How do I print out the contents of a vector?

Question

How do I print out the contents of a std::vector to the screen?

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A solution that implements the following operator<< would be nice as well:

template<container C, class T, String delim = ", ", String open = "[", String close = "]"> std::ostream & operator<<(std::ostream & o, const C<T> & x) {   // ... What can I write here? } 

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Here is what I have so far, without a separate function:

#include <iostream> #include <fstream> #include <string> #include <cmath> #include <vector> #include <sstream> #include <cstdio> using namespace std;  int main() {     ifstream file("maze.txt");     if (file) {         vector<char> vec(istreambuf_iterator<char>(file), (istreambuf_iterator<char>()));         vector<char> path;         int x = 17;         char entrance = vec.at(16);         char firstsquare = vec.at(x);         if (entrance == 'S') {              path.push_back(entrance);          }         for (x = 17; isalpha(firstsquare); x++) {             path.push_back(firstsquare);         }         for (int i = 0; i < path.size(); i++) {             cout << path[i] << " ";         }         cout << endl;         return 0;     } } 

Answer 1

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 do: 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.