From 444f2d796563501b68247ef80b03de125022649c Mon Sep 17 00:00:00 2001
From: Hennadii Stepanov <32963518+hebasto@users.noreply.github.com>
Date: Thu, 16 Jul 2026 15:05:52 +0100
Subject: [PATCH] fuzz, refactor: Remove `Serialize` overload

Serialization parameters should be embedded into the object being
serialized rather than passed as a separate argument. This works here
because only serialization is performed and no new object needs to be
constructed.

Github-Pull: #35679
Rebased-From: afab8d4225c720f8d8658d4ea134505b3d9531f9
---
 src/test/fuzz/deserialize.cpp | 10 +---------
 1 file changed, 1 insertion(+), 9 deletions(-)

diff --git a/src/test/fuzz/deserialize.cpp b/src/test/fuzz/deserialize.cpp
index cc8f59f4403..3162527e32d 100644
--- a/src/test/fuzz/deserialize.cpp
+++ b/src/test/fuzz/deserialize.cpp
@@ -57,14 +57,6 @@ namespace {
 struct invalid_fuzzing_input_exception : public std::exception {
 };
 
-template <typename T, typename P>
-DataStream Serialize(const T& obj, const P& params)
-{
-    DataStream ds{};
-    ds << params(obj);
-    return ds;
-}
-
 template <typename T, typename P>
 T Deserialize(DataStream&& ds, const P& params)
 {
@@ -103,7 +95,7 @@ void DeserializeFromFuzzingInput(FuzzBufferType buffer, T&& obj)
 template <typename T, typename P>
 void AssertEqualAfterSerializeDeserialize(const T& obj, const P& params)
 {
-    assert(Deserialize<T>(Serialize(obj, params), params) == obj);
+    assert(Deserialize<T>(Serialize(params(obj)), params) == obj);
 }
 template <typename T>
 void AssertEqualAfterSerializeDeserialize(const T& obj)