C++:将指针转换为 int 然后再转换回指针是否安全? [英] C++: Is it safe to cast pointer to int and later back to pointer again?

问题描述

将指针转换为 int 然后再转换回指针是否安全?

Is it safe to cast pointer to int and later back to pointer again?

如果我们知道指针是 32 位长，int 是 32 位长呢?

How about if we know if the pointer is 32 bit long and int is 32 bit long?

long* juggle(long* p) {
    static_assert(sizeof(long*) == sizeof(int));
    int v = reinterpret_cast<int>(p); // or if sizeof(*)==8 choose long here
    do_some_math(v); // prevent compiler from optimizing
    return reinterpret_cast<long*>(v);
}

int main() {
    long* stuff = new long(42);
    long* ffuts = juggle(stuff); 
    std::cout << "Is this always 42? " << *ffuts << std::endl;
}

这是否包含在标准中?

推荐答案

没有

例如，在 x86-64 上，一个指针是 64 位长，但 int 只有 32 位长.将指针转换为 int 并再次返回会使指针值的高 32 位丢失.

For instance, on x86-64, a pointer is 64-bit long, but int is only 32-bit long. Casting a pointer to int and back again makes the upper 32-bit of the pointer value lost.

如果你想要一个保证与指针一样长的整数类型，你可以在 中使用 intptr_t 类型.您可以安全地从指向 intptr_t 的指针重新解释转换并返回.

You may use the intptr_t type in <cstdint> if you want an integer type which is guaranteed to be as long as the pointer. You could safely reinterpret_cast from a pointer to an intptr_t and back.

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