Fix "CUDA Tensor __rsub__ breaks when device is not 0" #12956
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colesbury
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Oct 22, 2018
| {"__mul__", (PyCFunction)THPVariable_mul, METH_VARARGS | METH_KEYWORDS, NULL}, | ||
| {"__imul__", (PyCFunction)THPVariable_mul_, METH_VARARGS | METH_KEYWORDS, NULL}, | ||
| {"__sub__", (PyCFunction)THPVariable_sub, METH_VARARGS | METH_KEYWORDS, NULL}, | ||
| {"__rsub__", (PyCFunction)THPVariable_sub, METH_VARARGS | METH_KEYWORDS, NULL}, |
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October 26, 2018 11:40
This reverts commit dfe1f4d26ebff0a5a5543ddfb6a067e46becac69.
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Summary:
Currently, `a = 1 - torch.tensor([1]).to('cuda:1')` puts `a` in `cuda:1` but reports `a.device` as `cuda:0` which is incorrect, and it causes illegal memory access error when trying to access `a`'s memory (e.g. when printing). This PR fixes the error.
Fixes pytorch/pytorch#10850.
Pull Request resolved: pytorch/pytorch#12956
Differential Revision: D12835992
Pulled By: yf225
fbshipit-source-id: 5737703d2012b14fd00a71dafeedebd8230a0b04
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Currently,
a = 1 - torch.tensor([1]).to('cuda:1')putsaincuda:1but reportsa.deviceascuda:0which is incorrect, and it causes illegal memory access error when trying to accessa's memory (e.g. when printing). This PR fixes the error.Fixes #10850.