Question

我对C相当新，但我想知道，我们可以像无效指针一样使用像unsigned int这样的东西。我写了这个简短的例子来说明我想做什么，但我似乎无法让它工作，所以我想知道它是否有可能：

int main () {

    int i;

    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    int int_array[5] = {1, 2, 3, 4, 5};

    unsigned int unsigned_int;

    unsigned_int = (unsigned int) int_array;

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

    }

    unsigned_int = (unsigned int) char_array;

    for(i=0; i<5 ; i++) {
            printf("[unsigned_int] points to %p, which contains the char '%c'\n", unsigned_int, *((char *) unsigned_int));
            unsigned_int = unsigned_int + sizeof(char);
    }

}

上面的代码是我想要的工作，但下一个例子是我希望它的工作原理：

int main () {

int i;

int int_array[5] = {1, 2, 3, 4, 5};
char char_array[5] = {'a', 'b', 'c', 'd', 'e'};

void *void_pointer;

void_pointer = (void *) int_array;

for(i=0; i < 5; i++) {
printf("[integer pointer] has an adress of %p, and contains %d\n", void_pointer, *((int *)void_pointer));
void_pointer = (void *) ((int *)void_pointer + 1);
}

void_pointer = (void *)char_array;
for(i=0; i<5; i++) {
printf("[char pointer] has an adress of %p and contans %c\n", void_pointer, *((char *)void_pointer));
void_pointer = (void *) ((char *)void_pointer + 1);
}

}

我明白我试图做的事情可能是非传统的，但我真的想知道我是否能做到这一点。当我尝试编译第一个时，我得到一个错误，上面写着：

pointer_type5.c: In function ‘main’:
pointer_type5.c:12:24: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
         unsigned_int = (unsigned int) int_array;
                    ^
pointer_type5.c:15:92: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
         printf("[unsigned_int] points to %p, which contains the int %d\n", unsigned_int, *((int *) unsigned_int));
                                                                                        ^
 pointer_type5.c:20:24: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
         unsigned_int = (unsigned int) char_array;
                    ^
pointer_type5.c:23:103: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
                 printf("[unsigned_int] points to %p, which contains the char '%c'\n", unsigned_int, *((char *) unsigned_int));

当我跑第二个时，我得到了这个：

[integer pointer] has an adress of 0x7fffcba7a5f0, and contains 1
[integer pointer] has an adress of 0x7fffcba7a5f4, and contains 2
[integer pointer] has an adress of 0x7fffcba7a5f8, and contains 3
[integer pointer] has an adress of 0x7fffcba7a5fc, and contains 4
[integer pointer] has an adress of 0x7fffcba7a600, and contains 5
[char pointer] has an adress of 0x7fffcba7a5e0 and contans a
[char pointer] has an adress of 0x7fffcba7a5e1 and contans b
[char pointer] has an adress of 0x7fffcba7a5e2 and contans c
[char pointer] has an adress of 0x7fffcba7a5e3 and contans d
[char pointer] has an adress of 0x7fffcba7a5e4 and contans e

在考虑了您的回答后，我将原始代码更改为：

int main () {

    int i;

    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    int int_array[5] = {1, 2, 3, 4, 5};
    unsigned int *unsigned_int;

    unsigned_int = (unsigned int *) int_array;

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

    unsigned_int = (unsigned int *) char_array;

    for(i=0; i<5 ; i++) {
            printf("[unsigned_int] points to %p, which contains the char '%c'\n", unsigned_int, *((char *) unsigned_int));
           unsigned_int = (unsigned int *)((char *) unsigned_int + sizeof(char));
    }

}

这不完全是我想做的事情，但我想我想做的事情是不能完成的，所以这段代码服务于同样的输出。编辑代码让我想到了下一个小问题。我们为什么可以使用：

  unsigned_int = (unsigned int *)((char *) unsigned_int + sizeof(char));

当我们用int尝试它时，它一次跳过太多字节并给我们随机数。要解决这个问题，我只需将其更改为：

 unsigned_int = (unsigned int *) ((int *) unsigned_int + 1);

但为什么它不适用于sizeof（int）？

Answer 1

一般来说，没有。无法保证任何指针都可以在unsigned int中表示。从指针到整数的转换是实现定义的，可能毫无意义。实际上它通常是有意义的，但是你需要一个具有足够多的值位的整数类型来表示所有指针。 uintptr_t（来自stdint.h），如果已定义，则为此类型。

我们可以在C

1 个答案: