double pointer on a array and element of a table that points to a table

Question

I would like to know:

how to point the elements of a table to an array of elements?
How to make the operations on?

Here is a code that I wrote:

#include <stdio.h>
#include <stdlib.h>
#define N 4

int main(int argc, char const *argv[]) {
    int i;
    int *t[N];
    int tab1[N] = {1, 2, 3, 4},
        tab2[N] = {5, 6, 7, 8},
        tab3[N] = {9, 10,11, 12},
        tab4[N] = {13, 14, 15, 16};

    for (i = 0 ; i < N; i++) {
        *t[i] = tab1[i]; 
    }

    for (i = 0; i < N; i++) {
        printf("%d\n", *t[0]);
    }
}

When I run it, nothing is happening.

fogang@les-tatates:~/tp_304$ gcc -o test test.c
fogang@les-tatates:~/tp_304$ test
fogang@les-tatates:~/tp_304$

i want to implement this !

Just curious - why do you want to do that? Why not use `int t[N][N];` — Support Ukraine, Jun 15 '18 at 12:01
I have a non-deterministic automaton. To transform it into a deterministic automaton, I will have states that correspond to a list of states from the transition table. Example: State: [0, 1, 2, 3] state 0 -> [0, 1] state 1 ---> [2, 3] — K.J Fogang Fokoa, Jun 15 '18 at 12:16
Please [edit] your question and reveal what output you expect. — Jabberwocky, Jun 15 '18 at 12:18
@fogangfokoa Well, if you think you have a good reason then .... 1) The second loop has to be deleted and replaced by `t[0] = tab1; t[1] = tab2; ...` and to access elements use the form `t[0][0]`, `t[0][1]`, and so on The last loop could be `printf("%d\n", t[2][i]);` — Support Ukraine, Jun 15 '18 at 12:21
`*t[i] = tab1[i];` You did not allocate any memory for the pointers stored in `t`. Dereferencing and assigning something to it is UB. — Gerhardh, Jun 15 '18 at 12:22
tab = [1, 2, 3, 4] I want element "1" to point to a tab (eg t = [2,3]). so that to browse t I go through tab [0]. Do you understand me? Sorry if my english is bad! — K.J Fogang Fokoa, Jun 15 '18 at 12:28
@fogangfokoa no I don't understand. Just [edit] your question and tell us what output you expect. — Jabberwocky, Jun 15 '18 at 12:30
@fogangfokoa please tell us what exactly __should__ happen when you run the program you posted in your question. What do you think your program should print? — Jabberwocky, Jun 15 '18 at 12:40
@fogangfokoa still not clear, please simply do as I asked in my previous comment. — Jabberwocky, Jun 15 '18 at 12:59
Okay. i want that : t[0] contains tab1. Do you understand me @Jabberwocky — K.J Fogang Fokoa, Jun 15 '18 at 13:04
@fogangfokoa OK, I'm starting to understand, but still please tell me what you think your program should print — Jabberwocky, Jun 15 '18 at 13:05
`int t[N][N];` doesn't rule out non-deterministic data, unless the various dimensions change variably, in which case it is an array no longer. The picture on the hand-drawing shows the memory layout of a `int t[N][N];`. — Lundin, Jun 15 '18 at 13:20

score 1 · Answer 1 · answered Jun 17 '18 at 09:21

When I run it, nothing is happening.

fogang@les-tatates:~/tp_304$ gcc -o test test.c
fogang@les-tatates:~/tp_304$ test
fogang@les-tatates:~/tp_304$

test is a shell build-in.

To run a program called test being located inside the current work directory do:

fogang@les-tatates:~/tp_304$ ./test

Jabberwocky · Answer 2 · 2018-06-15T13:20:49.107

From the various comments you posted I guess (and I'm really guessing here) that you maybe want this:

#include <stdio.h>
#include <stdlib.h>
#define N 4

int main(int argc, char const *argv[]) {
  int *t[N];
  int tab1[N] = { 1, 2, 3, 4 },
    tab2[N] = { 5, 6, 7, 8 },
    tab3[N] = { 9, 10,11, 12 },
    tab4[N] = { 13, 14, 15, 16 };

  t[0] = tab1;
  t[1] = tab2;
  t[2] = tab3;
  t[3] = tab4;

  for (int j = 0; j < 4; j++)
  {
    for (int i = 0; i < N; i++) {
      printf("%d ", t[j][i]);
    }

    printf("\n");
  }
}

The for loop can also be written like this:

  for (int j = 0; j < 4; j++)
  {
    int *temp = t[j];
    for (int i = 0; i < N; i++) {
      printf("%d ", temp[i]);
    }

    printf("\n");
  }

Or even simpler:

int main(int argc, char const *argv[]) {
  int t[N][N] =
  {
    { 1, 2, 3, 4 },
    { 5, 6, 7, 8 },
    { 9, 10,11, 12 },
    { 13, 14, 15, 16 }
  };

  for (int j = 0; j < 4; j++)
  {
    int *temp = t[j];
    for (int i = 0; i < N; i++) {
      printf("%d ", temp[i]);
    }

    printf("\n");
  }
}

// printf("%d ", temp[i]); Don't work same for printf("%d ", *temp[i]); — K.J Fogang Fokoa, Jun 15 '18 at 13:53
@fogangfokoa sorry, your comment is unclear (as the whole question BTW). — Jabberwocky, Jun 15 '18 at 13:55
that's what i do, we understand already, The preblem is the printing data of the array. — K.J Fogang Fokoa, Jun 15 '18 at 14:03

Serge Ballesta · Answer 3 · 2018-06-15T14:08:13.230

0

As a complement to @Jabberwocky's answer, you can even do:

#include <stdio.h>
#include <stdlib.h>
#define N 4

int main(int argc, char const *argv[]) {
    int i;
    int tab1[N] = {1, 2, 3, 4},
        tab2[N] = {5, 6, 7, 8},
        tab3[N] = {9, 10,11, 12},
        tab4[N] = {13, 14, 15, 16};
    int *t[N] = {tab1, tab2, tab3, tab4};

  for (int j = 0; j < N; j++)
  {
    for (int i = 0; i < N; i++) {
      printf("%d ", t[i][j]);
    }

    printf("\n");
  }
}

edited Jun 15 '18 at 14:08

answered Jun 15 '18 at 13:16

Serge Ballesta

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1

this line : printf("%d ", t[i][j]); Don't works. It's exactly what i want to do ! – K.J Fogang Fokoa Jun 15 '18 at 13:59
2

Could you tell what *don't work* means here? **I** obtain values column wise here. What do **you** want? – Serge Ballesta Jun 15 '18 at 14:10
When i compile and run the .out, i do not get the values. – K.J Fogang Fokoa Jun 15 '18 at 14:42

double pointer on a array and element of a table that points to a table

3 Answers3