不使用命名空间 std 就不能编译的程序；

Question

我有一个编译器错误：

g++ main.cpp -o exec -Wall -std=c++17 -Wextra -pedantic -O2 -Wshadow -fsanitize=undefined
main.cpp: In instantiation of ‘std::__cxx11::string to_string(A) [with A = int; std::__cxx11::string = std::__cxx11::basic_string<char>]’:
main.cpp:68:25:   required from ‘void debug_out(Head, Tail ...) [with Head = int; Tail = {}]’
main.cpp:79:3:   required from here
main.cpp:43:3: error: ‘begin’ was not declared in this scope
   for (const auto &x : v) {
   ^~~
main.cpp:43:3: note: suggested alternative:
In file included from /usr/include/x86_64-linux-gnu/c++/7/bits/stdc++.h:95:0,
                 from main.cpp:1:
/usr/include/c++/7/valarray:1211:5: note:   ‘std::begin’
     begin(const valarray<_Tp>& __va)
     ^~~~~
main.cpp:43:3: error: ‘end’ was not declared in this scope
   for (const auto &x : v) {
   ^~~
main.cpp:43:3: note: suggested alternative:
In file included from /usr/include/x86_64-linux-gnu/c++/7/bits/stdc++.h:95:0,
                 from main.cpp:1:
/usr/include/c++/7/valarray:1231:5: note:   ‘std::end’
     end(const valarray<_Tp>& __va)
     ^~~
Makefile:7: recipe for target 'c++' failed
make: *** [c++] Error 1

使用命名空间 std; 代码编译得很好。我想知道为什么不使用命名空间 std; 就不能编译代码。

我错误地查看了给定的文件，但我害怕通过接触编译器文件而出错。我觉得它在某处缺少 std:: ，但我真的不明白编译器文件中怎么会出现错误，而且我没有看到它可能会在我的代码中丢失。

这是我用于编程比赛的代码。

#include <bits/stdc++.h>

template <typename A, typename B>
std::string to_string(std::pair<A, B> p);
template <typename A, typename B, typename C>
std::string to_string(std::tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
std::string to_string(std::tuple<A, B, C, D> p);
std::string to_string(const std::string& s) {
  return '"' + s + '"';
}
std::string to_string(const char* s) {
  return to_string((std::string) s);
}
std::string to_string(bool b) {
  return (b ? "true" : "false");
}
std::string to_string(std::vector<bool> v) {
  bool first = true;
  std::string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <std::size_t N>
std::string to_string(std::bitset<N> v) {
  std::string res = "";
  for (std::size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
std::string to_string(A v) {
  bool first = true;
  std::string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
std::string to_string(std::pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
std::string to_string(std::tuple<A, B, C> p) {
  return "(" + to_string(std::get<0>(p)) + ", " + to_string(std::get<1>(p)) + ", " + to_string(std::get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
std::string to_string(std::tuple<A, B, C, D> p) {
  return "(" + to_string(std::get<0>(p)) + ", " + to_string(std::get<1>(p)) + ", " + to_string(std::get<2>(p)) + ", " + to_string(std::get<3>(p)) + ")";
}
void debug_out() { std::cerr << std::endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  std::cerr << to_string(H) << ", ";
  debug_out(T...);
}
#define LOCAL
#ifdef LOCAL
#define debug(...) std::cerr << "[" << #__VA_ARGS__ << "]: ", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

int main(){
  debug(13);
}

Answer 1

template <typename A> std::string to_string(A v)将匹配所有类型，而不仅仅是可以迭代的容器。这包括to_string(13).

当你在时它似乎工作的原因using namespace std;是因为然后std::to_stringforint参数将是一个更好的匹配并且将被使用而不是你有缺陷的实现。