Program using malloc twice

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Program using malloc twice

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#1 by (2 votes)
#2 by (1 votes)
#3 by (1 votes)

3

Because this program, which divides a number into decimal notation, transforms it into binary notation and prints the number in the correct sequence (from the most significant to the least significant bit ). uses malloc twice?

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

int main(int argc, char *argv[]) {
    // Stack
    int decimal, q, r;
    int counter, i;
    char *binary = NULL;
    char *aux;

    printf("Digite um nÃºmero em base decimal: ");
    scanf("%d", &decimal);

    counter = 1;
    while (decimal >= 2) {
        q = decimal / 2;
        r = decimal - (q * 2);

        // Heap
        aux = (char *) malloc(counter * sizeof(char));
        if (binary != NULL) {
            memcpy(aux, binary, counter-1);
            free(binary);
        }
        binary = aux;

        if (r == 0) {
            binary[counter-1] = 48; //'0';
        } else {
            binary[counter-1] = 49; //'1';
        }

        //printf("resto %d = %d\n", counter, r);
        counter++;
        decimal = q;
    }
    //printf("ultimo quociente = %d\n", q);

    // Heap
    aux = (char *) malloc(counter * sizeof(char));
    if (binary != NULL) {
        memcpy(aux, binary, counter-1);
        free(binary);
    }
    binary = aux;

    if (decimal == 0) {
        binary[counter-1] = 48; //'0';
    } else {
        binary[counter-1] = 49; //'1';
    }

    printf("Resultado em binÃ¡rio = ");
    for (i = counter-1; i >= 0; i--) {
        printf("%c", binary[i]);
    }
    printf("\n");

    free(binary);

    return 0;
}

c

asked by anonymous 16.04.2015 / 02:56

3 answers

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score 2 · Answer 1

If the problem is to avoid making every calculation of one of the debris again after the loop ends, it is easy to solve. The problem is that it is stopping when it arrives at the rest 2 and correct is to do until the rest is 1. So a simple change in the condition of while solves this.

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

int main(int argc, char *argv[]) {
    // Stack
    int decimal, q, r;
    int counter, i;
    char *binary = NULL;
    char *aux;

    printf("Digite um nÃºmero em base decimal: ");
    scanf("%d", &decimal);

    counter = 1;
    while (decimal >= 1) { // <=============================== Mudei aqui de 2 para 1
        q = decimal / 2;
        r = decimal - (q * 2);
        aux = (char *) malloc(counter * sizeof(char));
        if (binary != NULL) {
            memcpy(aux, binary, counter-1);
            free(binary);
        }
        binary = aux;
        if (r == 0) {
            binary[counter-1] = 48; //'0';
        } else {
            binary[counter-1] = 49; //'1';
        }
        counter++;
        decimal = q;
    }
    printf("Resultado em binÃ¡rio = ");
    for (i = counter-1; i >= 0; i--) {
        printf("%c", binary[i]);
    }
    printf("\n");
    free(binary);
    return 0;
}

See running on ideone .

I did not analyze other aspects.

score 1 · Answer 2

Why does this program [...] use malloc twice?

For numbers with N bits, the while program cycle runs N - 1 time. This cycle does not run when decimal is 0 or 1 .

while (decimal >= 2) { /* ... */ }

Therefore, within the loop, the allocation made is not sufficient to store the entire binary representation.

So, at the end of the cycle, the program makes one more allocation to the last bit.

score 1 · Answer 3

This exercise they put you in is interesting: to understand the code of others and eventually modify it. The code of the statement has the virtue of: introducing several concepts (memory types, mallocs, memcpy, etc.) and still have "workable" things

In addition to your question, if it were me (keeping the spirit of it) in addition to the above improvements,

(1) it ended with dynamic memory

   char binary[32];

or at least allocate all at once (required size = 1 + log2 (decimal):

   binary=malloc((int)(1+log2(decimal)))

(2) replaced the

   if (r == 0) {
        binary[counter-1] = 48; //'0';
    } else {
        binary[counter-1] = 49; //'1';
    }

by

    binary[counter-1] = r + '0';

In summary ...

...
#include <math.h>

int main() {
    int decimal, q, r, counter;
    char *binary;

    printf("Digite um número em base decimal: ");
    scanf("%d", &decimal);
    binary=malloc((int)(1+log2(decimal)));

    counter = 0;               //from @pmg
    while (decimal > 0) {      //from @bigown
        q = decimal / 2;
        r = decimal % 2;
        binary[counter] = r + '0';
        counter++;
        decimal = q;
    }

    printf("Resultado em binário = ");
    while (counter--) { putchar(binary[counter]); }
    printf("\n");

    free(binary);
    return 0;
}