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I found minor misstake on examples/chapter02/checkDeviceInfor.cu.
The following code is written on line 40-42.
printf(" Total amount of global memory: %.2f MBytes (%llu "
"bytes)\n", (float)deviceProp.totalGlobalMem / pow(1024.0, 3),
(unsigned long long)deviceProp.totalGlobalMem);
I found minor misstake on
examples/chapter02/checkDeviceInfor.cu
.The following code is written on line 40-42.
https://github.com/deeperlearning/professional-cuda-c-programming/blob/master/examples/chapter02/checkDeviceInfor.cu#L40
Shouldn't
%.2f MBytes
be%.2f GBytes
?The reason is that this value is obtained by dividing deviceProp.totalGlobalMem by pow(1024.0, 3).
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