C中的内存映射地址（如何取消引用）

Question

我想假设 C 语言中的数组是微处理器中的一个内存区域，因此我可以在 PC 上编译一些代码。我编写了一个小程序来尝试使语法正确，但程序无法运行，当我更改访问变量的方式时，它要么崩溃，要么无法编译——太晚了，我不明白为什么.请问这是怎么回事？

// original code in microprocessor header that I need to change if I compile on the host
// BASE is simply a hex value that is later used as an address or a hex value
#define BASE (0x0000)
// used later in header like this (cannot change the way this is done)
#define OFFSET 0x0001
#define PERIPHERAL (BASE + OFFSET)
// also used like (also cannot change):
uint32_t var = PERIPHERAL | HEXMASK;

// here is how I intend to replace the uC specific code
// replace the BASE DEFINE with the next 2 lines of code:

// instead of writing to memory location, write to array of bytes instead, so declare it:    
uint8_t BASE_memory[4] = {0, 0, 0, 0};
// define BASE as hex value that can be used as drop-in replacement in either of the 2 uses shown above
#define BASE ((uint32_t)(BASE_memory))

// now test usage
// access contents of BASE_memory[0]
printf("contents of  BASE_memory[0] == %02x\n", *((uint32_t *)(BASE)));


// now I want to access PERIPHERAL, the second element of the array, i.e. BASE_memory[1]
printf("contents of  BASE_memory[1] == %02x\n", *((uint32_t *)(PERIPHERAL)));

Answer 1

我认为您使用的是 64 位系统。

#include <stdint.h>

uint8_t BASE_memory[4] = {1, 2, 3, 4};

int func1()
{
    return *(uint32_t *) (uint32_t) BASE_memory;
}

int func2()
{
    return *(uint32_t *) (uintptr_t) BASE_memory;
}

这是func1 的程序集输出：

leaq    _BASE_memory(%rip), %rax
movl    %eax, %eax
movl    (%rax), %eax

这是func2 的程序集：

movl    _BASE_memory(%rip), %eax

您可以看到，如果您将地址转换为uint32_t，那么还有一个额外的步骤，将高位设置为零。然后地址是错误的，你会得到一个分段错误。这就是为什么你使用uintptr_t 或intptr_t 而不是uint32_t。