I know they are different, I know how they are different and I read all questions I could find regarding char* vs char[]
But all those answers never tell when they should be used.
So my question is:
When do you use
const char *text = "text";
and when do you use
const char text[] = "text";
Is there any guideline or rule?
As an example, which one is better:
void withPointer()
{
const char *sz = "hello";
std::cout << sz << std::endl;
}
void withArray()
{
const char sz[] = "hello";
std::cout << sz << std::endl;
}
(I know std::string is also an option but I specifically want to know about char pointer/array)
Both are distinctly different, For a start:
The First creates a pointer.
The second creates an array.
Read on for more detailed explanation:
The Array version:
char text[] = "text";
Creates an array that is large enough to hold the string literal "text", including its NULL terminator. The array text is initialized with the string literal "text".The array can be modified at a later time. Also, the array's size is known even at compile time, so sizeof operator can be used to determine its size.
The pointer version:
char *text = "text";
Creates a pointer to point to a string literal "text". This is faster than the array version, but string pointed by the pointer should not be changed, because it is located in an read only implementation defined memory. Modifying such an string literal results in Undefined Behavior.
In fact C++03 deprecates use of string literal without the const keyword. So the declaration should be:
const char*text = "text";
Also,you need to use the strlen() function, and not sizeof to find size of the string since the sizeof operator will just give you the size of the pointer variable.
Which version is better?
Depends on the Usage.
If you do not need to make any changes to the string, use the pointer version.
If you intend to change the data, use the array version.
EDIT: It was just brought to my notice(in comments) that the OP seeks difference between:
const char text[] and const char* text
Well the above differing points still apply except the one regarding modifying the string literal. With the const qualifier the array test is now an array containing elements of the type const char which implies they cannot be modified.
Given that, I would choose the array version over the pointer version because the pointer can be(by mistake)easily reseated to another pointer and the string could be modified through that another pointer resulting in an UB.
Probably the biggest difference is that you cannot use the sizeof operator with the pointer to get the size of the buffer begin pointed to, where-as with the const char[] version you can use sizeof on the array variable to get the memory footprint size of the array in bytes. So it really depends on what you're wanting to-do with the pointer or buffer, and how you want to use it.
For instance, doing:
void withPointer()
{
const char *sz = "hello";
std::cout << sizeof(sz) << std::endl;
}
void withArray()
{
const char sz[] = "hello";
std::cout << sizeof(sz) << std::endl;
}
will give you very different answers.
In general to answer these types of questions, use the one that's most explicit.
In this case, const char[] wins because it contains more detailed information about the data within -- namely, the size of the buffer.
Just a note:
I'd make it static const char sz[] = "hello";. Declaring as such has the nice advantage of making changes to that constant string crash the program by writing to read-only memory. Without static, casting away constness and then changing the content may go unnoticed.
Also, the static lets the array simply lie in the constant data section instead of being created on the stack and copied from the constant data section each time the function is called.
If you use an array, then the data is initialized at runtime. If you use the pointer, the run-time overhead is (probably) less because only the pointer needs to be initialized. (If the data is smaller than the size of a pointer, then the run-time initialization of the data is less than the initialization of the pointer.) So, if you have enough data that it matters and you care about the run-time cost of the initialization, you should use a pointer. You should almost never care about those details.
I was helped a lot by Ulrich Drepper's blog-entries a couple of years ago:
so close but no cigar and more array fun
The gist of the blog is that const char[] should be preferred but only as global or static variable.
Using a pointer const char* has as disadvantages:
An additional variable
pointer is writable
an extra indirection
accessing the string through the pointer require 2 memory-loads
Just to mention one minor point that the expression:
const char chararr[4] = {'t', 'e', 'x', 't'};
Is another way to initialize the array with exactly 4 char.
