How to view contents of a .bin file generated in c

1

I had to create a program in C that reads a motion sensor activation file (.txt) and produces a binary file with the same information.

The .txt file has the following information:

   3B2 20051023 014857393 609f
   3B3 20051023 014857393 00ff
   3B4 20051023 014857503 609f
   3B5 20051023 014857503 00ff
   3B6 20051023 014857613 6093
   3B7 20051023 014857613 807f
   3B8 20051023 014857723 609f
   3B9 20051023 014857723 00ff
   3BA 20051023 014857834 609f
   3BB 20051023 014857864 00ff
   3BC 20051023 014904113 807f
   3BD 20051023 014904113 08f7
   3BE 20051023 014904223 807f
   3BF 20051023 014904223 00f7

I created the following code in C to read the file and save it inside a new .bin file, my doubt is in how to open this .bin file and check if what is written is correct, there is some program that converts .bin file for .txt?

Code:

#include <stdio.h>

int main() {
    int  data1,hora1;
    char sequencial1[4],ativacao1[5],nomeEntrada[50],nomeSaida[50];
    FILE *arqEntrada, *arqSaida;

    printf("Digite 2 nomes de arquivo: ");
    scanf("%s %s",&nomeEntrada,&nomeSaida);

    if ((arqEntrada = fopen(nomeEntrada,"r")) == NULL) {
        printf("Problema na abertura do arquivo %s.\n",nomeEntrada);
        return -1;      
    }
    if ((arqSaida = fopen(nomeSaida,"wb")) == NULL) {
        printf("Problema na abertura do arquivo %s.\n",nomeSaida);
        return -1;      
    }
    while (fscanf(arqEntrada,"%s %d %d %s",&sequencial1, &data1, &hora1, &ativacao1) > 0) {
            fwrite(&sequencial1,sizeof(char),3,arqSaida);
            fwrite(&data1,sizeof(int),1,arqSaida);
            fwrite(&hora1,sizeof(int),1,arqSaida);
            fwrite(&ativacao1,sizeof(char),4,arqSaida);
    } 

    fclose(arqEntrada);
    fclose(arqSaida);

    return 0;
}
    
asked by anonymous 16.10.2017 / 16:43

1 answer

2

Converter:

#include <stdio.h>

#define MAX_BUF_LEN 100

int main( int argc, char * argv[] )
{
    FILE * fin = NULL;
    FILE * fout = NULL;
    char buf[ MAX_BUF_LEN ] = {0};
    int nlinha = 1;
    int n = 0;
    int seq = 0;
    int ativ = 0;
    int dt = 0L;
    int hr = 0L;

    if(argc < 3)
    {
        printf( "Erro de Sintaxe:\n\t%s entrada.txt saida.bin\n", argv[0] );
        return 1;
    }

    if((fin = fopen( argv[1] , "r")) == NULL)
    {
        printf( "Erro abrindo arquivo para leitura: %s\n", argv[1] );
        return 1;
    }

    if((fout = fopen(argv[2], "wb")) == NULL)
    {
        printf( "Erro abrindo arquivo para gravacao: %s\n", argv[2] );
        fclose(fin);
        return 1;
    }

    while( fgets( buf, MAX_BUF_LEN, fin) )
    {

        if((n = sscanf( buf, "%X %d %d %x", &seq, &dt, &hr, &ativ )) != 4)
        {
            printf( "Erro de leitura: %s:%d\n", argv[1], nlinha );
            continue;
        }

        fwrite( &seq, sizeof(short), 1, fout );
        fwrite( &dt, sizeof(int), 1, fout );
        fwrite( &hr, sizeof(int), 1, fout );
        fwrite( &ativ, sizeof(short), 1, fout );

        nlinha++;
    }

    fclose(fout);
    fclose(fin);

    return 0;
}

Entry:

   3B2 20051023 014857393 609f
   3B3 20051023 014857393 00ff
   3B4 20051023 014857503 609f
   3B5 20051023 014857503 00ff
   3B6 20051023 014857613 6093
   3B7 20051023 014857613 807f
   3B8 20051023 014857723 609f
   3B9 20051023 014857723 00ff
   3BA 20051023 014857834 609f
   3BB 20051023 014857864 00ff
   3BC 20051023 014904113 807f
   3BD 20051023 014904113 08f7
   3BE 20051023 014904223 807f
   3BF 20051023 014904223 00f7

Output:

$ xxd -g1 -c12 teste.bin 
0000000: b2 03 4f f4 31 01 b1 b4 e2 00 9f 60  ..O.1......'
000000c: b3 03 4f f4 31 01 b1 b4 e2 00 ff 00  ..O.1.......
0000018: b4 03 4f f4 31 01 1f b5 e2 00 9f 60  ..O.1......'
0000024: b5 03 4f f4 31 01 1f b5 e2 00 ff 00  ..O.1.......
0000030: b6 03 4f f4 31 01 8d b5 e2 00 93 60  ..O.1......'
000003c: b7 03 4f f4 31 01 8d b5 e2 00 7f 80  ..O.1.......
0000048: b8 03 4f f4 31 01 fb b5 e2 00 9f 60  ..O.1......'
0000054: b9 03 4f f4 31 01 fb b5 e2 00 ff 00  ..O.1.......
0000060: ba 03 4f f4 31 01 6a b6 e2 00 9f 60  ..O.1.j....'
000006c: bb 03 4f f4 31 01 88 b6 e2 00 ff 00  ..O.1.......
0000078: bc 03 4f f4 31 01 31 6b e3 00 7f 80  ..O.1.1k....
0000084: bd 03 4f f4 31 01 31 6b e3 00 f7 08  ..O.1.1k....
0000090: be 03 4f f4 31 01 9f 6b e3 00 7f 80  ..O.1..k....
000009c: bf 03 4f f4 31 01 9f 6b e3 00 f7 00  ..O.1..k....
    
16.10.2017 / 19:05