How do you copy the contents of an array to a std::vector in C++ without looping?
If you can construct the vector after you've gotten the array and array size, you can just say:
std::vector<ValueType> vec(a, a + n);
...assuming a
is your array and n
is the number of elements it contains. Otherwise, std::copy()
w/resize()
will do the trick.
I'd stay away from memcpy()
unless you can be sure that the values are plain-old data (POD) types.
Also, worth noting that none of these really avoids the for loop--it's just a question of whether you have to see it in your code or not. O(n) runtime performance is unavoidable for copying the values.
Finally, note that C-style arrays are perfectly valid containers for most STL algorithms--the raw pointer is equivalent to begin()
, and (ptr + n
) is equivalent to end()
.
How do I print out the contents of a vector?
If you have a C++11 compiler, I would suggest using a range-based for-loop (see below); or else use an iterator. But you have several options, all of which I will explain in what follows.
Range-based for-loop (C++11)
In C++11 (and later) you can use the new range-based for-loop, which looks like this:
std::vector<char> path;
// ...
for (char i: path)
std::cout << i << ' ';
The type char
in the for-loop statement should be the type of the elements of the vector path
and not an integer indexing type. In other words, since path
is of type std::vector<char>
, the type that should appear in the range-based for-loop is char
. However, you will likely often see the explicit type replaced with the auto
placeholder type:
for (auto i: path)
std::cout << i << ' ';
Regardless of whether you use the explicit type or the auto
keyword, the object i
has a value that is a copy of the actual item in the path
object. Thus, all changes to i
in the loop are not preserved in path
itself:
std::vector<char> path{'a', 'b', 'c'};
for (auto i: path) {
i = '_'; // 'i' is a copy of the element in 'path', so although
// we can change 'i' here perfectly fine, the elements
// of 'path' have not changed
std::cout << i << ' '; // will print: "_ _ _"
}
for (auto i: path) {
std::cout << i << ' '; // will print: "a b c"
}
If you would like to proscribe being able to change this copied value of i
in the for-loop as well, you can force the type of i
to be const char
like this:
for (const auto i: path) {
i = '_'; // this will now produce a compiler error
std::cout << i << ' ';
}
If you would like to modify the items in path
so that those changes persist in path
outside of the for-loop, then you can use a reference like so:
for (auto& i: path) {
i = '_'; // changes to 'i' will now also change the
// element in 'path' itself to that value
std::cout << i << ' ';
}
and even if you don't want to modify path
, if the copying of objects is expensive you should use a const reference instead of copying by value:
for (const auto& i: path)
std::cout << i << ' ';
Iterators
Before C++11 the canonical solution would have been to use an iterator, and that is still perfectly acceptable. They are used as follows:
std::vector<char> path;
// ...
for (std::vector<char>::const_iterator i = path.begin(); i != path.end(); ++i)
std::cout << *i << ' ';
If you want to modify the vector's contents in the for-loop, then use iterator
rather than const_iterator
.
Supplement: typedef / type alias (C++11) / auto (C++11)
This is not another solution, but a supplement to the above iterator
solution. If you are using the C++11 standard (or later), then you can use the auto
keyword to help the readability:
for (auto i = path.begin(); i != path.end(); ++i)
std::cout << *i << ' ';
Here the type of i
will be non-const (i.e., the compiler will use std::vector<char>::iterator
as the type of i
). This is because we called the begin
method, so the compiler deduced the type for i
from that. If we call the cbegin
method instead ("c" for const), then i
will be a std::vector<char>::const_iterator
:
for (auto i = path.cbegin(); i != path.cend(); ++i) {
*i = '_'; // will produce a compiler error
std::cout << *i << ' ';
}
If you're not comfortable with the compiler deducing types, then in C++11 you can use a type alias to avoid having to type the vector out all the time (a good habit to get into):
using Path = std::vector<char>; // C++11 onwards only
Path path; // 'Path' is an alias for std::vector<char>
// ...
for (Path::const_iterator i = path.begin(); i != path.end(); ++i)
std::cout << *i << ' ';
If you do not have access to a C++11 compiler (or don't like the type alias syntax for whatever reason), then you can use the more traditional typedef
:
typedef std::vector<char> Path; // 'Path' now a synonym for std::vector<char>
Path path;
// ...
for (Path::const_iterator i = path.begin(); i != path.end(); ++i)
std::cout << *i << ' ';
Side note:
At this point, you may or may not have come across iterators before, and you may or may not have heard that iterators are what you are "supposed" to use, and may be wondering why. The answer is not easy to appreciate, but, in brief, the idea is that iterators are an abstraction that shield you from the details of the operation.
It is convenient to have an object (the iterator) that does the operation you want (like sequential access) rather than you writing the details yourself (the "details" being the code that does the actual accessing of the elements of the vector). You should notice that in the for-loop you are only ever asking the iterator to return you a value (*i
, where i
is the iterator) -- you are never interacting with path
directly itself. The logic goes like this: you create an iterator and give it the object you want to loop over (iterator i = path.begin()
), and then all you do is ask the iterator to get the next value for you (*i
); you never had to worry exactly how the iterator did that -- that's its business, not yours.
OK, but what's the point? Well, imagine if getting a value wasn't simple. What if it involves a bit of work? You don't need to worry, because the iterator has handled that for you -- it sorts out the details, all you need to do is ask it for a value. Additionally, what if you change the container from std::vector
to something else? In theory, your code doesn't change even if the details of how accessing elements in the new container does: remember, the iterator sorts all the details out for you behind the scenes, so you don't need to change your code at all -- you just ask the iterator for the next value in the container, same as before.
So, whilst this may seem like confusing overkill for looping through a vector, there are good reasons behind the concept of iterators and so you might as well get used to using them.
