push_back 来自同一个向量的元素是否安全?
vector<int> v;
v.push_back(1);
v.push_back(v[0]);
如果第二次 push_back 导致重新分配，则向量中第一个整数的引用将不再有效.所以这不安全?
If the second push_back causes a reallocation, the reference to the first integer in the vector will no longer be valid. So this isn't safe?
vector<int> v;
v.push_back(1);
v.reserve(v.size() + 1);
v.push_back(v[0]);
这样就安全了吗?
推荐答案
看起来像 http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-closed.html#526 解决了这个问题(或与它非常相似的问题)作为潜在缺陷在标准中:
It looks like http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-closed.html#526 addressed this problem (or something very similar to it) as a potential defect in the standard:
1) const 引用的参数可以在执行过程中改变函数的
1) Parameters taken by const reference can be changed during execution of the function
示例:
给定 std::vector v:
Given std::vector v:
v.insert(v.begin(), v[2]);
v.insert(v.begin(), v[2]);
v[2] 可以通过移动向量的元素来改变
v[2] can be changed by moving elements of vector
提议的解决方案是这不是缺陷:
The proposed resolution was that this was not a defect:
vector::insert(iter, value) 需要工作，因为标准不允许它不工作.
vector::insert(iter, value) is required to work because the standard doesn't give permission for it not to work.
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