C++中臨時對象的銷毀

Question

我有兩個類： Vote和Voter 。

Voter類有一個構造函數，它接收兩個字符串。

Vote類有一個構造函數，它從Voter類接收一個對象和一個字符串。

現在，我執行以下操作：

Voter vr6("Cyprus", "Regular");
eurovision += Vote(vr6, "USA");

其中 eurovision 是我重載+=運算符的類中的對象。

據我所知，在第二行中將創建一個臨時Vote對象。

我的問題是， vr6究竟如何受到臨時對象破壞的影響？

編輯： Vote構造函數和析構函數的定義：

Vote(Voter current_voter, string state1, string state2 = "", string state3 = "", string state4 = "", string state5 = "", string state6 = "", string state7 = "", string state8 = "", string state9 = "", string state10 = "") :
        voter(current_voter), voted_state(new string[VOTE_ARRAY_SIZE]){
        voted_state[0] = state1;
        voted_state[1] = state2;
        voted_state[2] = state3;
        voted_state[3] = state4;
        voted_state[4] = state5;
        voted_state[5] = state6;
        voted_state[6] = state7;
        voted_state[7] = state8;
        voted_state[8] = state9;
        voted_state[9] = state10;
    }
    ~Vote() {
        delete[] voted_state;
    }

Answer 1

由於您將vr6按值傳遞給Vote的構造函數，因此vr6的對象本身將不依賴於Vote對象的生命周期。 但是請注意，“按值調用”將創建vr6的臨時副本，一旦調用Vote構造函數的語句結束，該副本將被刪除。