Related
I'm trying to set a pointer array to a char array in class Tran, however it only applies the first letter of the string. I've tried many other ways but can't get the whole string to go into name.
edit: name is a private variable
char name[MAX_NAME + 1];
Trying to output it using cout << name << endl;
the input is:
setTran("Birth Tran", 1);
help would be appreciated, thank youu
namee[0] == NULL
name[0] = NULL;
These are bugs. NULL is for pointers. name[0] as well as namee[0] is a char. It may work (by work, I mean it will assign the first character to be the null terminator character) on some systems because 0 is both a null pointer constant and an integer literal and thus convertible to char, and NULL may be defined as 0. But NULL may also be defined as nullptr in which case the program will be ill-formed.
Use name[0] = '\0' instead.
name[0] = *namee;
however it only applies the first letter of the string.
Well, you assign only the first character, so this is to be expected.
If you would like to copy the entire string, you need to assign all of the characters. That can be implemented with a loop. There are standard functions for copying a string though; You can use std::strncpy.
That said, constant length arrays are usually problematic because it is rarely possible to correctly predict the maximum required size. std::string is a more robust alternative.
The underlying issue you are trying to assign a const char* to an char* const. When declaring
char name[MAX_NAME + 1];
You are declaring a constant memory address containing mutable char data (char* const). When you are passing a const char* to your function, you are passing a mutable pointer containing constant data. This will not compile. You should be doing a deep copy of the char array by using:
strcpy_s(dst, buffer_size, src);
This copy function will make sure that your array does not overflow, and that it is null terminated.
In order to be able to assign a pointer to a char array, it would need to be allocated on the heap with
char* name = new char[MAX_NAME + 1];
This would allow assigning a char* or char* const to it afterwards. You however need to manage the memory dynamically at this point, and I would advise against this in your case, as passing "Birth Tran" would lead to undefined behaviours as soon as char* const namee goes out of scope.
Is there any hidden feature of the sizeof keyword I'm missing here when it's applied to a literal and not a pointer initialized to another literal?
#include <iostream>
int main(int argc, char* argv[]) {
const char* string = "Hello, World!";
// sizeof(...) on literal; returns 14 (the actual length)
std::cout << "sizeof (with literal): " << sizeof("Hello, World!") << std::endl;
// sizeof(...) on initiated pointer; returns 4 (wonder why?)
std::cout << "sizeof (with variable): " << sizeof(string) << std::endl;
return 0;
}
Or is it just my compiler "optimizing" the source?
The question here may seem to answer my question, but only partially.
The answer provided makes note on C++ not being able to pass entire arrays as function parameters and as such receiving array arguments as pointers instead
(Which logically, argument pointers could be of any size).
To which I can guess here that the literal is probably optimized to be a fixed-sized array and the pointer has an arbitrary size during compile-time, but the question is: why?
It's easier to wrap my head around this if the variable had been uninitialized:
char* string; // Arbitrary size that's mutable (usually 4 bytes)
but not in this case scenario when its initiated to a fixed-size literal:
const char* string = "Hello, World!"; // Fixed-size of 14 allocated to variable?
After all the valid characters of the variable had been specified in initiation (which implicitly gives a length by the number of characters initialized), would that not make the variable inherit the allocated size of the literal?
To which I can guess here that the literal is probably optimized to be a fixed-sized array
A literal is not "optimized" to be a fixed-sized array. A string literal is a fixed-sized array: It is not a matter of optimization.
and the pointer has an arbitrary size during compile-time
It's unclear what you mean by "arbitrary". The size of a data pointer is determined by the target system. On a 64 bit x86 system, the size of a pointer is 8 bytes for example.
const char* string = "Hello, World!"; // Fixed-size of 14 allocated to variable?
What a pointer points to does not affect the size of the pointer.
would that not make the variable inherit the allocated size of the literal?
Only thing that affects the size of a variable is the type of the variable. If the type of the variable is const char*, then size of that variable is exactly same as sizeof(const char*).
A pointer is a pointer, and it’s size is the size of a pointer, regardless of what it points at. Initializing a pointer to point at a string literal doesn’t change that — it’s still a pointer.
Don’t conflate arrays and pointers. They are two different things. The name of an array in most contexts decays into a pointer to its first element, and that often confuses beginners. One of the situations where the name of an array does not decay into a pointer is when the name is the argument to the sizeof operator. That’s why sizeof “abcdef” is 7 and not the size of a pointer.
