How to pass a char vector * (*) to a function?

Question

Guys, I have this code:

#include <stdio.h>

void exibir_nomes (char *nome)
{
    for (int i=0; i<5; i++)
        printf("Nome[%d] = %s\n", i, *nome);
}

int main ()
{
    char *nomes[5]={"Maria","Joaquina","Leticia","Edivalda","Romelia"};

    exibir_nomes(*nomes);
       
    return 0;
}

I would like, to pass this vector names to the function display names, I have tried some ways and they all go wrong, follows below the attempts:

exibir_nomes(nomes);

Exit:

teste_pilha.c: In function 'main':
teste_pilha.c:13:18: warning: passing argument 1 of 'exibir_nomes' from incompatible pointer type [-Wincompatible-pointer-types]
   13 |     exibir_nomes(nomes);
      |                  ^~~~~
      |                  |
      |                  char **
teste_pilha.c:3:26: note: expected 'char *' but argument is of type 'char **'
    3 | void exibir_nomes (char *nome)
      |                    ~~~~~~^~~~

And I think I get it, he says that my vector "names" is a pointer pointer, probably because it’s an array of names in the background right, so I changed the function to receive a char parameter that was a pointer:

void exibir_nomes (char **nome)
{
    for (int i=0; i<5; i++)
        printf("Nome[%d] = %s\n", i, **nome);
}

And he compiles, but then that’s the way out:

Nome[0] =

I really don’t know how to do this manipulation, someone can give me a light?

Answer 1

So, first you need to understand that the char variable type is responsible for storing one character at a time. And another, when working with vector remember that the first position is always the index [0]. See, you defined as follows:

char *nomes[5]={"Maria","Joaquina","Leticia","Edivalda","Romelia"};

At first you thought that since there are 5 names, it would be necessary to create a vector of 5 positions. It is wrong to interpret this way, you should put 4 positions, because as the index [0] counts, then there are five elements: [0]; [1]; [2]; [3]; [4]. Another point worth mentioning is that vectors can be treated as pointers in C, because you realize that by creating a vector of n positions it will point to the beginning of the defined sequence, do you agree? Therefore, the asterisk symbol is not required in this case. As I mentioned, the char variable stores one character at a time, to be correct what you did should be like this:

char nomes[4]={"M","J","L","E","R"};

Or you could define a name only:

char nomes[5]={"Maria"}; //Strings sempre terminam com o /0 no final, o nome Maria tem 5 letras + o /0 no final, por isso 6 elementos. 

However, let’s get to the point, I know the point of the question is to show the generic names defined, so let’s work on that statement. Look, to store these five names you have to create a two-dimensional array of characters, like this:

char nomes[5][10]={"Maria","Joaquina","Leticia","Edivalda","Romelia"}; //cinco nomes cada um com 9 caracteres (lembre-se do /0 no final); 

From this, just pass the variable name as parameter to function displa_names. The error will continue because as it comes to a matrix you can replace:

that:

void exibir_nomes (char *nome)

therefore:

void exibir_nomes (char nome[5][10])

Your go inside the function is also incorrect, see:

  for (int i=0; i<5; i++)
        printf("Nome[%d] = %s\n", i, *nome);
}

You won’t display anything with this, if you want to work with pointers remember what I said that vectors and pointers behave the same way. Therefore, the correct one would present each of the positions on the screen using the indexes:

for (int i=0; i<5; i++)
            printf("Nome[%d] = %s\n", i, nome[i]);
    }

Still the program will not execute what you wanted from the beginning, I adjusted your code and it got like this:

#include <stdio.h>

void exibir_nomes (char nome[5][10])
{
    for (int i=0; i<5; i++)
        {
         printf("Nome[%d] = %s\n", i, nome[i]);
        }
}

int main ()
{
    char nomes[5][10]={"Maria","Joaquina","Leticia","Edivalda","Romelia"};
    exibir_nomes(nomes);
}

This way the names are presented on the screen, however if you intend to work strictly with pointers to solve this issue, I believe it would be more challenging and even more complex. Furthermore, I believe that some tips here will serve as a basis for you to discuss this and other exercises in the same style.