matrix m1(5,5);
matrix m2(5,5);
m1 = matrix(m2);
对于上述法典(关于任意类别,矩阵),如果将m2的信息抄送给谁,那么是否会要求仲裁员提供与m1有关的信息?
没有一个,转让经营人需要处理在进行转让之前可持有的任何资源<代码>m1。 只有当<代码>m1即将离开范围时,才会叫 des。
无,一一俟一物体在 st上被分配,就会被拆除,直至其脱离范围或明确叫走其 des子(你可能永远不会这样做)。 因此,在这种情况下,如果矩阵界定超负荷运营商=(最接近的矩阵和组件;电网)成员功能,则称为操作员=()并复制机。 否则,缺省转让就被用来简单地复制临时矩阵(m2)中的所有成员变数,而忽略了这些变数以前的数值。
我认为,这取决于矩阵是否妥善执行过反应堆,以及转让操作者是如何实施的。 如果矩阵具有操作的脱轨器和矩阵使用“信号-swap”(类似于复印件-swap idiom),那么M1应适当释放。
除此之外,你在用m1 = m2时实际上不需要汇总表。 这只是要打造影机,然后将临时复制件寄给1米。 因此,无用工作正在进行。
这个矩阵是什么集装箱? 如果它具有价值,那么就没有问题了。 如果说是聪明的,那么就没有问题了。
但是,是否涉及普通点? 然后完全要实施<代码>矩阵:operator=(const Format&)。 如果需要传唤任何逃兵,他们就应当叫他们。 事情没有发生,你确实必须思考这些问题。
一般说来,使用普通集装箱类别来持有普通电站是一种坏的想法。 如果 des子叫,你就会泄露记忆。 如果被传唤,如果你通过另一个点子接触同一物体,你就会坠毁。 (如果你有两个矩阵,两个矩阵都含有同一个标的点。)
由于使用普通集装箱类别处理普通点人的集装箱是困难和危险的。 应为此目的使用专用集装箱或专门站。
Boost, for example, contains specialized container classes specifically for holding pointers. Boost also has a shared pointer class that can be held by ordinary containers.
你们只需要决定应该怎样做。 如果有的话,a=b;? 还是应当为<代码>a在内部指明新的物体?
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