(int*) &var 是什么意思? [英] What does (int*) &var mean?

问题描述

(int*) &i是什么意思?

  char i;
  int* p = (int*) &i;
  ...
  *p = 1234567892;
  ...

如果是 * &i，我会理解.但在这种情况下，这是一个int".

If it was * &i, I would understand. But in this case, this an "int" in there.

推荐答案

&i : 表示取 i 的地址(这是一个字符*)

&i : means to take the address of i (which is a char*)

(int*)&i :将该指针强制转换为指向整数的指针(这样做不好/错误，但您告诉编译器这样做它甚至不会发出警告)

(int*)&i : casts that pointer to be a pointer to integer (which is bad/wrong to do, but you told the compiler to do it so it won't even give a warning)

int* p = (int*)&i; :声明将 i 的指针存储在 p 中(并且也将其强制转换:编译器甚至不会抱怨)

int* p = (int*)&i; : a statement that says to store the pointer of i in p (and cast it too: the compiler won't even complain)

*p = 1234567892; :将这个值写入 p 指向的 base 位置的几个字节(尽管 p认为它指向一个int，是char！).其中一个字节将在 i 中结束，但其他字节将覆盖与 i 相邻的字节.

*p = 1234567892; : write this value, which is several bytes to the base location pointed to by p (which although p thinks it points to an int, is to char!). One of those bytes will end up in i, but the others will over write the bytes neighboring i.

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