cout with pointers c++

Question

我不明白为什么cout不工作在这个代码:

#include<iostream>
using namespace std;
int main() {
int v = 65;
int* q = &v;
char** o = (char**)&q;
cout <<  o << endl;  // output: 012FFCAC
cout <<  *o << endl; // output: A
cout << **o << endl; // output: A
printf("%c",*o);     // cause an error
printf("%p",*o);    // works and the output=&v

cout在此代码中不起作用

#include<iostream>
using namespace std;
int main() {
char v = 65;
cout <<  &v << endl;  // output: A╠╠╠╠▀?╞«4°O 

Answer 1

因为int(通常)是4个字节，65将非常整齐地放在第一个字节中，而为int分配的其余内存将是0，这种字节模式恰好与字符串在内存中的存储方式非常接近。

所以当通过char*访问内存时，大多数情况下它会打印A，即使大多数打印都是错误的

int v = 65;  // Byte pattern on most machines [65,0,0,0] 
int* q = &v; 
char** o = (char**)&q; // *o now points to [65,0,0,0] ==> ['A','','',''] ==> "A"
std::cout <<  o << std::endl;  
// input:      char** 
// printed as: pointer 
// result:     012FFCAC
std::cout <<  *o << std::endl;  // Undefined Behaviour
// input:      char*
// printed as: null terminated string
// result:     "A"
std::cout <<  **o << std::endl; // Undefined Behaviour 
// input:      char
// printed as: single character
// result:     'A'
printf("%c",*o);    // %c expects a [char] type but is given [pointer to char]      ERROR
printf("%p",*o);    // %p expects a [pointer] type and is given a [pointer to char] OK

由于char(通常)是1字节，因此没有空终止，一旦开始打印就会停止打印，并且它会继续打印内存中的任何内容，直到遇到0或访问冲突

char v = 65;
std::cout << &v << std::endl;  // &v is not a well-formed null terminated string: Undefined Behaviour
// input:      char*
// printed as: null terminated string
// result:     "A<whatever other data is around v in memory>"