如何在结构内创建指针

Question

I have the following structs:

typedef struct bucket {
    unsigned int contador; 
    unsigned int * valor;
} Bucket;

typedef struct indice {
    unsigned int bs;            
    unsigned int valor;      
    unsigned int capacidade; 
    Bucket * bucket;
} Indice;

typedef struct tabela {
    unsigned int bs; 
    Indice * indice;
} Tabela;

And I want to do something like this:

tabela->indice[2].bucket = &tabela->indice[0].bucket;

But I get segmentation fault.

How can I get the tabela->indice[0].bucket address and associate to tabela->indice[2].bucket .

Thanks!

Answer 1

I'm probably going to get neg repped agian for attempting to answer ac question, but here goes nothing

have you tried getting rid of the & ?

tabela->indice[2].bucket = tabela->indice[0].bucket;

Answer 2

You have to initialize your pointers to point to something valid. Simply creating an instance of your struct does not do that for you. For example:

Tabela t = {};  /* t is a valid, zero-initialized object, but indice is not valid */

t.indice = malloc(sizeof(Indice) * some_count);
/* now t.indice is valid */

for(int i = 0; i < some_count; ++i)
    t.indice[i].bucket = malloc(sizeof(Bucket));

/* now t.indice[0-(some_count-1)].bucket are all valid */

---

By the way, your pointer copy is incorrect.

tabela->indice[2].bucket = tabela->indice[0].bucket;
/* assuming indices 0 and 2 are valid, this now works (but leaks memory)
   note that I removed the &.  It is incorrect and produces
   a Bucket** because t.indice[n].bucket is already a pointer */

However, that results in a memory leak. It's hard for me to figure out what you are actually trying to accomplish here.