关于指针的C代码

Question

我有这个简单的代码，它使用从 Jon Eriksen 的书中获取的指针，我正在尝试编译它，但是当我运行它时 gcc 在编译和段错误(核心转储)中给我警告。

#include<stdio.h>

int main(){
    int i;
    int int_array[5] = {1, 2, 3, 4, 5};
    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    unsigned int hacky_nonpointer;

    hacky_nonpointer = (unsigned int)  int_array;   

    for(i=0; i < 5; i++){
        printf("[hacky_nonpointer] points to %p which contains the integer %d\n", hacky_nonpointer, *((int *) hacky_nonpointer));
        hacky_nonpointer = hacky_nonpointer + sizeof(int); // hacky_nonpointer = (unsigned int) ((int *) hacky_nonpointer + 1);
    }

    printf("\n\n\n");

    hacky_nonpointer = (unsigned int) char_array;

    for(i=0; i < 5; i++){
        printf("[hacky non_pointer] points to %p which contains the char %c\n", hacky_nonpointer, *((char *) hacky_nonpointer));
        hacky_nonpointer = hacky_nonpointer + sizeof(char); // hacky_nonpointer = (unsigned int *) ((char *) hacky_nonpointer + 1);

     }
}

输出:

command line: "gcc -g -o pointer_types5  pointer_types5.c"

pointer_types5.c: In function ‘main’:

pointer_types5.c:16:24: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
     hacky_nonpointer = (unsigned int)  int_array;

pointer_types5.c:20:103: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
  points to %p which contains the integer %d\n", hacky_nonpointer, *((int *) hacky_nonpointer));

pointer_types5.c:20:47: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘unsigned int’ [-Wformat=]
         printf("[hacky_nonpointer] points to %p which contains the integer %d\n", hacky_nonpointer, *((int *) hacky_nonpointer));

pointer_types5.c:29:24: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
     hacky_nonpointer = (unsigned int) char_array;

pointer_types5.c:35:101: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
 er] points to %p which contains the char %c\n", hacky_nonpointer, *((char *) hacky_nonpointer));

pointer_types5.c:35:48: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘unsigned int’ [-Wformat=]
         printf("[hacky non_pointer] points to %p which contains the char %c\n", hacky_nonpointer, *((char *) hacky_nonpointer));



command line: "./pointer_types5"
Segmentation fault (core dumped)

some more info about my os: 
uname -a : Linux PINGUIN 4.10.0-33-generic #37-Ubuntu SMP Fri Aug 11 10:55:28 UTC 2017 x86_64 x86_64 x86_64 GNU/Linux

Answer 1

这个程序是错误的，书中给出的建议显然也是错误的。指针可以转换为整数类型，但结果是实现定义的。 pointer-to-void 可以转换为 intprt_t/uintptr_t，如果它们存在，并且然后再次返回到指向空的指针 并且您将取回原始指针。这是标准给出的唯一保证。其他东西可能在一个或另一个平台上工作，但它们根本不可移植 - 没有理由在大型程序中使用它们。

%p 需要一个指向 void 的指针作为参数。传递 int 具有未定义的行为。

在程序中尝试的东西根本无法用 C 语言可移植地表达，但可能更正确的程序是

#include <stdio.h>
#include <inttypes.h>

int main(void) {
    int i;
    int int_array[5] = {1, 2, 3, 4, 5};
    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    uintptr_t hacky_nonpointer;

    hacky_nonpointer = (uintptr_t)(void *)int_array;   

    for(i=0; i < 5; i++){
        printf("[hacky_nonpointer] points to %p which contains the integer %d\n", (void *)hacky_nonpointer, *((int *)(void *)hacky_nonpointer));
        hacky_nonpointer = hacky_nonpointer + sizeof(int);
    }

    printf("\n\n\n");

    hacky_nonpointer = (uintptr_t)(void *)char_array;

    for(i=0; i < 5; i++){
        printf("[hacky non_pointer] points to %p which contains the char %c\n", (void *)hacky_nonpointer, *((char *)(void *)hacky_nonpointer));
        hacky_nonpointer = hacky_nonpointer + sizeof(char); // hacky_nonpointer = (unsigned int *) ((char *) hacky_nonpointer + 1);

     }
}

然而，没有什么可以保证 hacky_nonpointer = hacky_nonpointer + sizeof(int); 在每个平台上都像作者预期的那样运行。然而，它可以在许多普通系统上运行，即 Raspberry Pi、普通 ARM 架构、现代 x86-32 和 x86-64 等等。

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这本书不教你 C，它教作者对他认为 C 应该是什么的错误解释。