But neither the Maoist leader, popularly known as Prachanda, nor the Nepali Congress candidate, Ram Chandra Poudel, could win a majority.
VOA: standard.2010.07.29
There are absolutely going to be times where you're running a program where the programmer, say you, didn't possibly know in advanced how much RAM the program was going to need.
它们总是在,程序员运行程序的地方,你可能预先不知道那个程序,需要多少内存。
I've been using the blackboard a little ad hoc here, but suppose I neaten things up here now and present this rectangle is let's say R RAM, the memory inside of your computer.
我在这里特别地使用这块黑板,假如我整理这些东西,把它们呈现,这个长方形,假如说就是内存,你电脑的内存。
So the heap is a chunk of memory in a computer's RAM that's conceptually allocated to what's called dynamic memory allocation.
堆是在计算机RAM中的一块内存,它可以进行,动态内存分配。
Only the operating system has controlling of byte zero NULL in the computer's RAM and so if a function ever returns null, aka zero, well, something must have gone wrong because that can't possibly belong to me that memory because by human convention zero is owned by the operating system; not by a program I wrote.
只用操作系统在内存中能够控制,字节0,并且如果一个函数返回,或者说0,好的,可能出错了,因为那可能是不属于我的内存,因为惯例上,0是由操作系统拥有的,而不是由我的程序拥有的。
So, conceptually if you've ever wondered why you get access in all of your functions to global variables that's because they're not down here, they're at the very top of RAM and any function can access that RAM way up there, but for now the interesting player in the story is this thing called the heap.
所以,如果你想知道为什么全局变量能在,所有的函数中使用,那是因为它们不在这下面,而是在内存的顶端,那样任意函数都可以在内存中使用它们,现在,这里面一个有用的角色是,叫做堆的东西。
To do that the data travels from RAM along with software signals that tell the hard drive how to store that data.
为了能让数据从RAM中,到硬盘是要使用软件信号告诉硬盘,如何存储这些数据。
So you can think of it as left or right, top to bottom, or whatever, the point is they come next in RAM.
你可以想象它们为left或者right,从上到下,要点是它们在内存中紧随其后。
Now that's changing the more years that pass the more of you have 64-bit computers and the more servers have many, many gigabytes of RAM and so you need actually 64-bits, but for now let's assume a common system whereby a pointer just by definition of the homework is 32 bits.
现在计算机也在改变,也有更多的人,使用64位的计算机,更多的服务器有很多,很多G的内存,那样你就需要64位的,现在我们假设通常的系统中,根据家庭作业所定义的,一个指针为32位。
Instead I'm going to be expecting the address of some int and the address of another int and thanks to this address I can literally find this address in RAM, do anything I want there, -- return and what I've just done is actually changed or mutate the values of those original variables.
我们所期望的是int型地址,和另外一个int型地址,幸亏这个地址,我可以找在RAM中找到它的地址,然后做我想做的事情,返回我处理后的东西,我可能改变了这些源变量。
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