Question 193128 in Ubuntu Question posted in
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Ubuntu – why cout &char operation same expresion not return the same answer

#include "iostream"

using namespace std;

int main() {
    char c = 'x';
    cout << &c << endl;
}

cout << &c << endlprint ‘x’

2.

#include "iostream"

using namespace std;

int main() {
    int x = 222222;
    char c = 'x';
    cout << &c << endl;
    cout << x << endl;
}

cout << &c << endl;print ‘xd’,seems the adress of the variable

os version

Linux vm 5.19.0-45-generic #46-Ubuntu SMP PREEMPT_DYNAMIC Wed Jun 7 09:08:58 UTC 2023 x86_64 x86_64 x86_64 GNU/Linux

gcc version

gcc version 12.2.0 (Ubuntu 12.2.0-3ubuntu1)

the expression are the same ,why cout different?

Tags:

3

Answers


  1. 0

    if you are trying to store the address of x in c, this would be right way to do that:

    int main() {
    int x = 222222; // creates an int varriable
    int *c = &x; //creates a pointer to hold address of an int and assings address of x to the pointer 
    cout << c << endl; // prints the pointer value which is address of x
    cout << x << endl; // prints the value of x

}

    now if you want to use the address to retreive the value of x then this line will do that:

    cout << *c << endl; //prints value stored in the address c is pointing at which is value of x

    Hope this helps to understand the concept.

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  2. 0

    The problem with this

    char c = 'x';
cout << &c << endl;

    is that, since &c is a char*, it will use this operator<<:

    template< class Traits >
basic_ostream<char, Traits>&
    operator<<( basic_ostream<char, Traits>& os, const char* s );

    That function will use std::char_traits<char>::length(reinterpret_cast<const char*>(s)) to figure out how many characters to insert into the stream. That function in turn

    • return the length of the character sequence pointed to by s, that is, the position of the terminating null character.

    Since you only have a single char (which is not ), the length function will read whatever lies after x in memory. This has undefined behavior, so it could crash or find a somewhere and insert a lot of garbage into the stream.

    If you want to print its address, cast to a void*:

    std::cout << static_cast<void*>(&c) << 'n';
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  3. 0

    You’re reading out of bounds of c and therefore your program’s behavior is undefined.

    When a char* is passed to std::cout‘s << operator, it treats that pointer as a c-style nul-terminated string. That means it will keep reading characters from successive bytes starting at the byte pointed to by the pointer until it finds a nul (0) byte.

    But a single char isn’t nul-terminated, so cout will try to read data from after the char object, resulting in undefined behavior. In practice it will end up wandering into the bytes that represent the int 222222 and interpreting them as characters. Eventually it will find a nul byte and stop (hopefully before it wanders out of your process’s currently mapped address space and crashes).

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