在C ++ 20之前将malloc用于int未定义行为 [英] Is using malloc for int undefined behavior until C++20

问题描述

有人告诉我，以下代码在C ++ 20之前具有未定义的行为：

I was told that the following code has undefined behavior until C++20:

int *p = (int*)malloc(sizeof(int));
*p = 10;

是真的吗？

论点是<$的生存期在为对象分配值之前，未启动c $ c> int 对象（ P0593R6 ）。要解决此问题，应使用 new 放置：

The argument was that the lifetime of the int object is not started before assigning the value to it (P0593R6). To fix the problem, placement new should be used:

int *p = (int*)malloc(sizeof(int));
new (p) int;
*p = 10;

我们真的必须调用一个琐碎的默认构造函数来启动对象的生存期吗？

Do we really have to call a default constructor that is trivial to start the lifetime of the object?

同时，该代码在纯C语言中没有未定义的行为。但是，如果我在C代码中分配 int 并将其用于C ++代码？

At the same time, the code does not have undefined behavior in pure C. But, what if I allocate an int in C code and use it in C++ code?

// C source code:
int *alloc_int(void)
{
    int *p = (int*)malloc(sizeof(int));
    *p = 10;
    return p;
}

// C++ source code:
extern "C" int *alloc_int(void);

auto p = alloc_int();
*p = 20;

它仍然是未定义的行为吗？

Is it still undefined behavior?

推荐答案

是真的吗？

Is it true?

是。从技术上讲，没有任何内容。

Yes. Technically speaking, no part of:

int *p = (int*)malloc(sizeof(int));

实际上创建类型为 int 的对象，因此取消引用 p 是UB，因为那里没有实际的 int 。

actually creates an object of type int, so dereferencing p is UB since there is no actual int there.

我们真的必须调用默认的构造函数来启动对象的生命周期吗？

Do we really have to call default constructor that is trivial to start the life time of the object?

您必须来避免C ++ 20之前的未定义行为？是。如果不这样做，编译器实际上会造成伤害吗？

Do you have to per the C++ object model to avoid undefined behavior pre-C++20? Yes. Will any compiler actually cause harm by you not doing this? Not that I'm aware of.

[...]仍然是未定义的行为吗？

[...] Is it still undefined behavior?

是。在C ++ 20之前的版本中，您实际上仍未在任何地方创建 int 对象，因此它是UB。

Yes. Pre-C++20, you still didn't actually create an int object anywhere so this is UB.

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