integer pointer to an array[][][]

Code:

---

int sarray[2][10][3];

int ***s;
s = sarray;      // warning: assignment from incompatible pointer type

sarray[0][0][0] = 2;

printf("Nb in array is %d",s[0][0][0]);

---

This gives me Segmentation fault (core dumped)



So how do I make this work?I
I need an integer pointer to an array[][][].

Thanks

Well, you can't just make a three-level pointer to an array. The way a three-level pointer works it that it contains an address of an address of an address, and when you do s[0][0][0] the compiler will read what s points to, then reads the address at that place, then read the next level address. Since you have only assigned the first level with the address of the array, you will not be able to do all three levels of indirection. When you have a 3D array, you would need a 2D set of pointers to get to each location in the array.

--
Mats

Well, it gives you a warning which explains the problem. The correct is:

Code:

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int (*s)[2][10][3];

---

the parenthesis are not optional.

If you want more information we can explain (again) the differences

edit: Have some doubts:) Is the above correct?

oh

So let me ask another question.

How can I make a global array of int? if it is defined like this:

Code:

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int[2][3][argc] a;

int main(int argc,char** argv)

{

}

void fct()
{
  //fct uses a[][][];
}

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That's because int ***s is a pointer to int while sarray is not, and your s pointer should look like

Code:

---

int (*s)[10][3];
s = sarray

---

Code:

---

int (*s)[2][10][3];

---

Thanks,
but could you explain it?
I would really appreciate it.

Code:

---

int (*s)[2][10][3];

---

by the way it works.

Quote:

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Originally Posted by jordanguyoflove

Code:

---

int (*s)[2][10][3];

---

Thanks,
but could you explain it?
I would really appreciate it.

---

It's a pointer to a 3D array of integers (with the sizes 2, 10 and 3 elements for the respective dimensions). The parenthesis is needed to avoid the compiler making it a 3D array of pointers.

The dimensions must match the original array in all but the last (right-most) dimension, as the size of each dimension determines how much the compiler needs to add on for each dimension when calculating the actual location of s[z][y][x].

--
Mats

Code:

---

int (*s)[10][3];
s = sarray

---

although it works i still get a warning: warning: assignment from incompatible pointer type

Quote:

---

It's a pointer to a 3D array of integers (with the sizes 2, 10 and 3 elements for the respective dimensions). The parenthesis is needed to avoid the compiler making it a 3D array of pointers.

The dimensions must match the original array in all but the last (right-most) dimension, as the size of each dimension determines how much the compiler needs to add on for each dimension when calculating the actual location of s[z][y][x].

---

Thanks

Quote:

---

Originally Posted by jordanguyoflove

Code:

---

int (*s)[10][3];
s = sarray

---

although it works i still get a warning: warning: assignment from incompatible pointer type

---

Because your array sarray is one dimension more than the pointer s. You need the same number of dimensions.

--
Mats

Quote:

---

Originally Posted by jordanguyoflove

Code:

---

int (*s)[10][3];
s = sarray

---

although it works i still get a warning: warning: assignment from incompatible pointer type

---

Post the dimensions of sarray; I thought that it was a 3D array; while int (*s)[2][10][3] is a pointer to a 4D array.

Quote:

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Originally Posted by itCbitC

Post the dimensions of sarray; I thought that it was a 3D array; while int (*s)[2][10][3] is a pointer to a 4D array.

---

No, it's a pointer to a 3D array. Of course, if the pointer itself points to multiple elements of such arrays, you have essentially resulted in a 4D array. But as such, the pointer is a pointer to a 3D array.

sarray IS posted above.

--
Mats

Quote:

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Originally Posted by matsp

No, it's a pointer to a 3D array. Of course, if the pointer itself points to multiple elements of such arrays, you have essentially resulted in a 4D array. But as such, the pointer is a pointer to a 3D array.

sarray IS posted above.

--
Mats

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Oops! got mixed up with the array dimensions, I meant if s is int (*s)[2][10][3] then s is a pointer to a 3D array while sarray is a pointer to a 2D array if it is declared as int sarray[2][10[3] then I am not sure how the OP was able to get s = sarray without the compiler complaining.

I am sorry I made a mistake while writing the post it's "int (*s)[2][10][3];" that is working.

but i still get the msg warning: assignment from incompatible pointer type