如何将std::vector的内容打印到屏幕上?
实现以下操作符<<的解决方案也很好:
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?
}
以下是目前为止我所做的，没有单独的函数:
#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;
}
}
如果boost是一个选项，那么你可以使用boost::algorithm::join。例如，打印std::string的向量:
#include <boost/algorithm/string/join.hpp>
std::vector<std::string> vs { "some", "string", "vector" };
std::cout << boost::algorithm::join(vs, " | ") << '\n';
对于其他类型的向量，首先需要转换为字符串
#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
#include <boost/algorithm/string/join.hpp>
#include <boost/range/adaptor/transformed.hpp>
int main()
{
using boost::adaptors::transformed;
using boost::algorithm::join;
// Generate the vector
std::vector<int> vi(10);
std::iota(vi.begin(), vi.end(), -3);
// Print out the vector
std::cout << join(vi |
transformed(static_cast<std::string(*)(int)>(std::to_string)),
", ")
<< '\n';
}
Godbolt演示
我认为最好的方法是通过在程序中添加这个函数来重载操作符<<:
#include <vector>
using std::vector;
#include <iostream>
using std::ostream;
template<typename T>
ostream& operator<< (ostream& out, const vector<T>& v) {
out << "{";
size_t last = v.size() - 1;
for(size_t i = 0; i < v.size(); ++i) {
out << v[i];
if (i != last)
out << ", ";
}
out << "}";
return out;
}
然后你可以在任何可能的向量上使用<<运算符，假设它的元素也定义了ostream&运算符<<:
vector<string> s = {"first", "second", "third"};
vector<bool> b = {true, false, true, false, false};
vector<int> i = {1, 2, 3, 4};
cout << s << endl;
cout << b << endl;
cout << i << endl;
输出:
{first, second, third}
{1, 0, 1, 0, 0}
{1, 2, 3, 4}
对于那些感兴趣的人:我写了一个通用的解决方案，它两全其美，更通用于任何类型的范围，并在非算术类型周围加上引号(适合于类似字符串的类型)。此外，这种方法不应该有任何ADL问题，也可以避免“意外”(因为它是根据具体情况明确添加的):
template <typename T>
inline constexpr bool is_string_type_v = std::is_convertible_v<const T&, std::string_view>;
template<class T>
struct range_out {
range_out(T& range) : r_(range) {
}
T& r_;
static_assert(!::is_string_type_v<T>, "strings and string-like types should use operator << directly");
};
template <typename T>
std::ostream& operator<< (std::ostream& out, range_out<T>& range) {
constexpr bool is_string_like = is_string_type_v<T::value_type>;
constexpr std::string_view sep{ is_string_like ? "', '" : ", " };
if (!range.r_.empty()) {
out << (is_string_like ? "['" : "[");
out << *range.r_.begin();
for (auto it = range.r_.begin() + 1; it != range.r_.end(); ++it) {
out << sep << *it;
}
out << (is_string_like ? "']" : "]");
}
else {
out << "[]";
}
return out;
}
现在它在任何范围都很容易使用:
std::cout << range_out{ my_vector };
类似字符串的检查留有改进的空间。
在我的解决方案中，我也有static_assert检查，以避免std::basic_string<>，但为了简单起见，我在这里省略了它。
