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copies: add test that makes both the merging csets dirty and fails...
copies: add test that makes both the merging csets dirty and fails This patch is a part of series which is about the case when both the merging csets are not descendant of merge base. The existing code assumes if c1 is dirty there shouldn't be any partial copies from c2 i.e both2['incomplete'] and same for c2, if c2 is dirty both1['incomplete'] should be empty, but this is not the right assumption. Now as we know we can have both c1 and c2 dirty at the same time, it is possible that c1 is dirty and both2['incomplete'] has some value. Or if c2 is dirty and both1['incomplete'] has some value. Added test shows that because of this assumption it could fail. Differential Revision: https://phab.mercurial-scm.org/D5962

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bdiff.cc
44 lines | 1.0 KiB | text/x-c | CppLexer
/*
* bdiff.cc - fuzzer harness for bdiff.c
*
* Copyright 2018, Google Inc.
*
* This software may be used and distributed according to the terms of
* the GNU General Public License, incorporated herein by reference.
*/
#include <memory>
#include <stdlib.h>
#include "fuzzutil.h"
extern "C" {
#include "bdiff.h"
int LLVMFuzzerTestOneInput(const uint8_t *Data, size_t Size)
{
auto maybe_inputs = SplitInputs(Data, Size);
if (!maybe_inputs) {
return 0;
}
auto inputs = std::move(maybe_inputs.value());
struct bdiff_line *a, *b;
int an = bdiff_splitlines(inputs.left.get(), inputs.left_size, &a);
int bn = bdiff_splitlines(inputs.right.get(), inputs.right_size, &b);
struct bdiff_hunk l;
bdiff_diff(a, an, b, bn, &l);
free(a);
free(b);
bdiff_freehunks(l.next);
return 0; // Non-zero return values are reserved for future use.
}
#ifdef HG_FUZZER_INCLUDE_MAIN
int main(int argc, char **argv)
{
const char data[] = "asdf";
return LLVMFuzzerTestOneInput((const uint8_t *)data, 4);
}
#endif
} // extern "C"