Problem with assigning 1D array to a 2D array

[bintom]

I am trying to write a program with a function int** One2Two(int *a, int size, int (*A)[10]) that takes the address of a1D int array (A1) , the size of A1 and the address of a 2D int array (A2). The function should copy the values of A1 and store them in A2 as suggested by the pattern below:

If A1 is 1, 2, 3:
then A2 should be 1 0 0
1 2 0
1 2 3

If A1 is 1, 2, 3, 4, 5:
then A2 should be 1 0 0 0 0
1 2 0 0 0
1 2 3 0 0
1 2 3 4 0
1 2 3 4 5

My program is given below:

#include <iostream>

using namespaces std;

int** One2Two(int *a, int size, int (*A)[10])
{ int i, j;

for(i=0; i<size; i++)
{ for(j=0; j<=i; j++)
**(A+i*size+j) = *(a+i);
for(; j<size; j++)
**(A+i*size+j) = *(a+i);
}
}

int main()
{ int a[10], (*A)[10], size, i, j;
a[0]=1; a[1]=2; a[2]=3; a[3]=4; a[4]=5;
size = 5;

One2Two(a, size, A);

for(i=0; i<size; i++)
{ for(j=0; j<size; j++)
cout << **(A+i*size+j) << " ";
cout << "\n";
}
}

However, my program hangs in the inner loop of the function and I have to close the program run. Can anybody help me figure out what the problem is???

Thanks in advance.

T. V. Thomas

 

[Victor Bazarov]

I am trying to write a program with a function int** One2Two(int *a,

int size, int (*A)[10]) that takes the address of a1D int array (A1) ,
the size of A1 and the address of a 2D int array (A2). The function
should copy the values of A1 and store them in A2 as suggested by the
pattern below:

If A1 is 1, 2, 3:
then A2 should be 1 0 0
1 2 0
1 2 3

If A1 is 1, 2, 3, 4, 5:
then A2 should be 1 0 0 0 0
1 2 0 0 0
1 2 3 0 0
1 2 3 4 0
1 2 3 4 5

My program is given below:

#include <iostream>

using namespaces std;

int** One2Two(int *a, int size, int (*A)[10])
{ int i, j;

for(i=0; i<size; i++)
{ for(j=0; j<=i; j++)
**(A+i*size+j) = *(a+i);
for(; j<size; j++)
**(A+i*size+j) = *(a+i);
}
}

int main()
{ int a[10], (*A)[10], size, i, j;

'A' is an array of 10 pointers to int. 'a' is an array of 10 ints.

a[0]=1; a[1]=2; a[2]=3; a[3]=4; a[4]=5;

Here you assign ints to ints. That's perfectly fine.

size = 5;

One2Two(a, size, A);

Here you pass an array of 10 pointers to int (10 pointers) to the
function. That function will dereference those pointers. Where do
those pointers point to?

for(i=0; i<size; i++)
{ for(j=0; j<size; j++)
cout << **(A+i*size+j) << " ";
cout << "\n";
}
}

However, my program hangs in the inner loop of the function and I
have to close the program run. Can anybody help me figure out what
the problem is???

You dereference uninitialized pointers. Your program has undefined
behavior. Anything is allowed to happen. Read up on "nasal demons".

V

 

[bintom]

I initialized A to 0, yet the problem persists. Can somebody else help with a solution? Thanks.

 

[Ike Naar]

int main()
{ int a[10], (*A)[10], size, i, j;

'A' is an array of 10 pointers to int. 'a' is an array of 10 ints.

'A' is a pointer to an array of 10 ints.

An array of 10 pointers to int is declared as 'int *A[10];',
or, with explicit parentheses, 'int *(A[10]);'.

 

[Ike Naar]

I am trying to write a program with a function int** One2Two(int *a,
int size, int (*A)[10]) that takes the address of a 1D int array (A1),
the size of A1 and the address of a 2D int array (A2).

You are confused.