I have to call a method with the following signature:
int sendTo(const void* buffer, int length, const SocketAddress& address, int flags=0);
My first question is:
What does const void* buffer exactly mean? My intention is: it means that it is a constant (unmodifiable) pointer which can point to anything. ist this somehow right?
Second question:
The purpose of this method is, obviously, to send data over a socket.
The first parameter is the data, the second is the length of that data.
If I want to pass the string "hello" as the first parameter, how would I do this?
My idea:
char hello_str[1024] = "hello"
socket.sendTo(hello_str, sizeof(hello_str),.....);
would this work? But this way I have a way too big char array.
How can I create the array with the right size?
Recall that const protect its left side unless there's nothing to it's left, then it protects its right side. Applying this to your code we'll get that const is protecting void* buffer. That means that the value it points to cannot be modified:
The first is a pointer to a const variable - The pointer can be changed.
The second is a const pointer to a variable - The value can be changed.
This will work, you can easily try it.
And as others already answered, creating it with the right size is done simply by:
char hello_str[] = "hello";
it means that it is a constant (unmodifiable) pointer which can point to anything
No, that would be void *const. It's rather a pointer-to-anything, where the pointee (the "anything" itself) can't be modified.
would this work?
yes, apart from the missing semi-colon.
How can i create the array with the right size?
char hello_str[] = "hello";
or even
const char hello_str[] = "hello";
First question: H2C03 is right, you should make the type be void * const to prevent the pointer from being modified.
Second question: You have some options here, depending on exactly what you are doing. Here are two examples that would work:
char hello_str[] = "hello"
socket.sendTo(hello_str, sizeof(hello_str)-1,...);
socket.sendTo(hello_str, strlen(hello_str),...);
In the first call to sendto, we are calculating the size of the string at compile time. We subtract 1 in order to avoid sending the null termination character at the end of the string. In the second case, we are computing it at runtime by calling the standard strlen function which is available in C and C++.
What does const void* buffer exactly mean?
It's an untyped pointer (that's another way of saying that it can point to anything). The value it points to cannot be modified (that's not entirely correct, but at least that is what you should think when you see that). The pointer itself can change however:
const void * buffer = &a;
buffer = &b; // this is valid!
Besides that, your function call is completely correct.
You can do :
const char hello_str[] = "hello"; // Don't forget the const
socket.sendTo(hello_str, sizeof(hello_str)-1,.....);
// or socket.sendTo(hello_str, strlen(hello_str),.....);
To answer to your questions :
What does const void* buffer exactly mean? My intention is: it means that it is a constant (unmodifiable) pointer which can point to anything. ist this somehow right?
No, That would be void* const. const void* means that it is the pointee that cannot be modified.
You may read this.
I have just done what appears to be a common newbie mistake:
First we read one of many tutorials that goes like this:
#include <fstream>
int main() {
using namespace std;
ifstream inf("file.txt");
// (...)
}
Secondly, we try to use something similar in our code, which goes something like this:
#include <fstream>
int main() {
using namespace std;
std::string file = "file.txt"; // Or get the name of the file
// from a function that returns std::string.
ifstream inf(file);
// (...)
}
Thirdly, the newbie developer is perplexed by some cryptic compiler error message.
The problem is that ifstream takes const * char as a constructor argument.
The solution is to convert std::string to const * char.
Now, the real problem is that, for a newbie, "file.txt" or similar examples given in almost all the tutorials very much looks like a std::string.
So, is "my text" a std::string, a c-string or a *char, or does it depend on the context?
Can you provide examples on how "my text" would be interpreted differently according to context?
[Edit: I thought the example above would have made it obvious, but I should have been more explicit nonetheless: what I mean is the type of any string enclosed within double quotes, i.e. "myfilename.txt", not the meaning of the word 'string'.]
Thanks.
So, is "string" a std::string, a c-string or a *char, or does it depend on the context?
