Ykhjuy

This program converts a decimal number to a binary number. This is one of my first C programs and I am wondering if I have used the elements of this language properly. Suggestions for improvement are welcome.

#include <stdio.h>

#include <string.h>



void print_out_reversed(char string)

{

    int index = strlen(string);



    while (string[index] != '')

        index--;



    for (int i = index; i >= 0; i--)

        putchar(string[i]);

    putchar('n');

}



void print_decimal_number_binary(int number)

{

    if (number == 0)

    {

        printf("0n");

        return;

    }



    char bits[sizeof(int) * 8 + 1] = {0};

    int index = 0;



    while (number > 0)

    {

        if (number % 2 == 0)

        {

            bits[index] = '0';

        }

        else

        {

            bits[index] = '1';

        }

        number = number / 2;

        index++;

    }



    print_out_reversed(bits);

}



int main()

{

    printf("enter number: ");

    int number;

    scanf("%i", &number);

    print_decimal_number_binary(number);

}

Terminology

It's important to be able to understand (and describe) what's actually going on. Your program

1. converts from an integer decimal string representation to an integer using scanf. This integer is then represented as a binary number in the processor.


2. converts from that integer back into a string representation, but rather than it being decimal, it's binary.

So yes - it technically converts from "decimal to binary", but really it's "decimal string to integer to binary string".

Use const

void print_out_reversed(char string)

doesn't modify string, so write const char string.

Simplify your strlen usage

This:

int index = strlen(string);



while (string[index] != '')

    index--;



for (int i = index; i >= 0; i--)

    putchar(string[i]);

can be

for (int i = strlen(string)-1; i >= 0; i--)

    putchar(string[i]);

It seems that you don't trust what strlen is doing, which is why you have that intermediate while loop. But that loop won't have any effect, because the null terminator will always be where strlen says it is.

Use math instead of if

This:

    if (number % 2 == 0)

    {

        bits[index] = '0';

    }

    else

    {

        bits[index] = '1';

    }

can be

bits[index] = '0' + (number & 1);

Use combined operation and assignment

This:

number = number / 2;

should be

number /= 2;

or, for speed (which the compiler will do for you anyway)

number >>= 1;

#include stdio.h

#include string.h



void get_bits(void * num, char * out, int bytes);



int main(void)

{

    int x = 0;



    printf("Enter an integer: ");

    scanf("%i", &x);



    char bits[65] = {0};

    get_bits(&x, bits, 4);



    printf("%d in binary %sn", x, bits);



    return 0;

}



//assumes char array of length 1 greater than 

//number of bits



void get_bits(void * num, char *out, int bytes)

{

    unsigned long long filter = 0x8000000000000000;

    unsigned long long temp;



    if(bytes <= 0) return;

    if(bytes > 8) bytes = 8;



    filter = filter >> (8*(sizeof(unsigned long long)-bytes));

    memcpy(&temp, num, bytes);

    int bits = 8*bytes;

    for(int i=0;i<bits;i++) {

        if(filter & temp)

            out[i] = '1';

        else

            out[i] = '0';



        temp = temp << 1;

    }

    out[bits] = '';

}

The code above stores characters representing the bits for char, short, int, or long long in a character array (up to 64 bits). It does not require reversing the order of bits and restoring them. Essentially, it creates a mask ("filter") with a single bit set. This is always the highest order bit. initially, it is the highest order bit for a 64 bit number. For a 32-bit integer, the mask will be shifted to have the 32 bit set, all others 0. The mask will be adjusted for whatever number of bytes the caller indicates are in the number.

A logical and is performed with the number. If the corresponding bit is 1 in the number, it stores a 1 character in the array passed to get_bits. The number, temporarily stored in the variable, temp, is shifted left in each iteration of the for loop. This allows comparison each iteration of the highest order bit in the number. If it is not set in the number, filter & temp is 0 and a 0 character is stored at that position in the output string.

The user of the function get_bits must pass a void pointer, which stores the address of the number, for which the bits are determined. The argument out is a pointer to a character array for the output string. This should be one character longer than the number of bits in the number. The argument bytes represents the bytes of data required to store the number. The output character array is null terminated at position corresponding to the number of bits in the number (that is the length of the string with null terminator is one greater than the number of bytes *8).