起草一个计算简单功能组成部分的方案。 在测试时,我发现,如果我使用了1 000多万件大面积,那就会产生错误答案。 我发现,一旦阵列在加澳新集团中被操纵,就会出现错误。 1 000万和以下人员做了出色的工作,产生了正确的结果。
是否可以对可转至万国邮联或根据万国邮联计算的内容数量设定一个规模限制?
P.S. using C style arrays containing floats.
你们可以使用多种不同的记忆。 尤其是,
cuMemAlloc)cuMemHostAlloc)cuMemAllocHost)cuMemAllocPitch)每种类型的记忆都与自己的硬件资源限额有关,其中许多可以通过使用<代码>cuDeviceGetAttribute<>。 1,000,000 浮标可能产生超过<条码>1 000 000(float)的体积。 您可以一度安排的最大区块数目也是一种限制:如果你超过,油轮将无法发射(你可以很容易地利用<代码>cuDeviceGetAttribute<>)。 你们可以找到使用“CUDA”司机APICA不同记忆量的统一要求,但就简单的方案而言,你可以进行合理的猜测,检查分配职能的价值,以确定分配是否成功。
There is no restriction on the amount of bytes that you can transfer; using asynchronous functions, you can overlap kernel execution with memory copying (providing that your card supports this). Exceeding the maximum number of blocks you can schedule, or consuming the available memory on your device means that you will have to split up your task so that you can use multiple kernels to handle it.
For compute capability>=3.0 the max grid dimensions are 2147483647x65535x65535, so for a that should cover any 1-D array of sizes up to 2147483647x1024 = 2.1990233e+12.
10亿个元素阵列肯定是罚款。
1 000 000 000 000 000 000 000/1024=976562.5, 共976563个区块。 仅能确保,如果read子X+blockIdx.*blockDim.x>=你未经加工从油轮返回的人数。
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