Neither C nor C++ have a built-in string data type, so any double-quoted strings in your code are essentially const char * (or const char [] to be exact). "C string" usually refers to this, specifically a character array with a null terminator.
In C++, std::string is a convenience class that wraps a raw string into an object. By using this, you can avoid having to do (messy) pointer arithmetic and memory reallocations by yourself.
Most standard library functions still take only char * (or const char *) parameters.
You can implicitly convert a char * into std::string because the latter has a constructor to do that.
You must explicitly convert a std::string into a const char * by using the c_str() method.
Thanks to Clark Gaebel for pointing out constness, and jalf and GMan for mentioning that it is actually an array.
"myString" is a string literal, and has the type const char[9], an array of 9 constant char. Note that it has enough space for the null terminator. So "Hi" is a const char[3], and so forth.
This is pretty much always true, with no ambiguity. However, whenever necessary, a const char[9] will decay into a const char* that points to its first element. And std::string has an implicit constructor that accepts a const char*. So while it always starts as an array of char, it can become the other types if you need it to.
Note that string literals have the unique property that const char[N] can also decay into char*, but this behavior is deprecated. If you try to modify the underlying string this way, you end up with undefined behavior. Its just not a good idea.
std::string file = "file.txt";
The right hand side of the = contains a (raw) string literal (i.a. a null-terminated byte string). Its effective type is array of const char.
The = is a tricky pony here: No assignment happens. The std::string class has a constructor that takes a pointer to char as an argument and this is called to create a temporary std::string and this is used to copy-construct (using the copy ctor of std::string) the object file of type std::string.
The compiler is free to elide the copy ctor and directly instantiate file though.
However, note that std:string is not the same thing as a C-style null-terminated string. It is not even required to be null-terminated.
ifstream inf("file.txt");
The std::ifstream class has a ctor that takes a const char * and the string literal passed to it decays to a pointer to the first element of the string.
The thing to remember is this: std::string provides (almost seamless) conversion from C-style strings. You have to look up the signature of the function to see if you are passing in a const char * or a std::string (the latter because of implicit conversions).
So, is "string" a std::string, a c-string or a char*, or does it depend on the context?
It depends entirely on the context. :-) Welcome to C++.
A C string is a null-terminated string, which is almost always the same thing as a char*.
Depending on the platforms and frameworks you are using, there might be even more meanings of the word "string" (for example, it is also used to refer to QString in Qt or CString in MFC).
The C++ standard library provides a std::string class to manage and represent character sequences. It encapsulates the memory management and is most of the time implemented as a C-string; but that is an implementation detail. It also provides manipulation routines for common tasks.
The std::string type will always be that (it doesn't have a conversion operator to char* for example, that's why you have the c_str() method), but it can be initialized or assigned to by a C-string (char*).
On the other hand, if you have a function that takes a std::string or a const std::string& as a parameter, you can pass a c-string (char*) to that function and the compiler will construct a std::string in-place for you. That would be a differing interpretation according to context as you put it.
Neither C nor C++ have a built-in string data type.
When the compiler finds, during the compilation, a double-quoted strings is implicitly referred (see the code below), the string itself is stored in program code/text and generates code to create even character array:
The array is created in static storage because it must persist to be referred later.
The array is made to constant because it must always contain the original data (Hello).
So at last, what you have is const char * to this constant static character array.
const char* v()
{
char* text = “Hello”;
return text;
// Above code can be reduced to:
// return “Hello”;
}
During the program run, when the control finds opening bracket, it creates “text”, the char* pointer, in the stack and constant array of 6 elements (including the null terminator ‘\0’ at the end) in static memory area. When control finds next line (char* text = “Hello”;), the starting address of the 6 element array is assigned to “text”. In next line (return text;), it returns “text”. With the closing bracket “text” will disappear from the stack, but array is still in the static memory area.
You need not to make return type const. But if you try to change the value in static array using non constant char* it will still give you an error during the run time because the array is constant. So, it’s always good to make return constant to make sure, it cannot be referred by non constant pointer.
