为何C++代码中cout打印char指针/数组输出字符串而非地址？

Ah, this is a classic gotcha with C++'s std::cout and C-style strings! Let me break down exactly what's going on here:

The core reason: std::cout has a special overload for char* #

Normally, when you pass a pointer (like an int* or double*) to cout, it will print the memory address that the pointer holds. But for char* specifically, the standard library provides an overloaded version of operator<< that treats the pointer as a C-style string.

What this means is: instead of printing the address, cout will start reading characters from the memory location the pointer points to, and keep printing until it hits a null terminator ('\0').

Why your code outputs "mani" #

Looking at your code:

int main(){ 
    char *q; 
    char b[5]={'m','a','n','i'}; 
    q=&b[0]; 
    cout<<b<<endl; 
    cout<<q<<endl;
}

• You declared b as a 5-element char array, but only initialized 4 characters. In C++, when you initialize an array with fewer elements than its size, the remaining elements are value-initialized — for char, that means they get set to '\0' (the null terminator). So your array b actually looks like this in memory: {'m','a','n','i','\0'}.
• q points to the first element of b (b[0]), which is the start of this valid C-style string. When you run cout << q, it reads from that address, prints 'm', 'a', 'n', 'i', then stops at the '\0' — hence you see "mani" instead of the address.

How to print the address instead #

If you want to see the actual memory address stored in q, you need to bypass the char* overload by casting the pointer to void*. cout doesn't have a special overload for void*, so it will print the address as expected:

cout << static_cast<void*>(q) << endl;

内容的提问来源于stack exchange，提问作者Manish Sharma