Double pointer & Two Dimensional Array

Why Below Program Produces Segmentation error

Code:

---

void print_arr(float **p)
{
printf(" 0 %f 1 %f 2 %f\n",p[0][0],p[0][1],p[0][2]);
}

void main()
{
float arr[2][3] = {{0,1,2},{3,4,5}};
float **fl_arr;
fl_arr = (float *)arr;
print_arr(fl_arr);
fl_arr++;
print_arr(fl_arr);
}

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A pointer to a pointer isn't the same as a two dimensional array. Compile with warnings on.

Read: Arrays and Pointers


Quzah.

fl_arr is a pointer to a pointer, but it only has space for one address.

Segmentation fault - Wikipedia, the free encyclopedia

You are referencing memory that does not exist when you refer to members of p in print_arr(). fl_arr does not have members, so p has an address, but p[0] does not.

arr is not a pointer. It is an array. You could refer to the address of a:

Code:

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fl_arr = &arr;

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In which case the compiler will warn you of an incompatible type.

fl_arr must itself be an array in order to be used as one.

Code:

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float *fl_arr[2] = {arr[0],arr[1]};

---

Now it is an array of two pointers, which the higher dimensions of multi-dimensional arrays can be considered pointers and arrays of pointers.

You cannot do this with a pointer array tho:

Code:

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fl_arr++;

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Now it has members you have to refer to them. fl_arr[0]++ is allowed, but will not have the effect I think you intend.