C 指针：它们具有相同的功能吗？
假设您有一个“void *a”和“void *ptr”指向代码中定义的不同地址。然后我想知道这两行是否等效且功能相同？
*((unsigned **)((char*)ptr+4)) = a;
第二
*((unsigned *)((char*)ptr+4)) = a;
个抛出警告“赋值从指针生成整数而无需强制转换”
此外，它是否也与上面的操作相同？：
*((char*)ptr+4) = a;
Let's say you have a "void *a" and "void *ptr" that point to different addresses defined in your code. Then I was wondering if these two lines were equivalent and functionally the same?
*((unsigned **)((char*)ptr+4)) = a;
and
*((unsigned *)((char*)ptr+4)) = a;
The second one throws a warning that "assignment makes integer from pointer without a cast"
Also, would it also be the same as the above to just do?:
*((char*)ptr+4) = a;
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这些并不等同。 #1 将
(char*)ptr+4解析为指向无符号 (*unsigned) 的指针，而 #2 将其解析为unsigned。a是一个 void 指针，因此它可以转换为*unsigned，但不能转换为unsigned（隐式），这就是您收到警告的原因。#3 将其解析为
char，这也会产生警告。These are not equivalent. #1 resolves
(char*)ptr+4to be a pointer to unsigned (*unsigned), while #2 resolves it to beunsigned.ais a void pointer, so it can be casted to*unsigned, but not tounsigned(implicitly), that's why you get the warning.The #3 resolves the same to a
char, which would also yield a warning.