如何确认指向常量字符的指针已被重新分配至新地址

问题描述 #

我正在学习C++，遇到以下问题：将指向常量的指针重新赋值给新字符变量后，无法显示预期的地址值。
在提供的代码中，我原本认为执行pointer_to_const = &c;后，指针应显示与指向字符变量b时不同的新地址，但运行代码后却未得到预期结果。
Constant Pointer部分的输出符合预期，但Pointer to Const部分存在两个疑问：

1. 执行pointer_to_const = &c;后，用cout输出pointer_to_const时看不到新地址，难道c没有唯一地址吗？
2. 在Pointer to Const部分可以用cout显示&pointer_to_const，但无法直接正确显示&b或&c。
   我参考C语言的实现方式，尝试在C++中完成该操作，以此学习相关知识。

代码示例 #

#include <iostream>

int main(){
    char a = 'a';
    char b = 'b';
    char c = 'c';
  
    // Constant Pointer 
    std::cout << "-------Constant Pointer---------" << std::endl;
    char *const constant_pointer = &a;
    std::cout << "Constant pointer a has an address of: " << static_cast<void*>(constant_pointer) << " and value of: " << *constant_pointer << std::endl;
    *constant_pointer = 'x';
    std::cout << "Constant pointer a has an address of: " << static_cast<void*>(constant_pointer) << " and a new value of: " << *constant_pointer << std::endl;

     // Pointer to Const 
    std::cout << std::endl;
    std::cout << "--------Pointer to Const--------" << std::endl; 
    const char *pointer_to_const = &b;
    // **Edit** - section updated to properly display address pointed to by pointer_to_const 
    std::cout << "Pointer to const has an address of: " << static_cast<const void*>(pointer_to_const) << " and value of: " << *pointer_to_const << std::endl;
    pointer_to_const = &c;
    std::cout << "Pointer to const has a new address of: " << static_cast<const void*>(pointer_to_const) << " and value of: " << *pointer_to_const << std::endl;
    std::cout << std::endl;
}

运行输出 #

-------Constant Pointer---------(works as expected)
Constant pointer a has an address of: 0x7ff7b1bc749f and value of: a
Constant pointer a has an address of: 0x7ff7b1bc749f and a new value of: x

--------Pointer to Const--------(not sure why I don't get a new address for c)
Pointer to const has an address of: 0x7ff7b1bc7488 and value of: b
Pointer to const has a new address of: 0x7ff7b1bc7488 and value of: c

--------Pointer to Const--(CORRECTED OUTPUT BASED ON USER FEEDBACK)
Pointer to const has an address of: 0x7ff7b54a844e and value of: b
Pointer to const has a new address of: 0x7ff7b54a844d and value of: c

问题解答 #

疑问1：地址未更新的假象 #

你最初看到地址没变化，是因为代码修改前没有对pointer_to_const做正确的类型转换。const char*类型的指针直接用cout输出时，会被当作C风格字符串处理，而非输出内存地址。你后来添加的static_cast<const void*>(pointer_to_const)才是正确写法——将指针转换为void*类型后，cout才会输出它指向的内存地址。

从修正后的输出可以看到，转换类型后，pointer_to_const指向b和c的地址确实不同（0x7ff7b54a844e和0x7ff7b54a844d），说明c有唯一的内存地址，只是之前的输出方式错误导致了误解。

疑问2：无法直接显示&b/&c的原因 #

&b和&c的类型是char*（或const char*），和pointer_to_const一样，直接用cout输出时会被解析为字符串指针。如果b或c所在内存后没有终止符\0，cout会持续输出内存内容直到遇到\0，导致乱码或错误，而非输出地址。

正确的做法是将它们强制转换为void*类型，示例代码如下：

std::cout << "Address of b: " << static_cast<void*>(&b) << std::endl;
std::cout << "Address of c: " << static_cast<void*>(&c) << std::endl;

而&pointer_to_const是指针自身的地址，类型为const char**，不会被cout当作字符串处理，所以能直接输出地址。

关键知识点 #

• char*/const char*类型指针，cout默认按C风格字符串输出，而非内存地址；
• 要输出指针的内存地址，必须强制转换为void*类型；
• 指向常量的指针（const char*）可修改指向，与常量指针（char* const，不可修改指向）是完全不同的概念。

内容的提问来源于stack exchange，提问作者Stevo