Base eight, by hand

Print a number in octal (base 8) without letting printf do the conversion: no %o or %lo anywhere in the program.

Write n as a sum of powers of 8 and the digits fall out: 493 is 7 * 64 + 5 * 8 + 5 * 1, so in octal it is 755. Start at the largest power of 8 that fits in n. Divide n by it: the quotient is the next digit, and the remainder carries on to the next power down. Stop after the power 1.

The starter reads n and prints the start of the line (493 in octal is ). You print the digits, one printf with fmt_digit each, and the starter ends the line. Print no leading zeros, but 0 itself prints as 0.

A negative n prints the way %lo would print it: the octal digits of the 64 bits the machine actually stores for it (its two's complement pattern), so -1 comes out as 1777777777777777777777. That means treating n's bits as unsigned throughout: udivUnsigned divide; divide-by-zero writes 0. rather than sdivSigned divide; divide-by-zero writes 0., and when you compare two such values, the unsigned conditions b.loConditional branch (write as B.EQ, B.NE, B.LT, ...)., b.lsConditional branch (write as B.EQ, B.NE, B.LT, ...)., b.hiConditional branch (write as B.EQ, B.NE, B.LT, ...). and b.hsConditional branch (write as B.EQ, B.NE, B.LT, ...)., not their signed partners.

Hint: to find the starting power, compare it with n / 8 rather than comparing power * 8 with n. Near the top of the 64-bit range, power * 8 no longer fits.

What is checked

• with the input 493 the program prints 493 in octal is 755 and returns 0
• no %o-style format string appears in the program
• the checker also tries values you do not see: 0, exact powers of 8, the largest 64-bit number, and negatives