“赋值使指针无需强制转换”是什么意思？意思是？

发布于 2024-12-10 19:39:35 字数 317 浏览 0 评论 0原文

早些时候，我问过类似的问题，但后来我更改了代码。现在编译器给了我一个不同的警告。这是我的代码现在的示例：

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

当我尝试编译时，我收到“警告：赋值使指针来自整数而不进行强制转换。”这是什么意思？我应该做什么来解决它？

澄清一下，我不希望“a”指向比“b”大4个字节的地址（即a！= b+4）。在我的程序中，我知道存储在 ((char *)b + 4) 处的值本身就是另一个指针，我想将该指针存储在“a”中。

原文

Earlier, I asked a similar question, but I've since changed my code. Now the compiler gives me a different warning. This is an example of what my code looks like now:

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

When I try to compile, I get "warning: assignment makes pointer from integer without a cast." What does this mean, and what should I do to fix it?

To clarify, I don't want 'a' to point to an address that is 4 bytes greater than 'b' (i.e., a != b+4). In my program, I know that the value stored at ((char *)b + 4) is itself another pointer, and I want to store this pointer in 'a'.

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聆听风音 2024-12-17 19:39:35

因为，“存储在 ((char *)b + 4) 的值本身就是另一个指针”，只需向结果添加显式转换：

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

因为您假设 sizeof(void *)==sizeof(int)，代码不可移植，但我相信您已经知道了。

一种可移植的替代方法是使用 intptr_t 而不是 int。

Since, "the value stored at ((char *)b + 4) is itself another pointer", simply add an explicit cast to the result:

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

Since you're assuming that sizeof(void*)==sizeof(int), the code is not portable, but I am sure you already knew that.

A portable alternative is to use intptr_t instead of int.

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半边脸i 2024-12-17 19:39:35

那么 b+4 中存储了一个指针值，您想将其加载到 a 中吗？然后将其视为 1，因此 (b+4) 指向一个指针 - 指向 int 指针的指针是 int**：

  *(int**)((char*)b+4)

So there's a pointer value stored at b+4 which you want to load into a? Then treat it as one, and so (b+4) points to a pointer - pointer to a pointer to int is int**.: