C++指针输出一致性疑问：学习指针时的代码运行问题求助

Hey there! Let's dig into why that last line of your code is giving you confusing results—this is a common pitfall when working with pointer type casts in C++.

First, let's recap what your code is doing step by step:

• You declare an int x = 10 and a pointer p pointing to its address.
• You declare a char c2 = 'c' (which stores the ASCII value 99 in 1 byte of memory) and a pointer p2 pointing to it.
• You cast p2 to a void* (this is correct for printing the address, since cout treats char* as a string pointer instead of an address).
• The problematic line is this one:

  cout<<"Content of address in p2/p3:"<< *(static_cast<int *>(p3))<<endl;
  

Why This Line Causes Unexpected Output #

There are two core issues here:

1. Mismatched Type Size & Memory Access
   A char typically takes up 1 byte of memory, while an int usually takes 4 bytes (this depends on your platform, but it's the most common case). When you cast p3 (which points to a 1-byte char) to an int* and dereference it, you're telling the compiler to read 4 consecutive bytes starting from c2's address. But only the first byte holds valid data (99 for 'c'); the other 3 bytes are uninitialized stack memory (could be leftover values from other variables, random garbage, or even invalid memory).

2. Undefined Behavior
   The C++ standard defines this kind of cross-type pointer cast and dereference as undefined behavior. That means there's no guarantee what your program will do—you might get a random number, a value that changes between runs, or even a crash in some cases. The compiler doesn't have to handle this scenario predictably.

The Fix #

If you want to correctly read the value stored in c2, you need to cast p3 back to the original type (char*) instead of int*:

cout<<"Content of address in p2/p3:"<< *(static_cast<char *>(p3))<<endl;

This will only read the 1 byte that actually holds c2's value, so you'll see c printed (or 99 if you cast the result to int like static_cast<int>(*(static_cast<char *>(p3)))).

A Quick Note on void* #

Your use of static_cast<void*>(p2) to print the address is totally correct—since cout will try to print a char* as a null-terminated string, casting to void* forces it to display the raw memory address instead.

内容的提问来源于stack exchange，提问作者N00bie