c++ - 对指针及其地址感到困惑

Question

我在代码中创建了一些指针，但结果不是我所期望的。

这是简单的代码:

int main(int argc, char const* argv[])
{
    int myInt = 23;
    int* ptr = &myInt;

    char* buffer = new char[8];
    memset(buffer, 0, 8);

    char** ptr2 = &buffer;

    std::cout << "ptr address is " << ptr << std::endl;
    std::cout << "buffer pointer is pointing to address " << buffer << std::endl;
    std::cout << "ptr2 pointer is pointing to address " << ptr2 << std::endl;
    std::cout << "Dereferencing ptr2 = " << *ptr2 << std::endl;
    return 0;
} 

下面是运行代码的结果:

ptr address is 0x7ffde215a14c

buffer pointer is pointing to address

ptr2 pointer is pointing to address 0x7ffde215a150

Dereferencing ptr2 =

我想知道为什么缓冲区指针地址不显示以及为什么对 ptr2 的取消引用也没有显示任何内容，但指向缓冲区指针的指针 (ptr2) 是显示那个地址。这些对我来说都毫无意义。

Answer 1

流<<运算符是 explicitly overloaded for all kinds of char* 将其打印为 null-terminated string。要打印指针，您需要对其进行转换:

std::cout << "buffer pointer is pointing to address " << reinterpret_cast<void*>(buffer) << std::endl;