Simplify check for identical shaders

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.

source/tools/spirv/compile.py

@@ -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)