Write a C program that inverts the bits of a non-negative number provided by the user through argument. Note, the program is to invert the order of the bits, i. e., move the lowest bit to the highest position, 2nd lowest bit to 2nd highest position, etc. It is not to flip the bits (0à1, 1à0). The program prints out the number, its binary representation, the value after the inversion, and the binary representation after the inversion. Your program can assume that the number is a 32-bit unsigned integer (i. e., a number in [0, 232 – 1]).

See explaination for code

Explanation:

Below is the program code.

#include <stdio.h>

#include <stdlib.h>

void toBinary( unsigned int num, int bits[]);

void print(unsigned int num);

unsigned int invert(unsigned int num);

int main(int argc, char *argv[]){

unsigned num, inverted;

if(argc != 2){

printf("ERROR: should pass a number as command line argument\n");

printf("e.g. Usage: %s 4891183\n", argv[0]);

return 1;

}

num = atoi(argv[1]);

inverted = invert(num);

print(num);

print(inverted);

return 0;

}

void toBinary(unsigned int num, int bits[])

{

int i;

//LSB in bits[31]

for(i = 31; i >= 0; i--){

bits[i] = num % 2;

num = num / 2;

}

}

void print(unsigned int num){

int bits[32];

int i = 0;

toBinary(num, bits);

printf("%-20u ", num);

for(i = 0; i < 32; i++){

if(i % 4 == 0)

printf(" "); //space after every 4 bits

printf("%d", bits[i]);

}

printf("\n");

}

unsigned int invert(unsigned int num){

int bits[32];

int i;

toBinary(num, bits);

unsigned int newnum = 0;

for(i = 31; i >= 0; i--){

newnum = (newnum << 1 ) | bits[i];

}

return newnum;

}#include <stdio.h>

#include <stdlib.h>

void toBinary( unsigned int num, int bits[]);

void print(unsigned int num);

unsigned int invert(unsigned int num);

int main(int argc, char *argv[]){

unsigned num, inverted;

if(argc != 2){

printf("ERROR: should pass a number as command line argument\n");

printf("e.g. Usage: %s 4891183\n", argv[0]);

return 1;

}

num = atoi(argv[1]);

inverted = invert(num);

print(num);

print(inverted);

return 0;

}

void toBinary(unsigned int num, int bits[])

{

int i;

//LSB in bits[31]

for(i = 31; i >= 0; i--){

bits[i] = num % 2;

num = num / 2;

}

}

void print(unsigned int num){

int bits[32];

int i = 0;

toBinary(num, bits);

printf("%-20u ", num);

for(i = 0; i < 32; i++){

if(i % 4 == 0)

printf(" "); //space after every 4 bits

printf("%d", bits[i]);

}

printf("\n");

}

unsigned int invert(unsigned int num){

int bits[32];

int i;

toBinary(num, bits);

unsigned int newnum = 0;

for(i = 31; i >= 0; i--){

newnum = (newnum << 1 ) | bits[i];

}

return newnum;

}

it does automatically do that fr , and idk what to do bout it , it must be automoderators

some have built technology like spoofed call blockers into their networks, so there's nothing you would even have to install.

this happens because the number is known for scamming, they don't want it to me known as "scam likely." perhaps it's just a telemarketer. maybe it's an identity thief. it might even be a scam artist.  

either way, "scam likely" is probably not an old friend whose number you forgot that you saved. it's your wireless carrier's way of letting you know: answer at your own risk.

explanation:

to combat the seemingly rising epidemic of spam callers in the u.s., the nation's four major wireless carriers have been trying out new methods to reduce the number of unsolicited calls americans receive each day.

some have built technology like spoofed call blockers into their networks, so there's nothing you would even have to install.

hope this . mark me brainliest.