I get really weird output from the last cout statement, no idea what would be causing this:
char ex1_val = 'A';
char* ex1;
ex1 = &ex1_val;
cout << *ex1 << endl;
char** ex2;
ex2 = &ex1;
cout << *ex2 << endl;
Here's my output:
A
A����
ex2 is declared as char**, which means it's a pointer to a pointer to char.
When you print the contents of *ex2, you're dereferencing it once, so type of *ex2 is a pointer to char.
Effectively, the compiler assumes, that you want to print a C-style string, which is a NULL-terminated string. So it prints all bytes stored in memory, starting from A stored in ex1_val, followed by random values stored directly after that, until a NULL character is encoutered.
This is actually a very bad situation, as you never know when or if there's a NULL in memory. You might end up trying to access a memory area you're not allowed to use, which may result in crashing the application. This is an example of an undefined behaviour (UB).
To print out the actual contents of ex1_val via the ex2 pointer, you'd need to make it like this:
cout << **ex2 << endl;
where typeof(**ex2) is an actual char.
When you dereference ex2 (in *ex2) you get a pointer to a char (i.e. char*), and the operator<< overload to handle char* outputs it as a C-style zero-terminated string.
Since *ex2 doesn't point to a zero-terminated string, only a single character, you get undefined behavior.
You want to do cout << **ex2 << endl;
ex2, as you have declared it, as a pointer to a pointer to a character. Doing *ex2 only gets you the address of the pointer it is pointing to.
Related
If cstring is a pointer, then why can it get a value directly? Secondly, Why wasn't the *cstring's result equal to whole of string? Third, cstring is a non-constant pointer to a constant character, so why change its value and not change its address?
#include <cstdio>
#include <conio.h>
#include <iostream>
using namespace std;
int main()
{
const char* cstring = "string";
cout << cstring << endl << *cstring << endl << &cstring << endl;
cstring = "foo";
cout << cstring << endl << *cstring << endl << &cstring << endl;
_getch();
return 0;
}
If cstring is a pointer, then why can it get a value directly?
The operator << of cout for const char* is specialized to behave that way. It will treat the pointer as a NULL terminated string and will print it instead of the pointer value. For different types you get different behaviour. For char* you have the whole string printed.
Why wasn't the *cstring's result equal to whole of string?
That is because the type of *cstring is char and again, operator << behaves correctly by just printing a single char. a const char* is essentially an array of char.An array is essentially a pointer to the first element of the array. If you use the * operator on a pointer you are accessing whatever the pointer points to. If it points to the first element, well, you get the first element.
Third, cstring is a non-constant pointer to a constant character, so why change its value and not change its address?
As you said, cstring is a non-constant pointer to constant data. You cannot change the place it points to (it is a constant pointer), but you can substitute the content of the pointed data with other stuff. You point to the same location but the content of that cell changes.
If cstring is a pointer, then why can it get a value directly?
Anyone can get the value pointed at by a pointer by dereferencing the pointer. That's what happens when you do std::cout << cstring. The proper overload gets chosen that prints the string represented by cstring assuming that is correctly formed, null-terminated C-style string.
Secondly, Why wasn't the *cstring's result equal to whole of
string?
cstring is a const char*, so *cstring is a const char. Pass that to std::cout and it will call an overload that prints one char. The function that is called internally doesn't even know that this is just one char in a string.
Third, cstring is a non-constant pointer to a constant character, so
why change its value and not change its address?
You can't change the address of a variable. cstring is in a fixed place on the stack. You change the value of cstring, which is the address of the string that it's pointing to (it is now pointing to a different string, which has a different address, "string" of course stil has the same address).
What you probably wanted to try is this:
const char* cstring = "string";
std::cout << (void*)cstring << std::endl;
cstring = "foo";
std::cout << (void*)cstring << std::endl;
Now you can see the different addresses. One is the address of "string" and one is the address of "foo".
To recreate this little problem of mine use this code (Yes it's bruteforced and could be slimmed down):
char hello[] = "John";
char *ptr1 = &hello[0];
char *ptr2 = &hello[1];
char *ptr3 = &hello[2];
char *ptr4 = &hello[3];
std::cout << ptr1 << "$";
std::cout << ptr2 << "$";
std::cout << ptr3 << "$";
std::cout << ptr4 << "$";
as you may see after trying this code, it returns the following:
John$ohn$hn$n$
I have no idea why it takes all elements after the specified one.
