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I am creating an action using Rust, and I found /bin/compile.py in the default image used by OpenWhisk, which is openwhisk/action-rust-v1.34:nightly. Upon inspecting the code, I noticed that the RELEASE environment variable must be set in os.environ for the compile option to use the release build. However, I don't know how to set this environment variable.
def build(tgt_dir):
# support RELEASE
cmd = ["cargo", "build"]
bin_dir = "debug"
if "RELEASE" in os.environ:
cmd += "--release"
bin_dir = "release"
How do I set this environment variable?
The text was updated successfully, but these errors were encountered:
On Tue, 16 Jul 2024 at 06:03, kh3654po ***@***.***> wrote:
I am creating an action using Rust, and I found /bin/compile.py in the
default image used by OpenWhisk, which is
openwhisk/action-rust-v1.34:nightly. Upon inspecting the code, I noticed
that the RELEASE environment variable must be set in os.environ for the
compile option to use the release build. However, I don't know how to set
this environment variable.
def build(tgt_dir):
# support RELEASE
cmd = ["cargo", "build"]
bin_dir = "debug"
if "RELEASE" in os.environ:
cmd += "--release"
bin_dir = "release"
How do I set this environment variable?
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I am creating an action using Rust, and I found /bin/compile.py in the default image used by OpenWhisk, which is openwhisk/action-rust-v1.34:nightly. Upon inspecting the code, I noticed that the RELEASE environment variable must be set in os.environ for the compile option to use the release build. However, I don't know how to set this environment variable.
How do I set this environment variable?
The text was updated successfully, but these errors were encountered: