Why does using a static allocated 2D table on a function with pointer** results a segmantation fault

Question

#include <stdio.h>

void init_tableau2D(int **t ,int ligne ,int colonne){                                                                                      
int i,j;       

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

int main()
{
int tab[3][2]={{5,8},{11,6},{37,45}}; 
/* Here I have allocated staticly a 2D table*/
init_tableau2D(tab,3,2); 

return 0;
}

`tab` does not decay to `int **`, it decays to `int (*)[2]`. — Osiris, Oct 15 '18 at 12:32
Please make sure you have enabled warnings on your compiler. You should see warning or error on mismatching types. — user694733, Oct 15 '18 at 12:33
You are probably interested in https://stackoverflow.com/questions/3911400/how-to-pass-2d-array-matrix-in-a-function-in-c — hellow, Oct 15 '18 at 12:37

score 1 · Accepted Answer · answered Oct 15 '18 at 12:37

1

Arrays decay to pointers if you pass them as parameter, but tab in your example decays to int (*)[2] and not to int **. To make it work you need to change the function definition to:

void init_tableau2D(int ligne ,int colonne, int t[ligne][colonne])

answered Oct 15 '18 at 12:37

Osiris

2,783
9
17

Can you explain more about why does the tab has been casted to an **int(\*)[2]** and not to an **int\*\*** ? – Mr.Jupiter Oct 15 '18 at 12:52
@RookedBishop In most cases arrays decay to pointers to its first element, like if you pass them as parameter. `int [n]` will become `int *`. But if you have an array of arrays, the first element is an array and therefore it decays to a pointer to an array. So `int [3][2]` will decay to `int (*)[2]`. – Osiris Oct 15 '18 at 12:56
@RookedBishop [This](https://stackoverflow.com/questions/4470950/why-cant-we-use-double-pointer-to-represent-two-dimensional-arrays) is maybe useful too. No problem. – Osiris Oct 15 '18 at 13:01

Why does using a static allocated 2D table on a function with pointer** results a segmantation fault

1 Answers1