trying to pass an array of structure with the index.

[momotaro]

hi every one im trying to pass to a function an array with its index:

//this is my struct
typedef struct{
    char word[20];
    int anagrams;
}Word;
 
//this is the prototype of the function:
int find_anagrams(char **, char **);
//code...
if(find_anagrams(word_arr[i].word, word_arr[j].word));
//code...
//the definition(where there is the problem!!!!)
int find_anagrams(char *word1, char *word2)
{
    word_arr[i]. // dont work i can not access my struct!!!
}

[ilikepython]

hi every one im trying to pass to a function an array with its index:

//this is my struct
typedef struct{
    char word[20];
    int anagrams;
}Word;
 
//this is the prototype of the function:
int find_anagrams(char **, char **);
//code...
if(find_anagrams(word_arr[i].word, word_arr[j].word));
//code...
//the definition(where there is the problem!!!!)
int find_anagrams(char *word1, char *word2)
{
    word_arr[i]. // dont work i can not access my struct!!!
}

Unless word_arr is global, you won't be able to access it through the function. Do you want to pass the struct as an arguement instead of just the char *?

if(find_anagrams(word_arr[i], word_arr[j])){}
 

Please explain what you are trying to do.

[weaknessforcats]

This code:

//this is my struct
typedef struct{
char word[20];
int anagrams;
}Word;

//this is the prototype of the function:
int find_anagrams(char **, char **);
//code...
if(find_anagrams(word_arr[i].word, word_arr[j].word));
//code...
//the definition(where there is the problem!!!!)
int find_anagrams(char *word1, char *word2)
{
word_arr[i]. // dont work i can not access my struct!!!
}

You do not pass a char*. You pass a Word or a Word*:

int find_anagrams(Word *word1, Word* word2)
{
     //in here you use word1->word to use the array
}
 

[Peterwkc]

int find_anagrams(struct nameofyourstruct word[], int size);

The previous notation is much more efficient. I just recommended another approach to do so without using pointer.

[weaknessforcats]

int find_anagrams(struct nameofyourstruct word[], int size);

The previous notation is much more efficient. I just recommended another approach to do so without using pointer.

This still uses a pointer.

struct nameofyourstruct word[]

is the same as

struct nameofyourstruct* word

[Savage]

This still uses a pointer.

struct nameofyourstruct word[]

is the same as

struct nameofyourstruct* word

There are some slight diferances,but it can be sayed that they are equal..

Savage

[weaknessforcats]

There are some slight diferances,but it can be sayed that they are equal..

What are the differences?

[Savage]

What are the differences?

You don't need to allocate memory for [],but you do need to allocate it for *.

Savage

[weaknessforcats]

You don't need to allocate memory for [],but you do need to allocate it for *.

char arr[];
char* arr1;
 

This compiles and no memory was allocated. Of course to use these things tou wiull need to allocate memory. I'm just saying that char arr[] and char* arr1 are equivalent since the name fo the array is the address of element 0. Therefore arr and arr1 are both addresses of char, ot char*

The most you might say is that char[] involves an array but the compiler has no way to check this. All this is, really, is initialization syntax where you have the compiler allocate an array based on the number of enumerated elements.

[JosAH]

There are some slight diferances,but it can be sayed that they are equal..

Savage

There are no difference in the value context it is used here (read: as a parameter
to a function) a T[] is the same as a T*

kind regards,

Jos