Thread: Birth 2 Parents 1 Son - LinkedList

Better way to get two animals on a linkedlist

I want to make a function that give “birth of a son”, frist i need to choose two parents.

What is the best approach ?

I already got one function working with just 1 parent.

With two parents the only approach is with 2 pointers ?

Code:

struct SPECIES {
    char especie[MAX];
    char idalphanumeric[MAX]; // Alphanumeric ID
    long int nranimais; // Se for 0 não aparece nas espécies existentes.   
    pno listadeanimais;
    pspecies prox;
};


struct ANIMAL {
    char especie[MAX]; // ID Unico
    int id; // id único por animal        
    char nome[MAX]; // Nome do animal
    int peso;
    char nomelocal[MAX]; // Area onde o animal se encontra.
    pno prox; // Ponteiro para proximo no da lista  


    // Ramos familaires 1 pai,1 mae, nFilhos ?
};

Code:

species * birth2animals(species * listoriginal, Area * listareas, int total) {


    species *list = listoriginal;
    species buffer;


    no *aux, *aux2, bufferanimal;


    char progenitor[MAX];
    char progenitor2[MAX];


    strncpy(progenitor, "ZFixe", MAX);
    strncpy(progenitor, "Simba", MAX);


    // 1 Progenitor
    while (list != NULL) {
        aux = list->listadeanimais;
        while (aux != NULL) {
            if (!strcmp(aux->nome, progenitor)) { // Already got the 1 parent

// Do some verification and stuff


            }
            aux = aux->prox;
        }
        list = list->prox;
    }
    return listoriginal;
}

What i got so far.

Code:

species * birth2animals(species * listoriginal, Area * listareas, int total) {


    species *list = listoriginal;
    species *list2 = listoriginal;
    species buffer;


    no *aux, *aux2, bufferanimal;


    char progenitor[MAX];
    char progenitor2[MAX];


    strncpy(progenitor, "Simba", MAX);
    strncpy(progenitor2, "ZFixe", MAX);


    int parentfind = 0;
    int parentfind2 = 0;


    // 1 Progenitor
    while (list != NULL && parentfind == 0) {
        aux = list->listadeanimais;
        while (aux != NULL) {
            if (!strcmp(aux->nome, progenitor)) { // Already got the 1 parent                
                parentfind = 1;
                break;
            }
            aux = aux->prox;
        }
        list = list->prox;
    }
    
    // 2 Progenitor
    while (list2 != NULL && parentfind2 == 0) {
        aux2 = list2->listadeanimais;
        while (aux2 != NULL) {
            if (!strcmp(aux2->nome, progenitor2)) { // Already got the 2 parent
                parentfind2 = 1;
                printf("Name 2 parent %s\n", aux2->nome);


                break;
            }
            aux2 = aux2->prox;
        }
        list2 = list2->prox;
    }


    printf("Name 1 parent %s\n", aux->nome);
    printf("Name 2 parent %s\n", aux2->nome);


    return listoriginal;
}

I got with working.

Sorry for my poor English.

My goal it was to choose two animals , and need to be the same specie to make a baby.

The baby will weight 20% of the total weight of both parents.

Parents stay with 80%.

Code:

species * birth2animals(species * listoriginal, Area * listareas, int total) {


    species *list = listoriginal;
    species *list2 = listoriginal;
    species buffer;


    no *aux, *aux2, bufferanimal;


    int id;
    int id2;
    char son[MAX];


    puts("Insert the ID of the parent");
    scanf("%d", &id);
    getchar();


    puts("Insert the ID of the parent 2");
    scanf("%d", &id2);
    getchar();


    puts("What is the name newborn ?");
    scanf(" %99{^\n]", son);
    getchar();




    int parentfind = 0;
    int parentfind2 = 0;


    // 1 Progenitor
    while (list != NULL && parentfind == 0) {
        aux = list->listadeanimais;
        while (aux != NULL) {
            if (aux->id == id) { // Already got the 1 parent                
                parentfind = 1;
                break;
            }
            aux = aux->prox;
        }
        // Not to count more species after i found the animal
        if (!parentfind) {
            list = list->prox;
        }
    }




    // 2 Progenitor
    while (list2 != NULL && parentfind2 == 0) {
        aux2 = list2->listadeanimais;
        while (aux2 != NULL) {
            if (aux2->id == id2) { // Already got the 2 parent
                parentfind2 = 1;
                break;
            }
            aux2 = aux2->prox;
        }
        // Not to count more species after i found the animal
        if (!parentfind2) {
            list2 = list2->prox;
        }
    }


    // Confirmation that you have 2 parents 
    if (aux == NULL || aux2 == NULL) {
        puts("One or two of the IDs does not exist");
        return listoriginal;
    }


    // Now i got the two parents i need to made some verifications.
    if (!strcmp(list->especie, list2->especie)) {// Same specie i can continue.
        if (!strcmp(aux->nomelocal, aux2->nomelocal)) { // Same area
            bufferanimal.peso = ((aux->peso * 0.20) + (aux2->peso * 0.20));
            strncpy(buffer.especie, list->especie, MAX);
            strncpy(bufferanimal.nomelocal, aux->nomelocal, MAX);
            strncpy(bufferanimal.nome, son, MAX);


            if (insereanimal(listoriginal, buffer, bufferanimal, listareas, total)) {
                aux->peso = aux->peso * 0.80;
                aux2->peso = aux2->peso * 0.80;
                return listoriginal;
            } else {
                puts("birth not possible");
                // TODO - Repor o peso em caso de falha
                return listoriginal;
            }
        } else {
            puts("not in the same local");
            return listoriginal;
        }
    } else {
        puts("Not the same specie");


        return listoriginal;
    }


}