How to copy vector?

[Immortal Nephi]

I create two vector variables. First variable: data is initialized
with integers. Second variable: copyData is empty. Both have same
sizes.
How do I write correct code to copy all elements from data to
copyData? I don’t have to go through iterator loop to copy each
element.
I want to know that vector data should be auto storage in the global
function’s stack frame. Then, the return type will be vector to copy
all elements.
If result size is different and is more than 100 elements, will
copyData be automatically resized?

vector< int > foo()
{
int numData = 1;
vector< int > result;
result.resize( 100 );

for( vector< int >::iterator it = result.begin();
it != result.end(); ++it )
*it = numData++;

return result;
}

int main()
{
vector< int > copyData;
copyData.resize( 100 );

copyData = foo();

return 0;
}

 

[Alf P. Steinbach]

* Immortal Nephi:

I create two vector variables. First variable: data is initialized
with integers. Second variable: copyData is empty. Both have same
sizes.

'copyData' contains 100 zeroes.

How do I write correct code to copy all elements from data to
copyData?

Just an assignment (for example), as you do below.

I don’t have to go through iterator loop to copy each
element.
I want to know that vector data should be auto storage in the global
function’s stack frame. Then, the return type will be vector to copy
all elements.
If result size is different and is more than 100 elements, will
copyData be automatically resized?
Yes.


vector< int > foo()
{
int numData = 1;
vector< int > result;
result.resize( 100 );

for( vector< int >::iterator it = result.begin();
it != result.end(); ++it )
*it = numData++;

return result;
}

You could write this more easily as

vector<int> foo()
{
vector<int> result;
for( int i = 1; i <= 100; ++i )
{
result.push_back( i );
}
return result;
}

int main()
{
vector< int > copyData;
copyData.resize( 100 );

This size specification is not necessary.

copyData = foo();

return 0;

And this 'return' isn't necessary either. 'main' returns 0 by default.

}

Cheers & hth.,

- Alf

 

[Victor Bazarov]

I create two vector variables. First variable: data is initialized
with integers. Second variable: copyData is empty. Both have same
sizes.

It is customary to describe what *type* your variables have. 'vector'
(or, rather, 'std::vector') is a template, which has to be instantiated
with a type (at least).

How do I write correct code to copy all elements from data to
copyData?

How do you copy a 'double'? You assign it. What about a 'bool'? You
assign it. Why do you think it's different with 'std::vector'? It's
not. You simply assign it.

> I don’t have to go through iterator loop to copy each
element.

No, you don't.

I want to know that vector data should be auto storage in the global
function’s stack frame.
Huh?

> Then, the return type will be vector to copy
all elements.
If result size is different and is more than 100 elements, will
copyData be automatically resized?

Uh... Assignment takes care of everything, if that's what you're asking.

vector< int > foo()
{
int numData = 1;
vector< int > result;
result.resize( 100 );

for( vector< int >::iterator it = result.begin();
it != result.end(); ++it )
*it = numData++;

return result;
}

Looks good. Or, you could simply do

int numData = 1;
vector<int> result;
for (int i = 0; i < 100; i++)
result.push_back(numData++);

(or drop 'numData' altogether and push 'i').

int main()
{
vector< int > copyData;
copyData.resize( 100 );

No need to resize.

copyData = foo();

That's just fine. Or, you could simply initialise it with the return
value of 'foo()':

return 0;
}

V

 

[Immortal Nephi]

It is customary to describe what *type* your variables have.  'vector'
(or, rather, 'std::vector') is a template, which has to be instantiated
with a type (at least).


How do you copy a 'double'?  You assign it.  What about a 'bool'?  You
assign it.  Why do you think it's different with 'std::vector'?  It's
not.  You simply assign it.

 >  I don’t have to go through iterator loop to copy each


No, you don't.


Huh?

 >  Then, the return type will be vector to copy


Uh...  Assignment takes care of everything, if that's what you're asking.






Looks good.  Or, you could simply do

    int numData = 1;
    vector<int> result;
    for (int i = 0; i < 100; i++)
       result.push_back(numData++);

(or drop 'numData' altogether and push 'i').




