• A+
Category：Languages

I came across an intriguing C code that prints `A + B`, but I have trouble understanding it.

## Input Format:

``A B ``

where `A`, `B` are integers between `0` and `10` separated by a single space.

## Code:

``main( n ) {     gets( &n );     printf("%d", n % 85 - 43); } ``

This was intended for short coding, please don't mind the warnings.

## What I understand so far:

`gets( &n )` stores the ASCII values of A, space, and B in the lower three bytes of `n`. For example, `A = 3` and `B = 8` would yield `n = 0x00382033`. Given conditions prevent `n` from overflowing. But I do not understand how `n % 85 - 43` yields `A + B`.

How do you come up with these numbers?

Assuming little-endian representation, and ignoring all the technically-wrong-in-modern-C stuff in the code, your "What I understand so far" is correct.

Storing those ASCII values into those bytes of `n` results in `n` taking the value `B*256*256 + space*256 + A`, where `B`, `space`, and `A` represent the corresponding ASCII values. 256 mod 85 is 1, so by the properties of modular arithmetic,

``(B*256*256 + space*256 + A) % 85 = (B + space + A) % 85 ``

Space is ASCII 32, and the ASCII value for a digit character is 48 + the value of the digit. Defining `a` and `b` as the numeric values of the digits entered (rather than the ASCII values of the digit characters), we have

``(B + space + A) % 85 = (b + 48 + 32 + a + 48) % 85                      = (a + b + 128) % 85                      = (a + b + 43) % 85  (B + space + A) % 85 - 43 = (a + b + 43) % 85 - 43                           = (a + b) % 85                           = a + b ``

where the last two equivalences rely on the fact that `a` and `b` take values from 0 to 9.