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Simplify check for identical shaders

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Previously when checking if two SPIR-V shaders are identical the
hashs of their file content would be compared and afterwards their
(unhashed) file contents as well. Comparing the file contents isn't
necessary, as the hash function used is a cryptographic one, which
guarantees the hash can be used as a representative of the hashed data.
This commit is contained in:
Dunedan 2024-09-19 11:49:21 +02:00
parent 668ae8a20e
commit 265ed76131
Signed by untrusted user: Dunedan
GPG Key ID: 885B16854284E0B2
1 changed files with 3 additions and 20 deletions
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23
source/tools/spirv/compile.py
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@ -49,14 +49,6 @@ def calculate_hash(path):
return hashlib.sha1(handle.read()).hexdigest()
def compare_spirv(path1, path2):
with open(path1, "rb") as handle:
spirv1 = handle.read()
with open(path2, "rb") as handle:
spirv2 = handle.read()
return spirv1 == spirv2
def resolve_if(defines, expression):
for item in expression.strip().split("||"):
item = item.strip()
@ -456,19 +448,10 @@ def build(rules, input_mod_path, output_mod_path, dependencies, program_name):
spirv_hash = calculate_hash(output_spirv_path)
if spirv_hash not in hashed_cache:
hashed_cache[spirv_hash] = [file_name]
hashed_cache[spirv_hash] = file_name
else:
found_candidate = False
for candidate_name in hashed_cache[spirv_hash]:
candidate_path = os.path.join(output_spirv_mod_path, candidate_name)
if compare_spirv(output_spirv_path, candidate_path):
found_candidate = True
file_name = candidate_name
break
if found_candidate:
os.remove(output_spirv_path)
else:
hashed_cache[spirv_hash].append(file_name)
file_name = hashed_cache[spirv_hash]
os.remove(output_spirv_path)
shader_element = ET.SubElement(program_root, shader["type"])
shader_element.set("file", "spirv/" + file_name)