But if the compiler finds a double-quoted strings is explicitly referred as an array, the compiler assumes that the programmer is going to (smartly) handle it. See the following wrong example:
const char* v()
{
char text[] = “Hello”;
return text;
}
During the compilation, compiler checks, quoted text and save it as it is in the code to fill the generated array during the runt time. Also, it calculate the array size, in this case again as 6.
During the program run, with the open bracket, the array “text[]” with 6 elements is created in stack. But no initialization. When the code finds (char text[] = “Hello”;), the array is initialized (with the text in compiled code). So array is now on the stack. When the compiler finds (return text;), it returns the starting address of the array “text”. When the compiler find the closing bracket, the array disappears from the stack. So no way to refer it by the return pointer.
Most standard library functions still take only char * (or const char *) parameters.
The Standard C++ library has a powerful class called string for manipulating text. The internal data structure for string is character arrays. The Standard C++ string class is designed to take care of (and hide) all the low-level manipulations of character arrays that were previously required of the C programmer. Note that std::string is a class:
You can implicitly convert a char * into std::string because the
latter has a constructor to do that.
You can explicitly convert a std::string into a const char * by using the c_str() method.
As often as possible it should mean std::string (or an alternative such as wxString, QString, etc., if you're using a framework that supplies such. Sometimes you have no real choice but to use a NUL-terminated byte sequence, but you generally want to avoid it when possible.
Ultimately, there simply is no clear, unambiguous terminology. Such is life.
To use the proper wording (as found in the C++ language standard) string is one of the varieties of std::basic_string (including std::string) from chapter 21.3 "String classes" (as in C++0x N3092), while the argument of ifstream's constructor is NTBS (Null-terminated byte sequence)
To quote, C++0x N3092 27.9.1.4/2.
basic_filebuf* open(const char* s, ios_base::openmode mode);
...
opens a file, if possible, whose name is the NTBS s
So, I'm wanting to get a better grasp on how string literals in C++ work. I'm mostly concerned with situations where you're assigning the address of a string literal to a pointer, and passing it around. For example:
char* advice = "Don't stick your hands in the toaster.";
Now lets say I just pass this string around by copying pointers for the duration of the program. Sure, it's probably not a good idea, but I'm curious what would actually be going on behind the scenes.
For another example, let's say we make a function that returns a string literal:
char* foo()
{
// function does does stuff
return "Yikes!"; // somebody's feeble attempt at an error message
}
Now lets say this function is called very often, and the string literal is only used about half the time it's called:
// situation #1: it's just randomly called without heed to the return value
foo();
// situation #2: the returned string is kept and used for who knows how long
char* retVal = foo();
In the first situation, what's actually happening? Is the string just created but not used, and never deallocated?
In the second situation, is the string going to be maintained as long as the user finds need for it? What happens when it isn't needed anymore... will that memory be freed up then (assuming nothing points to that space anymore)?
Don't get me wrong, I'm not planning on using string literals like this. I'm planning on using a container to keep my strings in check (probably std::string). I'm mostly just wanting to know if these situations could cause problems either for memory management or corrupted data.
String-literals have the type const char[N] (where N is the length + 1) and are statically allocated. You need not worry about memory issues; if a string is used in your program it is all handled for you, and resides somewhere in program memory (usually read-only).
That is, these are "the same":
static const char str[] = "a string";
"a string"
When you point to a string literal, you are pointing to the first character at the array. In fact, because the type is const char[], it's only safe to point to it via const char*. The conversion from string literal to char* is deprecated, and unsafe.
// the "same"
static const char str[] = "a string";
const char* strPtr = str; // decays
const char* s1 = "a string";
char* s2 = "a string"; // allowed, implicit const_cast
*s1 = 'A'; // not allowed, it's const
*s2 = 'B'; // allowed, it's not const (but leads to undefined behavior)
Firstly, declare the return value of foo as const, because string literals are constants that can't be changed without causing the dreaded "undefined behaviour". This will then force any pointers which use the return value of foo to also be declared as const, and potentially limiting the damage which can be (usually unintentionally) done. String literals are stored in the 'text' area of a binary executable - they're not created as such at run time.