbols: hexadecimal uses A..F for 10..15. Do not concatenate STR(15) for one hexadecimal digit; use an alphabet such as "0123456789ABCDEF" and CHAR.

Practice#

1. Convert 0, 1, 8, 15 to binary: 0, 1, 1000, 1111.
2. Explain why binary 1010 is decimal 10.
3. Add validation: 102 is not valid binary. Keep decoded numeric values within the exact integer range.