您必須在表格中存儲矢量&或矢量*。 後者應該工作,但我試圖管理第一個。 對此我發現SO: Why can't I make a vector of references?。 因此,我想別的東西:標準::的reference_wrapper ,它似乎是一個(/下)解決方案:
#include <functional>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main(int, char**)
{
// build contents of tables
vector<string> a { string{"A 0"}, string{"A 1"}, string{"A 2"} };
cout << "&a: " << hex << (size_t)&a << endl;
vector<string> b { string{"B 0"}, string{"B 1"}, string{"B 2"} };
cout << "&b: " << hex << (size_t)&b << endl;
vector<std::reference_wrapper<vector<string> > > tables {
a, b
};
// print contents of tables
for (size_t i = 0, n = tables.size(); i < n; ++i) {
const vector<string> &tbl = tables[i];
cout << "&tables[" << i << "]: " << hex << (size_t)&tbl << endl;
for (size_t j = 0, m = tbl.size(); j < m; ++j) {
cout << "tables[" << i << "][" << j << "]: '"
<< tbl[j] << "'" << endl;
}
}
// append another string to a
{ vector<string> &tbl = tables[0];
tbl[1] = string("A 1 modified");
tbl.push_back(string("A 3"));
tbl.emplace_back(string("A 4"));
}
// print contents of a
for (size_t i = 0, n = a.size(); i < n; ++i) {
cout << "a[" << i << "]: '" << a[i] << "'" << endl;
}
// append another table
vector<string> c { string("C 0"), string("C 1"), string("C 2") };
cout << "&c: " << hex << (size_t)&c << endl;
/* my 1st guess:
tables.emplace_back(std::ref(c));
* but even this works:
*/
tables.emplace_back(c);
// print contents of tables
for (size_t i = 0, n = tables.size(); i < n; ++i) {
const vector<string> &tbl = tables[i];
cout << "&tables[" << i << "]: " << hex << (size_t)&tbl << endl;
for (size_t j = 0, m = tbl.size(); j < m; ++j) {
cout << "tables[" << i << "][" << j << "]: '"
<< tbl[j] << "'" << endl;
}
}
// done
return 0;
}
在Cygwin上使用GCC編譯如下:
$ gcc --version
gcc (GCC) 5.4.0
Copyright (C) 2015 Free Software Foundation, Inc.
$ g++ -std=c++11 -o testVecRef testVecRef.cc
$ ./testVecRef
&a: 61cbec
&b: 61cbe0
&tables[0]: 61cbec
tables[0][0]: 'A 0'
tables[0][1]: 'A 1'
tables[0][2]: 'A 2'
&tables[1]: 61cbe0
tables[1][0]: 'B 0'
tables[1][1]: 'B 1'
tables[1][2]: 'B 2'
a[0]: 'A 0'
a[1]: 'A 1 modified'
a[2]: 'A 2'
a[3]: 'A 3'
a[4]: 'A 4'
&c: 61cbc8
&tables[0]: 61cbec
tables[0][0]: 'A 0'
tables[0][1]: 'A 1 modified'
tables[0][2]: 'A 2'
tables[0][3]: 'A 3'
tables[0][4]: 'A 4'
&tables[1]: 61cbe0
tables[1][0]: 'B 0'
tables[1][1]: 'B 1'
tables[1][2]: 'B 2'
&tables[2]: 61cbc8
tables[2][0]: 'C 0'
tables[2][1]: 'C 1'
tables[2][2]: 'C 2'
編譯,並開始與MS VS2013:
&a: cfba2ff2b8
&b: cfba2ff378
&tables[0]: cfba2ff2b8
tables[0][0]: 'A 0'
tables[0][1]: 'A 1'
tables[0][2]: 'A 2'
&tables[1]: cfba2ff378
tables[1][0]: 'B 0'
tables[1][1]: 'B 1'
tables[1][2]: 'B 2'
a[0]: 'A 0'
a[1]: 'A 1 modified'
a[2]: 'A 2'
a[3]: 'A 3'
a[4]: 'A 4'
&c: cfba2ff508
&tables[0]: cfba2ff2b8
tables[0][0]: 'A 0'
tables[0][1]: 'A 1 modified'
tables[0][2]: 'A 2'
tables[0][3]: 'A 3'
tables[0][4]: 'A 4'
&tables[1]: cfba2ff378
tables[1][0]: 'B 0'
tables[1][1]: 'B 1'
tables[1][2]: 'B 2'
&tables[2]: cfba2ff508
tables[2][0]: 'C 0'
tables[2][1]: 'C 1'
tables[2][2]: 'C 2'
Drücken Sie eine beliebige Taste . . .
btw .:如果我們在源代碼中嵌入數據表,我們更喜歡普通的舊數據(例如C數組的const char)。因此,編譯器可以在編譯時完全佈局存儲。沒有問題(由於施工順序等)可能發生。 – Scheff
你的例子的作品,但你能寫一個例子如何使用表追加「A 3」?簡單表[0] .push_back(「A 3」)在編譯時返回以下錯誤: 'push_back':不是'std :: reference_wrapper >>' 與 [ _Ty = std :: string ] –
@Math Guy:我編輯了我的樣本。請注意,我更喜歡使用「in-place」構造的較新vector <> :: emplace_back()。 – Scheff