operator new：void*到T*的转换

在C++14标准(C++98/11也一样)中，在Annex C Compatibility里有这么一条：

Change: Converting void* to a pointer-to-object type requires casting

char a[10];
void* b=a;
void foo() {
  char* c=b;
}

ISO C will accept this usage of pointer to void being assigned to a pointer to object type. C ++ will not.

但是为什么operator new()会返回void*且不用显式转换为T*就能赋值给T*呢？

global scope的operator new基本原型为：

1 2	void* operator new ( std::size_t count ); void* operator new[]( std::size_t count );

这么写不会有任何警告：

1
2
3

struct A{};
A* ap=new A;
delete ap;

自己在类内重载一个operator new返回void*也不会有问题：

struct A{
  void* operator new(std::size_t x){ return malloc(x); }
};

A* ap=new A; // OK
delete ap;

但是使用std::malloc来分配再赋值就有问题：

1 2	// error: cannot initialize a variable of type 'A ' with an rvalue of type 'void ' A* ap=std::malloc(sizeof(A));

编译器也是支持C++14标准的Clang 3.9…

我通过使用LLVM-IR分析发现，调用operator new时编译器会自动将void*转换为T*:

struct A{};

int main()
{
  A *z=new A;
  delete z;
}

其LLVM-IR为：

define i32 @main() #4 {
  %1 = alloca i32, align 4
  %2 = alloca %struct.A*, align 8
  store i32 0, i32* %1, align 4
  %3 = call i8* @_Znwy(i64 1) #7
  %4 = bitcast i8* %3 to %struct.A*
  store %struct.A* %4, %struct.A** %2, align 8
  %5 = load %struct.A*, %struct.A** %2, align 8
  %6 = icmp eq %struct.A* %5, null
  br i1 %6, label %9, label %7

; <label>:7:                                      ; preds = %0
  %8 = bitcast %struct.A* %5 to i8*
  call void @_ZdlPv(i8* %8) #8
  br label %9

; <label>:9:                                      ; preds = %7, %0
  %10 = load i32, i32* %1, align 4
  ret i32 %10
}
; Function Attrs: nobuiltin
declare noalias i8* @_Znwy(i64) #5

; Function Attrs: nobuiltin nounwind
declare void @_ZdlPv(i8*) #6

其中最重要的就是%4的操作：

# 调用new获得内存地址
%3 = call i8* @_Znwy(i64 1) #7
# 将上一步获得的地址转换为struct.A*
%4 = bitcast i8* %3 to %struct.A*
# 将转换后的地址存储为我们声明的指针
store %struct.A* %4, %struct.A** %2, align 8

经过测试，自己在类内重载operator new也会具有转换。

struct A{
  void* operator new(std::size_t x){ return malloc(x); }
};

int main()
{
  A *z=new A;
  delete z;
}

其LLVM-IR代码为：

define i32 @main() #4 {
  %1 = alloca i32, align 4
  %2 = alloca %struct.A*, align 8
  store i32 0, i32* %1, align 4
  %3 = call i8* @_ZN1AnwEy(i64 1)
  %4 = bitcast i8* %3 to %struct.A*
  store %struct.A* %4, %struct.A** %2, align 8
  %5 = load %struct.A*, %struct.A** %2, align 8
  %6 = icmp eq %struct.A* %5, null
  br i1 %6, label %9, label %7

; <label>:7:                                      ; preds = %0
  %8 = bitcast %struct.A* %5 to i8*
  call void @_ZdlPv(i8* %8) #6
  br label %9

; <label>:9:                                      ; preds = %7, %0
  %10 = load i32, i32* %1, align 4
  ret i32 %10
}

; Function Attrs: uwtable
define linkonce_odr i8* @_ZN1AnwEy(i64) #0 comdat align 2 {
  %2 = alloca i64, align 8
  store i64 %0, i64* %2, align 8
  %3 = load i64, i64* %2, align 8
  %4 = call i8* @malloc(i64 %3)
  ret i8* %4
}

; Function Attrs: nobuiltin nounwind
declare void @_ZdlPv(i8*) #5

declare i8* @malloc(i64) #1

可以看到%4处仍然具有void*到struct.A*的转换。
我没有在标准中找到operator new应该会转换void*到T*类型的显式要求，但是标准中提到了下面这样的描述：

The pointer returned shall be suitably aligned so that it can be converted to a pointer of any complete object type with a fundamental alignment requirement (3.11) and then used to access the object or array in the storage allocated (until the storage is explicitly deallocated by a call to a corresponding deallocation function).

循迹研究室

operator new：void到T的转换