C++


A peculiar way to get the biggest (max/maximum) value between two variables using bitwise operations 3

Recently, we wanted to make a test and see how we could find the maximum value between two variables using bitwise operations.

We ended up with the following peculiar way to get the biggest value between two variables using bitwise operations


r = a ^ ((a ^ b) & -(a < b));

The above formula has two modes:

  1. When a < b
  2. When a >= b

 

When a < b then the formula will change as follows:


r = a ^ ((a ^ b) & 0xFFFFFFFF);

As we all (should) know, when one of the operators on a bitwise AND operation is composed only from 1s, then the result is whatever value the other operator was holding.
So, the formula then simplifies as follows:


r = a ^ (a ^ b);

which is equal to


r = b;

because we when we apply twice the same value using XOR on another value, we revert back to the original value (so the second ^a nullifies the first ^a)

 

When a >= b then the formula will change as follows:


r = a ^ ((a ^ b) & 0x00000000);

When one of the operators on a bitwise AND operation is composed only from 0s, then the result is always 0 no matter what value the other operator was holding.
So, the formula then simplifies as follows:


r = a ^ (0x00000000);

which is equal to


r = a;

because when one of the operators in a XOR operation is only composed from 0s then the result will be the value of the other operator, no matter what it was.

 

Full example

Below you will find a full example that compares the execution speed of the two methods by executing each several thousands of time on the same random data.

[download id=”3875″]


#include <stdio.h>
#include <time.h>
#include <stdlib.h>

int main() {
    {
        const clock_t start = clock();

        srand(10);
        unsigned long int i;
        unsigned int max = 0;
        for (i = 0; i < 1000000000; i++) {
            const int a = rand();
            max = max < a ? a : max;
        }
        const clock_t end = clock();
        const float seconds = (float) (end - start) / CLOCKS_PER_SEC;
        printf("Seconds elapsed %f\tIf statement. Overall max value = %u\n", seconds, max);
    }

    {
        const clock_t start = clock();

        srand(10);
        unsigned long int i;
        unsigned int max = 0;
        for (i = 0; i < 1000000000; i++) {
            const int a = rand();
            max = a ^ ((a ^ max) & -(a < max));
        }
        const clock_t end = clock();
        const float seconds = (float) (end - start) / CLOCKS_PER_SEC;
        printf("Seconds elapsed %f\tBitwise operation. Overall max value = %u\n", seconds, max);
    }
    return 0;
}

Results

Our results show that using the traditional if statement with assignment is faster than using our formula as expected.
Which makes sense as there is an if statement in the formula as well and then additional operations to get the result, instead of just the assignment.

Seconds elapsed 5.770000 If statement. Overall max value = 2147483647
Seconds elapsed 6.180000 Bitwise operation. Overall max value = 2147483647

10 times bigger input

Seconds elapsed 57.450001 If statement. Overall max value = 2147483647
Seconds elapsed 63.869999 Bitwise operation. Overall max value = 2147483647

CentOS 7: C++: static linking cannot find -lstdc++ -lm and -lc

Recently, we were trying to compile a C++ application with the following compilation command on a CentOS 7 64bit :


g++ -static -O2 -lm -Wall -Wno-unused-result -std=c++11 -DCS_ACADEMY -DONLINE_JUDGE 510152025.cpp -o 510152025;

unfortunately, we got the following errors:

 /usr/bin/ld: cannot find -lstdc++
 /usr/bin/ld: cannot find -lm
 /usr/bin/ld: cannot find -lc
 collect2: error: ld returned 1 exit status

To resolve the issues, we performed the following installations to install the static versions of the glibc and libstdc libraries:


sudo yum install glibc-static libstdc++-static -y;

 


Fedora 26: C++: static linking cannot find -lstdc++ -lm and -lc

Recently, we were trying to compile a C++ application with the following compilation command on a Fedora 26 64bit :


g++ -static -O2 -lm -Wall -Wno-unused-result -std=c++14 -DCS_ACADEMY -DONLINE_JUDGE 510152025.cpp -o 510152025;

unfortunately, we got the following errors:

 /usr/bin/ld: cannot find -lstdc++
 /usr/bin/ld: cannot find -lm
 /usr/bin/ld: cannot find -lc
 collect2: error: ld returned 1 exit status

To resolve the issues, we performed the following installations to install the static versions of the glibc and libstdc libraries:


sudo dnf install glibc-static libstdc++-static -y;

 


C/C++: Get a random number that is in a specific range 1

Assuming you need to generate a random number that is in a specified range, you can do the following:


//int rand(void) creates a pseudo-random number in the range of 0 to RAND_MAX
//RAND_MAX is defined in stdlib.h and is the largest number rand will return (same as INT_MAX).
const int new_number = (rand() % (maximum_number + 1 - minimum_number)) + minimum_number;

The above code first creates a pseudo-random number that is in the range of [0, RAND_MAX].
Then it will divide it with the width (+1) of the range we want to use (maximum_number + 1 - minimum_number) and get the remainder (modulo).
The modulo will be in the range of [0, maximum_number - minimum_number], so we add to it the value of minimum_number to shift the result to the proper range.
This solution, as demonstrated in the example below, works for negative ranges as well.

Full example of generating 100000 random numbers that are all in the range [-31, 32].

const int maximum_number = 31;
const int minimum_number = -32;
unsigned int i;
for (i = 0; i <= 100000; i++) {
	const int new_number = (rand() % (maximum_number + 1 - minimum_number)) + minimum_number;
	printf("%d\n", new_number);
}