Any help?
It's because you're printing the pointer, not the thing it points to. And char pointers in C programs often point to the start of a char[] which represents a '\0'-terminated string. That is why std::ostream (of which cout is an instance) has an overloaded << operator that outputs the entire string, up to the terminator.
To print just the character, dereference the pointers first:
std::cout << *ptr1 << "$";
You could also use references instead, which are treated largely as a substitute for the value itself:
char &ptr1 = hello[0];
ptr1 through ptr4 are also pointers to characters, therefore << will treat them as strings, and print them as such, giving you what you saw.
If you just want the characters at each location, lines of the form:
char ch1 = hello[0];
will work, as will changing the output to:
std::cout << *ptr1 << "$";
Your question seems to be based on some completely unfounded expectations. It looks like you expected your code to print a single character for each cout << ... line. But why did you expect it to do that? What you send to cout in your code is actually a char * pointer. Why did you expect cout to dereference that pointer for you and print the character? A normal thing to expect would be to see the pointer value in the output (the address), not the character value.
So, the question you should really be asking is not "why it prints several characters from my array?". The question you should really be asking is "why it prints any of my characters at all?". You never dereferenced your pointer. Yet, instead of pointer value the output contains a sequence of char objects that pointer was pointing to. I.e. apparently cout decided to dereference your pointer for you, even though you never explicitly asked it to. That is the question you should be asking: why did my pointer get dereferenced at all?
And the answer is that [const] char * pointers receive special treatment when sent to cout using << operator. There's a dedicated overloaded version of << operator introduced specifically for that purpose. That version of << operator will treat char * pointers as pointing to the beginning of a C-string. And that operator will output the entire C-string all the way to the zero terminator character. That is exactly what is happening in your experiment.
Note that if you use any other pointer type (like int *) for example, you will not see the pointed value (or values) in the output. You will see the implementation-dependent representation of the pointer itself.
if you want to print value at particular address then :
std::cout << *ptr1 << "$";
std::cout << *ptr2 << "$";
std::cout << *ptr3 << "$";
std::cout << *ptr4 << "$";
I am not entirely sure but if it cout receives a char pointer, it will print, everything until it reaches a blank in memory, (C style character string), and if you just want to print a character, you should print only the content of the variable, that means just pass the content not the memory location...
Why does the following happen?
char str[10]="Pointers";
char *ptr=str;
cout << str << "\n"; // Output : Pointers
int abc[2] = {0,1 };
int *ptr1 = abc;
cout <<ptr1 << "\n"; // But here the output is an address.
// Why are the two outputs different?
As others have said, the reason for the empty space is because you asked it to print out str[3], which contains a space character.
Your second question seems to be asking why there's a difference between printing a char* (it prints the string) and int* (it just prints the address). char* is treated as a special case, it's assumed to represent a C-style string; it prints all the characters starting at that address until a trailing null byte.
Other types of pointers might not be part of an array, and even if they were there's no way to know how long the array is, because there's no standard terminator. Since there's nothing better to do for them, printing them just prints the address value.
1) because str[3] is a space so char * ptr = str+3 points to a space character
2) The << operator is overloaded, the implementation is called depending on argument type:
a pointer to an int (int*) uses the default pointer implementation and outputs the formatted address
a pointer to a char (char*) is specialized, output is formated as a null terminated string from the value it points to. If you want to output the adress, you must cast it to void*
The empty space is actually Space character after "LAB". You print the space character between "LAB" and "No 5".
Your second question: You see address, because ptr1 is actually address (pointer):
int *ptr1;
If you want to see it's first member (0), you should print *ptr1
This might be a stupid question, but I'm new to C++ so I'm still fooling around with the basics. Testing pointers, I bumped into something that didn't produce the output I expected.
