Put and take whole file?

Question

I have two whole variables a rp and rc, and I have to record them in a file as a way to save their value, and then take them and put them into variables like inteiro to use in my program.

To write to a file, I’m just using an escape character to let me know if the number is over or not. This character is the dot and comma ;.

I’m trying with the following code, where arq is my file:

To pick up value:

cc = 1;
while(! feof(arq))
{
    get = fgetc(arq);
    if(get != ';')
    {
        if(cc == 1)
        {
            rc = (int)get;
            cc ++;
        }
        else
            rp = (int)get;
        }
}
fclose(arq);

To place:

fputc(rc, arq);
fputc(';', arq);
fputc(rp, arq);
fputc(';', arq);

I know where the error is, but I can’t fix it. It’s taking the value ASCII character - when I put an integer it takes the value ASCII and not the number, and at the time of putting happens the same, it converts the whole to char and uses the ASCII.

Another problem is that I’m taking into consideration only numbers of a digit; how do I take larger numbers into account?

Answer 1

The practical solution

You should know a little bit about file manipulation and IO, and their syntax in C. To read, you should just check where appear the ';', and at these intervals use fscanf with integer syntax. To write, use fprintf with string syntax, for each printed number, print the break character.

The manual solution

The first step is to know the size of the number. For this you iterate character by character until you find your break character (which in this case is ';'). For this, you will pick up characters (using fgetc), one by one and compare with the break character. If it is not, you should remove the value equivalent to the character '0'. This is what you forgot to do. And then add to a fifo structure (page in English). When you find the break character, you instantiate it and do the same thing (this kind of thing is easier to do with object orientation). After you have all these queues, you use a concept of our numerical system. For example, 21 = 2*10^1 + 1*10^0, 432 = 4*10^2 + 3*10^1 + 2*10^0. If you notice that the values are multiplied by powers of 10, where the largest digit is multiplied by 10 (position - 1). Then, you use this logic:

enquanto o arquivo não estiver no fim:
    caractere = arquivo.le_caractere()
    enquanto caractere não for ';'
        queue.push(caractere - '0')
        caractere = arquivo.le_caractere()
    lista_de_queues += queue

para cada queue em lista_de_queues:
    enquanto houver itens no queue:
        numero += queue.pop() * 10^queue.size()
    lista_de_numeros += numero

Note that Queue.size() is necessarily the old size - 1, because by precedence, pop is called first. That way you can read the numbers.

Fortunately, to save is even easier! You must do the opposite operation, to extract a digit, and then save it as a char, added to '0'. The number of digits in a number is its logarithm in base 10. It transforms this into integer, and goes on to do the following using a lifo structure:

for(unsigned i = 1; i <= digitos; i++)
    stack.push( int( (numero % 10^i) / 10^(i-1) ) )
// nesse ponto, estarão na pilha, na ordem inversa. como é uma estrutura LIFO,
// ele vai nos devolver os dígitos na ordem correta
enquanto houver itens na pilha:
arquivo.escreve_caractere(stack.pop() + '0') 

Negative numbers, just check if the first character is '-'. If it is, multiply the final number by -1.