类型uint64_t的参数与void*类型的参数不兼容。

I have a function foo(void* pBuf). I need to pass it a 64 bit address but I can't seem to get the right typecast when I'm passing by value.

我有一个函数foo(void* pBuf)我需要传递它一个64位的地址，但是当我传递值时，我似乎不能得到正确的类型广播。

Example: foo(address). Where- uint64_t address=0x00000000DEADBEEF

例如:foo(地址)。- uint64_t地址= 0 x00000000deadbeef哪里

EDIT: Compiling using an ARM compiler.

编辑:使用ARM编译器编译。

uint64_t foo(void *pBuf){
    uint64_t retAddr = (uint64_t) pBuf;
    retAddr += 0x100000;
    return retAddr;
}

I'm on a 32-bit ARM and sizeof(void *) is 4

我的手臂是32位的，sizeof(void *)是4

Clarification: Why I needed a 64-bit address on a 32-bit ARM? Because my memory map uses 36-bit addressing.

说明:为什么我需要一个64位的地址在一个32位的ARM上?因为我的内存映射使用36位寻址。

3 个解决方案

#1

Call it this way:

叫它:

uint64_t address = 0xDEADBEEF;
foo((void*)address);

That is, you cast the address to a void-pointer to be compatible with the function signature.

也就是说，您将地址转换为与函数签名兼容的void指针。

#2

You should not use a 64-bits type for an address, as it is undefined behavior for 32-bits (or any non-64 bits) systems.

您不应该使用64位类型的地址，因为它是32位(或任何非64位)系统的未定义行为。

Rather, prefer using uintptr_t, which is standard C. See this question for more details or this page for references.

相反，更喜欢使用uintptr_t，它是标准的c。

Then a solution could be :

那么解决方案可以是:

uintptr_t address = 0xDEADBEEF;   /* will trigger a warning if the constant is > max possible memory size */
foo((void*)address);

Note : if uintptr_t is not available on your system, size_t is usually a good second choice.

注意:如果uintptr_t在您的系统上不可用，size_t通常是一个很好的第二选择。

Part 2 :

Looks like, in your rephrased question, you want to convert an address into a 64-bits integer.

看起来，在您的重新措辞的问题中，您希望将一个地址转换成一个64位整数。

In which case, a direct cast from ptr to integer is likely to trigger a compiler warning, due to potential differences in wideness.

在这种情况下，从ptr到integer的直接转换很可能会触发编译器警告，这是由于wirant中的潜在差异造成的。

Prefer a double cast : uint64_t value = (uint64_t)(size_t) ptr;

选择双铸:uint64_t值= (uint64_t)(size_t) ptr;

#3

I can think of two ways to get this right. Got a solution to my problem by calling foo the first way

我可以想出两种方法来解决这个问题。用第一种方法调用foo来解决我的问题

foo((void*)(uint32_t)address)
foo((void *)(uint32_t)地址)

This works only because my input to foo is always a 32-bit value. The returned value can be 64-bit.

这是因为我对foo的输入总是一个32位的值。返回的值可以是64位的。

Of course, a proper fix would be to change foo itself, if I could modify it. I could just pass foo(&address). Inside foo, retAddr = *pBuf.
当然，如果我可以修改foo，一个合适的修正就是修改foo本身。我可以传递foo(&address)。在foo中，retAddr = *pBuf。

Thanks for all the suggestions!

谢谢你的建议!

#1