Question

这是我守则认为的:

void *a = NULL;
void *b = //something;
a = (void *)(*((char *)b + 4));

上文(b+4)指出的价值是我要储存的地址。当我试图汇编时,我收到“警告:从不同面积的分类中抽取点”。这意味着什么,以及我应该做些什么来纠正?

EDIT:为了澄清,我不想指出4个字塔大于b。在我的节目中,我知道,储存在(果*+4)上的价值本身是另一个点,我想在一中储存这个点。

Answer 1

achar*:

(char *)b + 4

这意味着:

*((char *)b + 4)

页: 1 然后,请将<代码>char 至void*>和汇编者投诉。您不必在C中人工投放避免*,这样就应当:



a = (char *)b + 4;




www.un.org/Depts/DGACM/index_spanish.htm 最新评论: 页: 1

a = *(char **)((char *)b + 4)

Answer 2

it would improve the readability, if you wrote the statements apart I did this for you and get:

char *c = (char *)b;
char *d = c + 4;
a = (void *)
        (*d); // <- this is c char and you want to cast it to a pointer

似乎你写的星号太高。

Answer 3

What it sounds like: something is wrong in your pointers.

在这种情况下,重新审查你的表述:实际上,是<条码>地址<>char/code>?

(void *)(*((char *)b + 4));

so

(char *)b

代表:

(char *)b+4

但随后,*(char *b+4)> 不适用(char *)b+4,以便-,然后由您尝试将这件事(可能是一整件)变成4或8个逐个地址。 Ergo,不匹配的规模。

Answer 4

void *a = NULL;
char jig[10]="jigarkumar";
void *b = jig;
printf("b is %p",b);
a = (void *)(((char *)b + 4));
printf("
a is %p",a);

这里的问题是:

a = (void *)(*((char *)b + 4));
             ^
             |

this * fetch the value stored at b+4 address & it is assigned to a which is wrong so just remove that & your code will work fine