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test: fix creation of std::string objects with \0s

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A string literal `"abc"` contains a terminating `\0`, so that is 4
bytes. There is no need to write `"abc\0"` unless two terminating
`\0`s are necessary.

`std::string` objects do not internally contain a terminating `\0`, so
`std::string("abc")` creates a string with size 3 and is the same as
`std::string("abc", 3)`.

In `"\01"` the `01` part is interpreted as one number (1) and that is
the same as `"\1"` which is a string like `{1, 0}` whereas `"\0z"` is a
string like `{0, 'z', 0}`. To create a string like `{0, '1', 0}` one
must use `"\0" "1"`.

Adjust the tests accordingly.
This commit is contained in:
Vasil Dimov
2020-09-23 08:40:38 +02:00
parent b1291b2e8f
commit ecc6cf1a3b
6 changed files with 59 additions and 44 deletions
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6
src/test/net_tests.cpp
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@@ -23,6 +23,8 @@
#include <memory>
#include <string>
using namespace std::literals;
class CAddrManSerializationMock : public CAddrMan
{
public:
@@ -104,8 +106,8 @@ BOOST_AUTO_TEST_CASE(caddrdb_read)
BOOST_CHECK(Lookup("250.7.1.1", addr1, 8333, false));
BOOST_CHECK(Lookup("250.7.2.2", addr2, 9999, false));
BOOST_CHECK(Lookup("250.7.3.3", addr3, 9999, false));
BOOST_CHECK(Lookup(std::string("250.7.3.3", 9), addr3, 9999, false));
BOOST_CHECK(!Lookup(std::string("250.7.3.3\0example.com", 21), addr3, 9999, false));
BOOST_CHECK(Lookup("250.7.3.3"s, addr3, 9999, false));
BOOST_CHECK(!Lookup("250.7.3.3\0example.com"s, addr3, 9999, false));
// Add three addresses to new table.
CService source;