C++字符串变成指针

Question

我有一个函数与以下声明:

void playCard(string card);

以及以下实施:

void Player::playCard(string card)
{
    cout << "Playing " << card << "!" << endl;

    // Find iterator representing the card to be played
    vector<Card*>::iterator iter;
    for(iter = hand.begin(); iter != hand.end(); iter++)
    {
        if( (*iter)->getName() == card)
            continue;
    }

    // ERROR - Card not found in hand
    if(iter == hand.end())
        assert(false);

    // more stuff
}

该函数从以下代码块调用:

// Divide string into 2 words
istringstream iss(in, istringstream::in);
string command, target;
iss >> command >> target;

if(command == "play")
{
    players.at(currentTurn)->playCard(target);
}

球员被宣布为:

vector<Player*> players;

我遇到的问题是,无论"卡"字符串是什么,我都会找到卡未找到的断言.根据gdb:

Breakpoint 1, Player::playCard (this=0xb3a010, card=0x28ca00) at Player.cpp:138
138         cout << "Playing " << card << "!" << endl;
(gdb) print card
$1 = (string *) 0x28ca00

因此在playCard()函数中,由于某种原因,card变量是一个指针.有趣的是,cout语句仍然正确显示字符串的内容.在调用playCard()之前,变量不是根据gdb的指针,只是一个普通的字符串.

此外,这很有趣:

(gdb) print iter
$1 = {_M_current = 0xb9a328}
(gdb) print iter->getName()
Couldn't find method __normal_iterator<Card**,std::vector<Card*, std::allocator<Card*> > >::getName
(gdb) print *iter
$2 = (class Card *&) @0xb9a328: 0xb9a160
(gdb) print *iter->getName()
Couldn't find method __normal_iterator<Card**,std::vector<Card*, std::allocator<Card*> > >::getName
(gdb) print (*iter)->getName()

Program received signal SIGSEGV, Segmentation fault.
0x61111178 in memcpy () from /usr/bin/cygwin1.dll

因此,当我尝试打印出最后一个时,gdb会导致段错误,但是相同的代码在没有任何段错误的情况下执行.

我有一种感觉,我正在处理一些奇怪的内存问题,这些问题与一个充满指向对象的向量有关,但我不能完全指责它.

这里发生了什么？

Answer 1

你想要break而不是continue在循环内.

他们有点相似; continue意味着继续循环的下一次迭代,break意味着突破循环.