Question

我有以下（过于简化）的例子，其中valgrind抱怨未初始化的内存：

struct B {
  virtual ~B() {}
};
struct BB : public virtual B {
  virtual ~BB() = default;
};
struct BBB : public virtual B {
  virtual ~BBB() {}
};

struct X : public virtual B, public virtual BBB/*, public virtual BB*/ {
  X() { a = 0.0; b = 0.0; c = false;}
  double a;
  double b;
  bool c;
};

并像这样使用它：

auto y = std::make_shared<X>();
HexDump(y.get(), sizeof(X)); // simple hexdump function - similar to gtest output - in real application this is an "uninteresting mock call" where gtest prints binary representation of an object using its size

这里的工作示例：http://coliru.stacked-crooked.com/a/91a54b697010d84b

用gcc 4.9.2 c ++ 14标志编译并运行valgrind我得到以下输出：

==58887== Conditional jump or move depends on uninitialised value(s)
==58887==    at 0x93A065: __vfprintf (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x95E2DA: __v2printf (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x9441DC: vsprintf_l (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x93464C: sprintf (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x100165310: HexDump(void*, unsigned int) (in xxxx)
==58887== Use of uninitialised value of size 8
==58887==    at 0x93A69B: __ultoa (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x93A0CC: __vfprintf (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x95E2DA: __v2printf (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x9441DC: vsprintf_l (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x93464C: sprintf (in /usr/lib/system/libsystem_c.dylib)
==58887==    by 0x100165310: HexDump(void*, unsigned int) (in xxxxx)

这是由于编译器在对象末尾添加了填充（请注意＆＃34;跨越＆＃34; 8字节的bool）？如果是这样 - 检查可以做些什么不失败？

Answer 1

您的示例是未定义的行为，因为X不是POD。因此，以这种方式从随机字节开始读取是非法的。如果要转储对象，实际转储对象，并且不要假装转储字节是一回事。

无论是否gmock正在进行转储而不是你，都是如此。

使用std :: make_shared和虚拟继承时valgrind报告的未初始化值

1 个答案: