c++ - 在声明期间将值初始化为二维 vector

Question

我已经这样声明了我的二维数组。但因此出现错误

vector <vector <int> > plain(vector <int>(4,0))  = {{23,43,45,56},
                                                    {67,85,13,59},
                                                    {48,23,9,57},
                                                    {24,52,90,12}};


rijndael.cpp:12:51: error: expected ‘,’ or ‘;’ before ‘=’ token
rijndael.cpp:57:1: error: expected ‘}’ at end of input

所有前面的分号都已正确分配。你能告诉我这个声明哪里出了问题吗？

我删除了构造函数调用，现在我的声明是

vector <vector <int> > plain/*(vector <int>(4,0))*/  = {{23,43,45,56},
                                                        {67,85,13,59},
                                                        {48,23,9,57},
                                                        {24,52,90,12}};

但是现在的错误是

rijndael.cpp:15:19: error: in C++98 ‘plain’ must be initialized by constructor, not by ‘{...}’
rijndael.cpp:15:19: error: deducing from brace-enclosed initializer list requires #include <initializer_list>
rijndael.cpp:15:19: error: deducing from brace-enclosed initializer list requires #include <initializer_list>
rijndael.cpp:15:19: warning: extended initializer lists only available with -std=c++0x or -std=gnu++0x
rijndael.cpp:15:19: error: could not convert ‘{{23, 43, 45, 56}, {67, 85, 13, 59}, {48, 23, 9, 57}, {24, 52, 90, 12}}’ to ‘std::vector<std::vector<int> >’

我该如何纠正这个问题？非常感谢任何帮助。

Answer 1

你不能同时一个构造函数调用和一个列表初始化，坚持一个，例如:

vector<vector<int>> plain{{23,43,45,56},
                             {67,85,13,59},
                             {48,23,9,57},
                             {24,52,90,12}};

(你可以省略列表初始化的=。)

关于您的编辑:好吧，错误说明了一切，阅读它会有所帮助。列表初始化需要 C++11 模式。如果你不想这样，你必须用 ctor 复制以准备 2D 数组，然后在构建后填充它。