多次元配列に関する問題で，エラーの原因が不明

Question

前提・実現したいこと

https://atcoder.jp/contests/apg4b/tasks/APG4b_ce

AtCorder の基本問題の質問です.

発生している問題・エラーメッセージ

エラーメッセージ

./Main.cpp: In function ‘int main()’:
./Main.cpp:19:37: error: no match for ‘operator=’ (operand types are ‘std::basic_ostream<char>’ and ‘char’)
   19 |           cout << vec.at(i).at(j) = 'o';
      |                                     ^~~
In file included from /usr/include/c++/9/istream:39,
                 from /usr/include/c++/9/sstream:38,
                 from /usr/include/c++/9/complex:45,
                 from /usr/include/c++/9/ccomplex:39,
                 from /usr/include/x86_64-linux-gnu/c++/9/bits/stdc++.h:54,
                 from ./Main.cpp:1:
/usr/include/c++/9/ostream:408:7: note: candidate: ‘std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator=(std::basic_ostream<_CharT, _Traits>&&) [with _CharT = char; _Traits = std::char_traits<char>]’
  408 |       operator=(basic_ostream&& __rhs)
      |       ^~~~~~~~
/usr/include/c++/9/ostream:408:33: note:   no known conversion for argument 1 from ‘char’ to ‘std::basic_ostream<char>&&’
  408 |       operator=(basic_ostream&& __rhs)
      |                 ~~~~~~~~~~~~~~~~^~~~~
./Main.cpp:22:37: error: no match for ‘operator=’ (operand types are ‘std::basic_ostream<char>’ and ‘char’)
   22 |           cout << vec.at(i).at(j) = 'x';
      |                                     ^~~
In file included from /usr/include/c++/9/istream:39,
                 from /usr/include/c++/9/sstream:38,
                 from /usr/include/c++/9/complex:45,
                 from /usr/include/c++/9/ccomplex:39,
                 from /usr/include/x86_64-linux-gnu/c++/9/bits/stdc++.h:54,
                 from ./Main.cpp:1:
/usr/include/c++/9/ostream:408:7: note: candidate: ‘std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator=(std::basic_ostream<_CharT, _Traits>&&) [with _CharT = char; _Traits = std::char_traits<char>]’
  408 |       operator=(basic_ostream&& __rhs)
      |       ^~~~~~~~
/usr/include/c++/9/ostream:408:33: note:   no known conversion for argument 1 from ‘cha...

### 該当のソースコード



C++

ソースコード
include <bits/stdc++.h>
using namespace std;

int main() {
  int N, M;
  cin >> N >> M;
  vector<int> A(M), B(M);
  for (int i = 0; i < M; i++) {
    cin >> A.at(i) >> B.at(i);
  }
  // ここにプログラムを追記
  // (ここで"試合結果の表"の2次元配列を宣言)
  vector<vector<char>> vec(N, vector<char>(N));
  
  for (int i=0; i < M; i++) {
    for (int j = 0; j < M; j++) {
      for (int k = 1; k <= M; k++){
        if(i == A.at(M) && j == B.at(M)){
          cout << vec.at(i).at(j) = 'o';
        }
        else if (i == B.at(M) && j == A.at(M)) {
          cout << vec.at(i).at(j) = 'x';
        }
        else {cout << vec.at(i).at(j) = '-';}
      }
    }
    for (int i = 0; i < M; i++){
      for (int j = 0 ; j < M; j++){
        cout << vec.at(i).at(j) ;
        if(j = N-1) {
          cout << endl;
          else(cout << "　")
        }
      }
    }
  }

Answer 1

cout << vec.at(i).at(j) = 'o';

これだと
( cout<<vec.at(i).at(j) ) = 'o'; という話になるので．

cout << ( vec.at(i).at(j) = 'o' ); のようにすれば良いかと．