How to check whether the list is empty or with an element in C

Question

typedef struct {
    int tamanho;
    nodo *inicio;
} lista;

typedef struct {
    lista **vet;
    int tamTabHash;
    int colisoes;
} tabHash; 

tabHash *criaTabelaHash(int tamanho){
    if(tamanho < 1) return NULL;
    tabHash* hash = (tabHash*)malloc(sizeof(tabHash));
    hash->tamTabHash = tamanho;
    hash->colisoes = 0;  
    hash->vet = (lista**)calloc(tamanho, sizeof(lista*));
    return hash;
}

void gerarRelatorioHash(tabHash *th) {
    int i;

    int listocupada = 0, listvazia = 0;

    for (i = 0; i < th->tamTabHash ; i++) {
        if(th->vet[i]->tamanho >= 0) { // essa linha me da falha de segmentacao
    }
}

Answer 1

Let’s see that code:

    hash->vet = (lista**)calloc(tamanho, sizeof(lista*));

The hash->vet will point to a memory area that contains an amount of initially null pointers. This amount is given by tamanho.

Assuming that these pointers are not all overwritten with something else, it means that when accessing if(th->vet[i]->tamanho >= 0), if the th->vet[i] produce a null pointer by ->tamanho the result will be a segmentation failure.

To know what is the best solution, you would need to inform more about the context in which the functions criaTabelaHash and gerarRelatorioHash are used. However, supposing that in fact it is possible that th->vet[i] could be null, a possible solution would be to change the if to look like this:

        if (th->vet[i] && th->vet[i]->tamanho >= 0) {