Convert signed 64-bit int

print_I64.cpp

#include <iostream>
#include <chrono>

void print_I64(long long num) {
    // Try NOT using `if` statement, since it will make the program slower
    char str[32] = {0};
    short idx = 31;
    long long sign = num >> 63; // -1 for negative, 0 for positive
    // by doing this we assume that the spaces are filled with num's sign bit
    num ^= sign;
    num -= sign;
    // now we are sure that num is positive, 
    // and this seems to be faster than using normal ways to get absolute value ...
    // possble ways can be: int is_postive = (x > 0); x *= (2 * is_positive - 1); 
    do {
        --idx;
        str[idx] = num % 10 + '0';
        // but you can choose NOT to get a absolute value and 
        // acquire the character by using a predefined string 
        // like "9876543210123456789", and index it by `num % 10 + 9`
    } while (num /= 10);
    // if (sign) str[--idx] = '-'; // if we're going to eliminate all `if` statement ...
    str[idx] = '-' * (-sign) + str[idx] * (1 - sign); // using this also works
    std::cout << &str[idx]; // You can comment this line out when counting time cost
}

// above we are preforming directly on a 64-bit 'int', I'm wondering whether it would be 
// faster to split the 64-bit 'int' into two 32-bit 'int's on 32-bit machine

void print_I64_split(long long num) {
    // 'long long int' has 8 Bytes, which is 64 bit
    char str[32] = {0};
    short idx = 31;
    int * ptr = (int *) &num; // points to the block which stores the first 32 bit of 'num'
    int sign = *(ptr + 1); sign >>= 31; // -1 for negative, 0 for positive
    *(ptr)     ^= sign;
    *(ptr + 1) ^= sign;
    num -= sign;
    do {
        --idx;
        str[idx] = num % 10 + '0';
    } while (num /= 10);
    if (sign) str[--idx] = '-';
    std::cout << &str[idx];
}


int main()
{
    long long num; 
    std::cin >> num;
    std::chrono::steady_clock sc;
    auto start = sc.now();
    for (int i = 0; i < 10000; ++i) 
        print_I64(num);
    auto end = sc.now();
    auto time_span = static_cast<std::chrono::duration<double>>(end - start);
    std::cout << "operation took: " << time_span.count() << " seconds." << std::endl;
    return 0;
}