C - charArray 总是与 &charArray 相同吗？

Question

我正在阅读 an article about whole program optimization .链接时代码生成部分的最后一段说，将分配在堆栈上的数组归零可能不会产生相同的效果，具体取决于它的归零方式:

Turning on whole program optimization did uncover several bugs that had undefined behavior. Without WPO, these had somehow not crashed. With WPO, they did. In one case, a member function call was being made through a pointer to uninitialized memory. In several other cases, it was assumed that arrays on the stack were identical to their own addresses. That is, it was assumed that memset(&charArray, 0, sizeof(charArray)) would have the same effect as memset(charArray, 0, sizeof(charArray)). This is not guaranteed by the standard, and appears to change under WPO.

我认为如果我执行 char foo[1] foo 将始终是 == 到 &foo .有人可以解释这里发生了什么吗？谢谢

Answer 1

foo 是一个数组，在表达式中 foo 将转换为指向其第一个元素的指针，除非一元操作数 & 和 sizeof 运算符。因此，在这种情况下 foo == &foo[0]。 &foo是数组foo的地址，不是foo的第一个元素的地址。
foo 和&foo 的值虽然等价，但类型不同。 foo 衰减后是char * 类型，而&foo 是char (*)[1] 类型。