Convert "Floating" Fractional Binary to Decimal

Sorry for the delay to post the reply .. If someone else can only send it, it would be of help to many. In case the program ai also converts decimal float to binary, binary float to decimal ...

/*Trabalho de Calculo numerico  - Converter numero binario para numero decimal*/





#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<math.h>
#define max 30

int fraci[max],inteiro[max];

//Funcoes
void decimalparabinario();
void binarioparadecimal();
float pow();
//Fim Funcoes



int main()
{
    int op;
    float aux,num,f;
    int i,cnum,ch;
    do
    {
        printf("===== Menu ======\n");
        printf("1. Binario para Decimal\n");
        printf("2. Decimal para Binario\n");
        printf("3. Sair\n");
        printf("Digite uma opcao: \n");
        scanf("%d",&op); //
        system("clear"); //Comando utilizado para limpar tela


        switch(op)
        {
        case 1:

            binarioparadecimal();
            system("PAUSE");


            break;
        case 2:
            decimalparabinario();

            system("PAUSE");


            break;
        case 3://Sair do programa
            printf("Saindo..\n");

            break;
        default://Caso não seja nenhuma das anteriores...
            printf("Opcao Invalida\n");
            break;



        }



    }
    while(op != 3);


    system("PAUSE");

    return 0;
}

void decimalparabinario(){

    float numero,f;
    int i,cnum,t;

    printf("Digite um numero Decimal  : ");
    scanf("%f",&numero);
    printf("\n");

    f=numero-(int)numero;

    for(i=0; i<max; i++)
    {
        f=f-(int)f;
        fraci[i]=(int)(f*2);
        f*=2;
    }

    cnum=(int)numero;
    for(i=max-1; i>=0; i--)
    {
        if(cnum%2==0)
        {
            inteiro[i]=0;
        }
        else
            inteiro[i]=1;
        cnum/=2;
    }

    printf("O numero em Binario é igual   = ");
    t=0;

    for(i=0; i<max; i++)
    {
        if(inteiro[i]||t)
        {
            t=1;
            printf("%d",inteiro[i]);
        }
    }
    printf(".");

    for(i=0; i<7; i++)
        printf("%d",fraci[i]);

    printf("\n");



}


void binarioparadecimal()
{
    char num[50];
    int antesDec[50], depoisDec[50];
    int antesDecnum, depoisDecnum, tint=0, l, j=0, k=0;
    float fract=0, floatValue;
    char p = 'd';

    printf("******* Insira o binario para converter  para decimal *********\n");

    //
    printf("Digite o numero binario : ");
    scanf("%s",&num);

    //Separa o inteiro do fracionado

    for(l=0; l<strlen(num); l++)
    {
        if(num[l]=='.')
        {
            p='u';
        }
        else if(p=='d')
        {

            antesDec[l] = (int)num[l]-48;
            k++;
        }
        else
        {
            depoisDec[j] = (int)num[l]-48;
            j++;
        }
    }

    // Armazena o tamanho do inteiro e do fracionario
    antesDecnum = k;
    depoisDecnum= j;

    //Converter parte inteira para decimal
    j=0;
    for(l = antesDecnum-1; l>=0; l--)
    {
        tint = tint + (antesDec[l] *(int) pow(2,j));
        j++;
    }

    //Converter parte fracionaria para decimal
    j = -1;
    for(l = 0; l<depoisDecnum; l++)
    {
        fract = fract + (depoisDec[l]*pow(2,j));
        j--;
    }

    //Juntando as duas partes
    floatValue = tint + fract;
    printf("Numero em decimal é  = %f\n\n\n\n\n",floatValue);


}
float pow(int c, int d)
{
    float pow=1;
    if (d >= 0)
    {
        int i = 1;
        while (i <= d)
        {
            pow = pow * c;
            i++;
        }
    }
    else
    {
        int i = 0;
        while (i > d)
        {
            pow = pow/c;
            i--;
        }
    }
    return pow;
}


// intnum = numero inteiro.
// fracnume = numero fracionario

This is easy to solve with basic math:

1.01010101

is the same as

101010101 / 100000000

(just count the boxes after the comma)

 Binário       Decimal
 101010101 ÷   314 ÷
 100000000     256
────────────  ─────────────
 1.01010101    1.332031251

In short, just convert normally, and work in decimal.

For simplicity, count the boxes after the comma, make 2 ^ Número de casas 1 and use as a divider:

1.01010101
  └───┬──┘
      └── 8 casas

101010101 = 314
    2 ^ 8 = 256
    │   └───────── 8 casas
    └───────────── base 2, binário. Se fosse decimal, seria 10, hexadecimal 16 etc.

314/256 = 1.332031251

1. I am using a ^ b to indicate a raised to b .

What if it was hexadecimal?

Same thing:

  F,1BC
    └┬┘
     └── 3 casas

  F1BC = 61884
16 ^ 3 =  4096
 │   └───────── 3 casas
 └───────────── base 16

61884/4096 = 15,1083984375

Note: I saw the other question here on the site about showing number with decimal places in binary. I think it's much simpler to use this same technique here in that case too:

Go multiplying the number by 2 until you have no more decimals. Write down how many times you have multiplied. The number of times is the position of the comma in the binary. (if you want to advance this step, you can deduce with logarithm, but as it is outside the current question, I will not go into details).

One thing I've already mentioned in another question and applies in this case: regardless of the base you work on, the math is the same .

Running away from the theory and going straight to practice.

#include <string.h>
#include <stdio.h>

int bin2des(unsigned long long bin){
    int x;
    int y;
    char num[18];
    sprintf(num,"%d",bin);
    x = strlen(num);
    y = 1;
    int res;

    while(x--){
        if(num[x] == '-')
            res *= -1;
        else
            res += (num[x]-48)*y;
        y *= 2;
    }

    return res;
}

void main(){
    printf("%i\n",bin2des(101));
}