I have a segmentationfault at the line :
cout << b[0][0];
Could someone tell me what should I do to fix my code?
#include <iostream>
using namespace std;
int** gettab(int tab[][2]){
return (int**)tab;
}
int main() {
int a[4][2] = {{0, 0}, {1, 0}, {2, 0}, {2, 1}};
int ** b = gettab(a);
cout << b[0][0];
return 0;
}
A 2-dimensional array is not the same thing as an array of pointers, which is how int** is interpreted. Change the return type of gettab.
int* gettab(int tab[][2]){
return &tab[0][0];
}
int main() {
int a[4][2] = {{0, 0}, {1, 0}, {2, 0}, {2, 1}};
int* b = gettab(a);
cout << b[0]; // b[row_index * num_cols + col_index]
cout << b[1 * 2 + 0]; // the 1 from {1, 0}
}
Or:
int (*gettab(int tab[][2]))[2] {
return tab;
}
// or:
template<class T> struct identity { typedef T type; };
identity<int(*)[2]>::type gettab(int tab[][2]) {
return tab;
}
int main() {
int a[4][2] = {{0, 0}, {1, 0}, {2, 0}, {2, 1}};
int (*b)[2] = gettab(a);
cout << b[0][0];
}
Being c++, rather than c, there are much better ways of handling arrays of all sorts, and passing them around.
The type of tab without square brackets is not actually int **. It is actually int (*)[2]. When you apply two [] operators to the resulting pointer, you end up dereferencing the first value in your array, 0, as a NULL pointer. Try this instead:
#include <iostream>
using namespace std;
typedef int (*foo)[2];
foo gettab(int tab[][2]){
return tab;
}
int main() {
int a[4][2] = {{0, 0}, {1, 0}, {2, 0}, {2, 1}};
foo b = gettab(a);
cout << b[0][0];
return 0;
}
Your seg fault us because you pass in an "int*" effectively. A 2D array is not a double pointer ...
You are best off using a pointer that is "x*y" in size and addressing it without the 2 dimensions ... the code will end up the same anyway as the compiler will just generate the same code you would have to write more explicitly anyway :)
a 2 diminsional array isn't the same thing as an array of pointers. a 2 dimensional array is just a pointer to a hunk of memory that you have told the compiler to let you access as a 2 dimensional array
int* gettab(int tab[][2]) {
return (int*)tab;
}
int main() {
int a[4][2] = {{0, 0}, {1, 0}, {2, 0}, {2, 1}};
int* b = gettab(a);
cout << b[0 + 2*0];
return 0;
}
will do what you want. But I wonder if you really need to be trying to return a 2 dimensional array from a function in the first place. Perhaps a less made-up example if what you are trying to do would be helpful?
edit: fixed missing *2 in the calculation [0 + (sizeof(int)*2)*0]. edit again: well that was dumb. the multiplication of the column size of 2 by the size of an int is automatic in this case. sizeof(int) removed.
sizeof(int) doing in b[0 + sizeof(int)*0]?