Working with 128-bit Integers in C++

What this code does (short)

The code defines a 128-bit integer type for GNU C++ (__int128) and provides two stream operators so you can use cin >> and cout << with that type. This is useful because the standard iostream library does not know how to read or print __int128.

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Line-by-line explanation

// for: GNU C++20 (64)
using i128 = __int128;
const i128 ONE_128 = i128(1);
const i128 ZERO_128 = i128(0);

• using i128 = __int128; creates an alias i128 for the builtin GNU type __int128.

• __int128 is a 128-bit signed integer available in GNU compilers on 64-bit systems. It stores much bigger values than long long.

• These two lines define constants ONE_128 and ZERO_128 with values 1 and 0 typed as i128.

• They make the code clearer when you need literal 128-bit constants.

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Input operator: operator >>

std::istream &operator>>(std::istream &is, i128 &n) {
    n = 0;
    std::string s; is >> s;
    for (auto c : s) {
        n = 10 * n + c - '0';
    }
    return is;
}

• This overload lets you write cin >> my_i128_variable;.

• Steps it performs:

  1. Set n to zero.
  2. Read the whole token from the stream into a std::string s. This skips leading whitespace and reads until the next whitespace.
  3. For each character c in the string, it updates n = 10*n + (c - '0'). That converts the decimal digits to a numeric value.
  4. Return the input stream reference so chaining (e.g., cin >> a >> b;) works.

• Important limitations:

  • The code assumes the string contains only digits ('0'–'9'). It does not handle a leading - sign, a leading +, or invalid characters.
  • It does not check for overflow. If the number in the input is larger than 128 bits can hold, the behavior is undefined (it will wrap or give wrong results).

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Output operator: operator <<

std::ostream &operator<<(std::ostream &os, i128 n) {
    if (n == 0) {
        return os << 0;
    }
    std::string s;
    while (n > 0) {
        s += '0' + n % 10;
        n /= 10;
    }
    std::reverse(s.begin(), s.end());
    return os << s;
}

• This overload lets you write cout << my_i128_variable;.

• Steps it performs:

  1. If n is zero, it prints 0 and returns.
  2. Otherwise it builds a string s with the digits of n in reverse order: take n % 10 to get the last digit, append the corresponding character, and then divide n by 10.
  3. Reverse the string to get the correct digit order.
  4. Output the string and return the stream.

⚠️ Warning: The provided code only works correctly for positive integers. It does not support negative numbers or the sign -, and printing a negative i128 will also produce incorrect output.

• Important limitations:

  • The code assumes n is non-negative. If n is negative, the while (n > 0) loop never runs and the result is wrong. Typical fix: check if (n < 0) { os << '-'; n = -n; } before converting digits.
  • Converting a 128-bit integer to decimal takes time proportional to the number of digits (about 39 digits max for 128-bit), which is fine for competitive programming.

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How to use

i128 a;
cin >> a;
cout << a << '\n';