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Interpret a blob of memory as a rar file for fuzzing. #1090

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Jan 18, 2018
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Interpret a blob of memory as a rar file for fuzzing. #1090

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2e767d0
Interpret a blob of memory as a rar file for fuzzing. (#4)
aawc Jan 18, 2018
c67f731
Interpret a blob of memory as a rar file for fuzzing. (#5)
aawc Jan 18, 2018
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10 changes: 1 addition & 9 deletions projects/unrar/unrar_fuzzer.cc
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Original file line number Diff line number Diff line change
Expand Up @@ -9,19 +9,13 @@ extern "C" int LLVMFuzzerTestOneInput(const uint8_t* data, size_t size) {
std::stringstream ss;
ss << "temp-" << getpid() << ".rar";
static const std::string filename = ss.str();
std::ofstream file(filename,
std::ios::binary | std::ios::out | std::ios::trunc);
if (!file.is_open()) {
return 0;
}
file.write(reinterpret_cast<const char *>(data), size);
file.close();

std::unique_ptr<CommandData> cmd_data(new CommandData);
cmd_data->ParseArg(const_cast<wchar_t *>(L"-p"));
cmd_data->ParseArg(const_cast<wchar_t *>(L"x"));
cmd_data->ParseDone();
std::wstring wide_filename(filename.begin(), filename.end());
cmd_data->SetArcInMem(const_cast<unsigned char *>(data), size);
cmd_data->AddArcName(wide_filename.c_str());
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Can the filename be entirely removed now, or is it still required?

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Unfortunately, it is still required. Once we decide that we are definitely going down this path (in-memory), I can convince the maintainer to support it better i.e. not have to set filename at all,

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Does filename have to be unique, otherwise, we can just set it to a default like temp.rar ?

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It doesn't have to be but I agree that having a fixed name now is possible and faster so let me change that.

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Done.


try {
Expand All @@ -30,7 +24,5 @@ extern "C" int LLVMFuzzerTestOneInput(const uint8_t* data, size_t size) {
} catch (...) {
}

unlink(filename.c_str());

return 0;
}