Indexing
You can also use a integer type to index through the elements of the vector in the for-loop explicitly:
for (int i=0; i<path.size(); ++i)
std::cout << path[i] << ' ';
If you are going to do this, it's better to use the container's member types, if they are available and appropriate. std::vector
has a member type called size_type
for this job: it is the type returned by the size
method.
typedef std::vector<char> Path; // 'Path' now a synonym for std::vector<char>
for (Path::size_type i=0; i<path.size(); ++i)
std::cout << path[i] << ' ';
Why not use this in preference to the iterator
solution? For simple cases, you can do that, but using an iterator
brings several advantages, which I have briefly outlined above. As such, my advice would be to avoid this method unless you have good reasons for it.
std::copy (C++11)
See Joshua's answer. You can use the STL algorithm std::copy
to copy the vector contents onto the output stream. I don't have anything to add, except to say that I don't use this method; but there's no good reason for that besides habit.
std::ranges::copy (C++20)
For completeness, C++20 introduced ranges, which can act on the whole range of a std::vector
, so no need for begin
and end
:
#include <iterator> // for std::ostream_iterator
#include <algorithm> // for std::ranges::copy depending on lib support
std::vector<char> path;
// ...
std::ranges::copy(path, std::ostream_iterator<char>(std::cout, " "));
Unless you have a recent compiler (on GCC apparently at least version 10.1), likely you will not have ranges support even if you might have some C++20 features available.
Overload std::ostream::operator<<
See also Chris's answer below. This is more a complement to the other answers since you will still need to implement one of the solutions above in the overloading, but the benefit is much cleaner code. This is how you could use the std::ranges::copy
solution above:
#include <iostream>
#include <vector>
#include <iterator> // for std::ostream_iterator
#include <algorithm> // for std::ranges::copy depending on lib support
using Path = std::vector<char>; // type alias for std::vector<char>
std::ostream& operator<< (std::ostream& out, const Path& v) {
if ( !v.empty() ) {
out << '[';
std::ranges::copy(v, std::ostream_iterator<char>(out, ", "));
out << "\b\b]"; // use two ANSI backspace characters '\b' to overwrite final ", "
}
return out;
}
int main() {
Path path{'/', 'f', 'o', 'o'};
// will output: "path: [/, f, o, o]"
std::cout << "path: " << path << std::endl;
return 0;
}
Now you can pass your Path
objects to your output stream just like fundamental types. Using any of the other solutions above should also be equally straightforward.
Conclusion
Any of the solutions presented here will work. It's up to you (and context or your coding standards) on which one is the "best". Anything more detailed than this is probably best left for another question where the pros/cons can be properly evaluated, but as always user preference will always play a part: none of the solutions presented are objectively wrong, but some will look nicer to each coder.
Addendum
This is an expanded solution of an earlier one I posted. Since that post kept getting attention, I decided to expand on it and refer to the other excellent solutions posted here, at least those that I have personally used in the past at least once. I would, however, encourage the reader to look at the answers below because there are probably good suggestions that I have forgotten, or do not know, about.
How to insert std::vector or array to a std::forward_list without using any loop?
You could use a std::copy
with std::front_inserter
copy(arr.rbegin(), arr.rend(), front_inserter(listOne));
Working demo
Setting all items in an array to a number without for loop c++
If you want to set all the values to 0
then you can use:
int array[100] = {0}; //initialize array with all values set to 0
If you want to set some value other than 0
then you can use std::fill
from algorithm as shown below:
int array[100]; //not intialized here
std::fill(std::begin(array), std::end(array), 45);//all values set to 45
Appending a C-array to a vector in reverse order in C++98/03 without a for-loop
There you go. Use std::reverse_copy
with std::back_inserter
to add elements to the vector:
std::vector<int> vec;
int array[5] = { 1 , 2 , 3 , 4 , 5};
std::reverse_copy(array , array + 5, std::back_inserter(vec));
If you're concerned about performance, don't forget to reserve()
before copying.
Assigning to std::array element in std::vector of arrays fails
You are working on a copy of the array, you need a reference to the array in the vector
int main() {
vector<array<int, 1>> v;
array<int, 1> a = { 99 };
v.push_back(a);
cout << v[0][0];
auto& ar = v[0];
ar[0] = 42;
cout << v[0][0];
}
gives
99 42
Is it possible to copy part of a vector using the array copy?
Yes, use iterators:
vector<int> v2(v.begin() + x, v.begin() + x + y);
You can make it a bit more generic if you wish:
vector<int> v2(std::next(std::begin(v), x), std::next(std::begin(v), x + y));
The reason the pointer version (which the array decays into for both arguments) works in the first place is that pointers can be treated as random-access iterators.
iterating through an array to transfer its elements to a vector with certain conditions (c++)
The vectors passingGrades
and failingGrades
have no elements, so any access to their "elements" are invalid.
You can use std::vector::push_back()
to add elements to a std::vector
.
Also note that the loop using y
looks meaningless because the code inside the loop doesn't use y
and executes break;
in the first iteration.
In conclusion, the part:
for (int x = 0; x < 6; x++) {
for (int y = 0; y < 6; y++) {
if (userGrades[x] >= 60) {
passingGrades.at(x) = (userGrades[x]);
break;
}
else {
failingGrades.at(x) =(userGrades[x]);
break;
}
}
}
should be:
for (int x = 0; x < 6; x++) {
if (userGrades[x] >= 60) {
passingGrades.push_back(userGrades[x]);
}
else {
failingGrades.push_back(userGrades[x]);
}
}
Copying an array into another array in C++
If b
is an int[]
(that is, a C array) then you can do:
std::vector<int> data(b + 0, b + n);
If b
is also a std::vector
then you can just do:
std::vector<int> data = b;
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