Printing a pointer with <iostream> [duplicate]

问题

This question already has answers here:

cout << with char* argument prints string, not pointer value (6 answers)

Closed 2 years ago.

Why does

#include <iostream>
using namespace std;

int main() {
  cout << (char*)0x10 << endl; 
}

segfault, but

#include <iostream>
using namespace std;

int main() {
  cout << (void*)0x10 << endl; 
}

seems to work just fine?

回答1:

Because

cout::operator <<(void*) 

prints a memory address, and

cout::operator <<(char*)

prints a null-terminated character array, and you run into undefined behaviour when you attempt to read the char array from 0x10.

回答2:

The ostream::operator<< is overloaded, there is a version for char* which interprets the given pointer as a null-terminated string.

回答3:

There's a special overload for << with char*, so that C-style strings can be output easily.

Thus

cout << (char*)0x10 << endl; 

tries to print out the string located at (char*)0x10 which is not memory it's supposed to look at.

来源：https://stackoverflow.com/questions/11749725/printing-a-pointer-with-iostream