No need to resize.




That's just fine.  Or, you could simply initialise it with the return
value of 'foo()':

     vector<int> copyData = foo();

Victor,

Thank you for your assistance. I have another question. First
variable has 640 elements. Second variable has only 64 elements.
They look like ten 2D images with 64 area of elements.

I can't use assignment to copy one out of ten images. Do I need to
use memory address with offset? Or I have to use three dimension
array?

vector< int > foo()
{
static const int TwoDimensionArray = 10;
static const int Width = 8;
static const int Height = 8;

vector< int > result;
result.resize( TwoDimensionArray * Width * Height );
// or vector< int > result[ TwoDimensionArray ][ Width ][ Height ];

for( int i = 0; i != result.size(); ++i )
result.push_back( i );

return result; // I want to copy only one image with 64 elements with
memory address
// or return result[5][0][0];
}

 

[Victor Bazarov]

[.. huge unnecessary quote snipped..]

Victor,

Thank you for your assistance. I have another question.

Please don't quote what isn't necessary to continue the conversation.

> First
variable has 640 elements. Second variable has only 64 elements.
They look like ten 2D images with 64 area of elements.

So, the first one *logically* consists of the ten "areas", but
programmatically simply has 640 elements. The "boundary" between
"areas" is simply *assumed*, not necessarily programmed, correct?

I can't use assignment to copy one out of ten images.

Not directly, no. You can extract part using 'std::copy'.

> Do I need to
use memory address with offset?

No. Just a pair of iterators would do:

std::vector<int> all_images(640); // assuming it's filled in
...
std::vector<int> one_image;

int from = 260, to = from + 64; // whatever your range def is

assert(to <= all_images.size()); // or check and don't copy

std::copy(&all_images[from], &all_images[to],
std::back_inserter(one_image));

Or you can construct your 'one_image' from that range:

> Or I have to use three dimension
array?

I don't know. It might help. But it means you need to redesign your
program. It is up to you.

vector< int > foo()
{
static const int TwoDimensionArray = 10;
static const int Width = 8;
static const int Height = 8;

vector< int > result;
result.resize( TwoDimensionArray * Width * Height );
// or vector< int > result[ TwoDimensionArray ][ Width ][ Height ];

for( int i = 0; i != result.size(); ++i )
result.push_back( i );

return result; // I want to copy only one image with 64 elements with
memory address
// or return result[5][0][0];

You need to extract the result before you can copy it. Here you could
simply do

}

V

 

[James Kanze]

* Immortal Nephi:

[...]

This size specification is not necessary.

And this 'return' isn't necessary either. 'main' returns 0 by
default.

It's not necessary according to the formal rules of the
language, but it's necessary for readability and
maintainability. (Not to mention the fact that I've never yet
written a real program where return 0 was acceptable. There's
always some sort of possible error.)

 

[Alf P. Steinbach]

* James Kanze:

* Immortal Nephi:
[...]

int main()
{
vector< int > copyData;
copyData.resize( 100 );

This size specification is not necessary.

And this 'return' isn't necessary either. 'main' returns 0 by
default.

It's not necessary according to the formal rules of the
language, but it's necessary for readability and
maintainability. (Not to mention the fact that I've never yet
written a real program where return 0 was acceptable. There's
always some sort of possible error.)

He he, your second sentence trashes, trumps and generally stomps on the first
one. When the return is misleading, as it is here, and you do mention that, ,
how can it help maintainability instead of hindering maintainability?


Cheers,

- Alf

 

[James Kanze]

* James Kanze:

[...]

He he, your second sentence trashes, trumps and generally
stomps on the first one. When the return is misleading, as it
is here, and you do mention that, , how can it help
maintainability instead of hindering maintainability?

Because we're writing simple examples here; the "return 0"
serves as a placeholder for the "return whatever;" that will be
found in the final program. In general, if a function returns
an int, it should have a return statement; making an exception
for main doesn't buy you anything but confusing.