Question

I have code as such:

typedef intptr_t ptr_t;

const int num_elements = 100;
ptr_t *pstr = (ptr_t *)malloc(sizeof(ptr_t) * num_elements);
std::array<ptr_t,num_elements> *parray = new (pstr) std::array<ptr_t,num_elements>;

我愿把第1至2号要素uff起来,因此,我想用的是:sh。

auto s = parray->begin()++;
auto e = parray->end()--;
std::random_shuffle ( s, e );

I get a complaint that there is no overloaded function for this. I m feeling really stupid at my inability to see what I m doing wrong. How do I do this right?

EDIT:由于答复和反馈,已经改为

auto s = parray->begin();
s++;
auto e = parray->end();
std::random_shuffle ( s, e );

然而,在汽车上,我发现:汽车在间接程度上不同于 in。

Answer 1

回答你的直接问题: 我认为,你的错误是利用员额加薪操作员,在加薪之前将原值回去。自std:array以来,探测器基本上都是点击器。

auto s = parray->begin() + 1;
auto e = parray->end() - 1;

Edit: Now, as for the rest. Why on earth are you doing it that way? Have you considered std::vector<int> arr(100) to create a dynamic array of 100 elements? It has similar capabilities, without all of the direct manipulation of pointers?

第2号:在阅读你的评论后,我认识到,问题在于,你再次试图把你作为点子 array起来。在这种情况下,我根本不做新安排。假设你在<代码>pstr上有点,就应当这样做。

std::random_shuffle(pstr +1, pstr + num_elements - 1);

这样做是因为一个阵列中的简单点子将作为算法图书馆的随机存取器。

Answer 2

即使汇编者允许,你的法典也不会做你想要的东西。您再尝试使用从<条码>以下现值中恢复的增减/减值:<>条码> > <条码>end(> > ,因此,您仍然将原值分配到<条码>和<条码> ,然后(如果可能) 增减/减退的温室。你显然想要改变所分配的价值。

auto s= parray->begin();
auto e= parray->end();

++s;
--e;

std::random_shuffle(s, e);

Or, since you apparently have random access iterators anyway:

 std::random_shuffle(parray->begin()+1, parray->end()-1);

我不敢肯定,你为什么重新积极分配<代码>std:array <>/code>——这似乎在很大程度上打败了使用<代码>的点:array来开始。

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