The parameter 'a' (or 'A1') is a pointer-to-int, or, likewise, the
address of an int; it's not the address of a 1D int array (but it
could be the address of the first element of such an array).

'A' (or 'A2') is a pointer to a 1D array (more precisely, a pointer
to an array of 10 ints), or, likewise, the address of such an array;
it's not the address of a 2D array.

Why is the return type of One2Two pointer-to-pointer-to-int when
the function does not return anything?

Have a look chapter 6 of the C language FAQ,

http://c-faq.com/aryptr/index.html

 

[Victor Bazarov]

int main()
{ int a[10], (*A)[10], size, i, j;

'A' is an array of 10 pointers to int. 'a' is an array of 10 ints.

'A' is a pointer to an array of 10 ints.

You're right, of course. I totally missed the parentheses; shows how
much I've been lately using naked pointers and arrays of them.

An array of 10 pointers to int is declared as 'int *A[10];',
or, with explicit parentheses, 'int *(A[10]);'.

V

 

[bintom]

I too am coming to the conclusion that I AM confused. But given the problem I amim to solve, it would be great if u could hold my hand and show me the correct way of coding in this case.

And if u could explain why you chose that style of coding, that would be a bonus.

 

[Luca Risolia]

I am trying to write a program with a function int** One2Two(int *a,
int size, int (*A)[10]) that takes the address of a1D int array (A1)
, the size of A1 and the address of a 2D int array (A2). The function
should copy the values of A1 and store them in A2 as suggested by the
pattern below:

If A1 is 1, 2, 3: then A2 should be 1 0 0 1 2 0 1 2 3

If A1 is 1, 2, 3, 4, 5: then A2 should be 1 0 0 0 0 1 2 0 0 0 1 2 3 0
0 1 2 3 4 0 1 2 3 4 5

Use valarrays for numerical calculations:

#include <iostream>
#include <valarray>

using namespace std;

template<class T>
valarray<T> One2Two(const valarray<T>& v) {
const auto n = v.size();
valarray<T> m(n * n);
for (decltype(v.size()) i = 0; i < n;
m[slice(n * i + i, n - i, n)] = v, ++i);
return m;
}

int main() {
valarray<int> a = {1, 2, 3, 4, 5};
auto m = One2Two(a);
for (const auto& e : m)
cout << e << (((&e - &m[0]) + 1) % a.size() ? ' ' : '\n');
return 0;
}

1 0 0 0 0
1 2 0 0 0
1 2 3 0 0
1 2 3 4 0
1 2 3 4 5

 

[Juha Nieminen]

Although to be more in the spirit of C++, you should use C++
vectors rather than arrays.

There's nothing in the "spirit of C++" that would require using vectors
instead of static arrays. If anything, std::array would be the one to be
used instead of static C arrays.

 

[Luca Risolia]

There's nothing in the "spirit of C++" that would require using vectors
instead of static arrays. If anything, std::array would be the one to be
used instead of static C arrays.

std::valarray would be even more in the "spirit of C++" with regard to
the OP problem.

 

[Juha Nieminen]

std::valarray would be even more in the "spirit of C++" with regard to
the OP problem.

Why is std::valarray better as a substitute for a static array given
that it's a container that allocates the memory dynamically, while
std::array allocates it statically (thus being more efficient for
fixed-sized arrays)?

 

[Luca Risolia]

Why is std::valarray better as a substitute for a static array given
that it's a container that allocates the memory dynamically, while
std::array allocates it statically (thus being more efficient for
fixed-sized arrays)?

...because efficiency was not the concern of the OP. He asked how to make
his program correct. The easy way to write a program correctly is to
choose the right data structure: as opposite to std::static_array,
std::valarray offers the right interface to compute a subset of its
elements, via slices, to produce the kind of matrix expected by the OP.
Even in the case efficiency is a matter, std::valarray has been designed
to allow compiler/implementation optimizations during its operations,
and, depending on the kind/complexity of the numerical algorithms, that
might be a benefit if most of the time is not spent for memory allocations.