When I ran the following:
char r ('m');
cout << r << endl;
cout << &r << endl;
cout << (void*)&r << endl;
I expected this:
m
0042FC0F
0042FC0F
..but I got this:
m
m╠╠╠╠ôNh│hⁿB
0042FC0F
I was thinking that perhaps since r is of type char, cout would interpret &r as a char* and [for some reason] output the pointer value - the bytes comprising the address of r - as a series of chars, but then why would the first one would be m, the content of the address pointed to, rather than the char representation of the first byte of the pointer address.. It was as if cout interprets &r as r but instead of just outputting 'm', it goes on to output more chars - interpreted from the byte values of the subsequent 11 memory addresses.. Why? And why 11?
I'm using MSVC++ (Visual Studio 2013) on 64 bit Win7.
Postscript: I got a lot of correct answers here (as expected, given the trivial nature of the question). Since I can only accept one, I made it the first one I saw. But thanks, everyone.
So to summarize and expand on the instinctive theories mentioned in my question:
Yes, cout does interpret &r as char*, but since char* is a 'special thing' in C++ that essentially means a null terminated string (rather than a pointer [to a single char]), cout will attempt to print out that string by outputting chars (interpreted from the byte contents of the memory address of r onwards) until it encounters '\0'. Which explains the 11 extra characters (it just coincidentally took 11 more bytes to hit that NUL).
And for completeness - the same code, but with int instead of char, performs as expected:
int s (3);
cout << s << endl;
cout << &s << endl;
cout << (void*)&s << endl;
Produces:
3
002AF940
002AF940
A char * is a special thing in C++, inherited from C. It is, in most circumstances, a C-style string. It is supposed to point to an array of chars, terminated with a 0 (a NUL character, '\0').
So it tries to print this, following on in to the memory after the 'm', looking for a terminating '\0'. This makes it print some random garbage. This is known as Undefined Behaviour.
There is an operator<< overload specifically for char* strings. This outputs the null-terminated string, not the address. Since the pointer you're passing this overload isn't a null-terminated string, you also get Undefined Behavior when operator<< runs past the end of the buffer.
Conversely, the void* overload will print the address.
Because operator<< is overloaded based on the data type.
If you give it a char, it assumes you want that character.
If you give it a void*, it assumes you want an address.
However, if you give it a char*, it takes that as a C-style string and attempts to output it as such. Since the original intent of C++ was "C with classes", handling of C-style strings was a necessity.
The reason you get all the rubbish at the end is simply because, despite your assertion to the compiler, it isn't actually a C-style string. Specifically, it is not guaranteed to have a string-terminating NUL character at the end so the output routines will just output whatever happens to be in memory after it.
This may work (if there's a NUL there), it may print gibberish (if there's a NUL nearby), or it may fall over spectacularly (if there's no NUL before it gets to memory it cannot read). It's not something you should rely on.
Because there's an overload of operator<< which takes a const char pointer as it's second argument and prints out a string. The overload that takes a void pointer prints only the address.
A char * is often - usually even - a pointer to a C-style null-terminated string (or a string literal) and is treated as such by ostreams. A void * by contrast unambiguously indicates a pointer value is required.
The output operator (operator<<()) is overloaded for char const* and void const*. When passing a char* the overload for char const* is a better match and chosen. This overload expects a pointer to the start of a null terminated string. You give it a pointer to an individual char, i.e., you get undefined behavior.
If you want to try with a well-defined example you can use
char s[] = { 'm', 0 };
std::cout << s[0] << '\n';
std::cout << &s[0] << '\n';
std::cout << static_cast<void*>(&s[0]) << '\n';
I understand from here that the name of an array is the address of the first element in the array, so this makes sense to me:
int nbrs[] = {1,2};
cout << nbrs << endl; // Outputs: 0x28ac60
However, why is the entire C-string returned here and not the address of ltrs?
char ltrs[] = "foo";
cout << ltrs << endl; // Outputs: foo
Because iostreams have an overload for char * that prints out what the pointer refers to, up to the first byte that contains a \0.
If you want to print out the address, cast to void * first.
cout has operator<<() overloaded for char* arrays so that it outputs every element of the array until it reaches a null character rather than outputting the address of the pointer
cout, and generally, C++ streams, can handle C strings in a special way. cout operators <<, >> are overloaded to handle a number of different things, and this is one of them.