c - 为什么带有 %p 指针格式说明符的 printf 在 AVR 上打印垃圾字符而不是地址？

Question

我见过How to find the address of a variable when using AVR? - 结论基本上是:

printf()-debugging on embedded targets is a bit complicated, you need to dig through and figure out if there's code to connect printf() with a serial port, and so on. Also, a full printf() can be a bit heavy, so perhaps you're already using a trimmed-down standard library. It's not the compiler's responsibility, you need to figure out how it's implemented.

...但我真的无法弄清楚为什么会出现这个问题。

我有一个变量，我想将其用作“字符串数组”，并且我想对其进行 malloc 和 realloc:

char** my_array = malloc(sizeof(char*)*1);

然后，我将 printf 重定向到“打印”到 USB 串行端口，然后在串行终端中读取打印输出。

据我所知，printf 的 %p 格式说明符应该将变量的地址打印为十六进制数。我有这样的声明:

printf("my_array %p\r\n", (void *)&my_array);

我也尝试过:

printf("my_array %p\r\n", &my_array);
printf("my_array %p\r\n", my_array);

在所有情况下，我得到的打印输出都是这样的:

my_array  ▒▒▒ꓣ▒▒▒▒/▒▒f'w'▒`$▒▒W*▒▒X▒f'w'▒`$▒▒Y*▒▒Z▒f'w'▒`▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒#▒▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒j▒{▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒i▒z▒▒`$▒▒

...这显然不是一个十六进制数。

为什么会发生这种情况，我做错了什么？如何获得 AVR 中打印为十六进制数的变量地址？如果重要的话，我使用 CodeVisionAVR 编译器...

<小时/>

编辑:发现这个:

https://www.avrfreaks.net/forum/format-specifier-pointers-0

But do people actually use %p and especially on an AVR. On an AVR a pointer is a 16bit unsigned number. I'd generally just use %04X or something myself.

(OK just realised this forum might mean 32bit ARM or UC3 where I guess the pointers are 32 bit? Even so %08X in that case)

...所以我只是用:

printf("my_array %08X\r\n", (void *)&my_array);

...现在我得到打印输出:

my_array 00003FFB

...但现在我不知道这是否是一个实际的地址:)

Answer 1

您可以将指针转换为 uintptr_t 并打印:

printf("my_array %llu\r\n", (unsigned long long)(uintptr_t)&my_array);