Malloc function doesn´t work properly

Question

I'm getting a problem whit malloc(), don't know what's happening. I'm compiling with MinGW.

Cadena is just a typedef for char *, and leer_dato() is a function that returns a string.

Cadena leer_con_formato(const char * formato)
{
    int i = 0;
    Cadena dato = leer_dato();
    if (strlen(dato) != strlen(formato))
        return NULL;
    char * nuevo_dato = (char *)malloc(strlen(dato)); // Here's the problem
    if (!nuevo_dato)
        return NULL;
    while (dato[i] != '\0')
    {
        switch (formato[i])
    {
        case '0':
            if (!isdigit(dato[i]))
                return NULL;
            nuevo_dato[i] = dato[i];
            i++;
            break;
        case 'C':
            if (!isalpha(dato[i]))
                return NULL;
            nuevo_dato[i] = dato[i];
            i++;
            break;
        case 33 ... 47:
        case 58 ... 64:
        case 91 ... 96:
        case 123 ... 126:
            if (!ispunct(dato[i]))
                return NULL;
            nuevo_dato[i] = dato[i];
            i++;
            break;
        default:
            return NULL;
    }
}
nuevo_dato[i] = NULO;
return nuevo_dato;
}

"I'm getting a problem". What exactly is the "problem"? Don't you think it would make sense to tell us what you are observing? And please provide a [mcve]. We can't debug incomplete code (e.g what does `leer_dato` do?). — kaylum, Oct 16 '16 at 22:12
so what is the problem? What do you expect and what does actually happen? — rfreytag, Oct 16 '16 at 22:13
you should always check the return value of functions so that they are what you expect them to be — AndersK, Oct 17 '16 at 04:49
[Don't cast the result of `malloc` in C](http://stackoverflow.com/q/605845/995714) — phuclv, Oct 17 '16 at 04:52

score 0 · Accepted Answer · answered Oct 16 '16 at 22:30

0

You are not allocating enough memory. If strlen returns for example 5, then that means that your string contains 5 chars + 1 for the terminating 0-byte. So when allocating memory for a string you'll have to do it like so:

char * nuevo_dato = (char*)malloc(sizeof(char) * (strlen(dato) + 1));

The sizeof(char) is optional.

answered Oct 16 '16 at 22:30

jule

96
3

1

Note: Cast `(char*)` not needed either. – chux - Reinstate Monica Oct 16 '16 at 22:32

Malloc function doesn´t work properly

1 Answers1