I understand that
char *s = "Hello World!" ;
is stored in read only memory and the string literal cannot be modified via the pointer.
How is this different from
const char *s = "Hello World!";
Also is the type of 'string' char * or const char * ?
The difference is that the latter is legal and the former is not. That's a change that was made in C++11. Formally, "Hello World!" has type const char[13]; it can be converted to const char*. In the olden days, its type could be char[13], which could be converted to char*. C++ changed the type of the array by adding the const, but kept the conversion to char* so that existing C code that used char* would work in C++, but modifying the text that the pointer pointed to produced undefined behavior. C++11 removed the conversion to char*, so now you can only legally do
const char *s = "Hello world!";
By giving the type as const char *, it makes it harder to accidentally overwrite the memory, since the compiler will give an error if you try:
const char *s = "Hello World!";
s[0] = 'X'; // compile error
If you don't use const, then the problem may not be caught until runtime, or it may just cause your program to be subtly wrong.
Related
I have a code like this but I keep receiving this error :
A value of type "const char*" cannot be used to initialize an entity of type "char *"
What is going on?
I have read up on the following threads but have not been able to see any result to my answer as all of them are either from char to char* or char* to char:
Value type const char cannot be used to initialize an entity of type char*
Value of type char* cannot be used to initialize an entity of type "char"
#include <iostream>;
using namespace std;
int main() {
int x = 0; //variable x created
int cars (14);//cars is created as a variable with value 14
int debt{ -1000 };//debt created with value 1000
float cash = 2.32;
double credit = 32.32;
char a = 'a';//for char you must use a single quote and not double
char* sandwich = "ham";
return 0;
}
I am using Visual Studio Community 2017
That is correct. Let’s say you had the following code:
const char hello[] = "hello, world!";
char* jello = hello; // Not allowed, because:
jello[0] = 'J'; // Undefined behavior!
Whoops! A const char* is a non-const pointer to const char. If you assign its value to a non-const char*, you’ve lost its const property.
A const pointer to non-const char would be a char* const, and you can initialize a char* from that all day if you want.
You can, if you really want, achieve this with const_cast<char*>(p), and I occasionally have, but it’s usually a sign of a serious design flaw. If you actually get the compiler to emit instructions to write to the memory aliased by a string constant, you get undefined behavior. One of the many things that might go wrong is that some implementations will store the constant in read-only memory and crash. Or the same bytes of memory might be re-used for more than one purpose, because after all, we warned you never to change it.
By the way, the rules in C are different. This is solely for backward-compatibility with early versions of C that did not have the const keyword, and you should never write new code that uses a non-const alias to a string constant.
You need to make your string literal type const because in C++ it is a constant array of char, unlike C where it is just an array of char. You cannot change a string literal, so making it const is preferred in C++ for extra safety. It is the same reason you have to use an explicit cast when going from const char* to char*. It's still technically "allowed" in C++ since it is allowed in C, which is why it's just a warning. It's still bad practice to do so. To fix the warning, make it const.
const char* sandwich = "ham";
Your code (and underlying assumption) is valid pre C++11 standard.
String literals (e.g. "ham") since C++11 are of type const char* (or const char[]) if you will instead of char * they used to be. [Always read specs for breaking changes!!!]
Hence the warning in VS 2017. Change the compiler version to pre C++11 version and you will be amazed.
This has subtle nuances and can cause frustrating debug sessions
// C++11 or later
auto c = "Rowdie";
// c has type const char*, can't use c to modify literal
c[0] = 'H'; // illegal - CTE
// -vs-
char * d = "Rowdie";
d[0] = 'H';
cout << d; // outputs "Howdie"
Also another example is auto return type from functions
auto get_literal() {
// ... function code
return "String Literal";
}
// and using value later
char* lit = get_literal(); // You get same error as const char* cannot be init to char*
im realy confused about const char * and char *.
I know in char * when we want to modify the content, we need to do something like this
const char * temp = "Hello world";
char * str = new char[strlen(temp) + 1];
memcpy(str, temp, strlen(temp));
str[strlen(temp) + 1] = '\0';
and if we want to use something like this
char * str = "xxx";
char * str2 = "xts";
str = str2;
we get compiler warning. it's ok I know when i want to change char * I have to use something memory copy. but about const char * im realy confused. in const char * I can use this
const char * str = "Hello";
const char * str2 = "World";
str = str2; // and now str is Hello
and I have no compiler error ! why ? why we use memory copy when is not const and in const we only use equal operator ! and done !... how possible? is it ok to just use equal in const? no problem happen later?
As other answers say, you should distinguish pointers and bytes they point to.
Both types of pointers, char * and const char *, can be changed, that is, "redirected" to point to different bytes. However, if you want to change the bytes (characters) of the strings, you cannot use const char *.
So, if you have string literals "Hello" and "World" in your program, you can assign them to pointers, and printing the pointer will print the corresponding literal. However, to do anything non-trivial (e.g. change Hello to HELLO), you will need non-const pointers.
Another example: with some pointer manipulation, you can remove leading bytes from a string literal:
const char* str = "Hello";
std::cout << str; // Hello
str = str + 2;
std::cout << str; // llo
However, if you want to extract a substring, or do any other transformation on a string, you should reallocate it, and for that you need a non-const pointer.
BTW since you are using C++, you can use std::string, which makes it easier to work with strings. It reallocates strings without your intervention:
#include <string>
std::string str("Hello");
str = str.substr(1, 3);
std::cout << str; // ell
This is a confusing hangover from the days of early C. Early C didn't have const, so string literals were "char *". They remained char * to avoid breaking old code, but they became non-modifiable, so const char * in all but name. So modern C++ either warns or gives an error (to be strictly conforming) when the const is omitted.
Your memcpy missed the trailing nul byte, incidentally. Use strcpy() to copy a string, that's the right function with the right name. You can create a string in read/write memory by use of the
char rwstring[] = "I am writeable";
syntax.
That is cause your variables are just a pointers *. You're not modifiying their contents, but where they are pointing to.
char * a = "asd";
char * b = "qwe";
a = b;
now you threw away the contents of a. Now a and b points to the same place. If you modify one, both are modified.
In other words. Pointers are never constants (mostly). your const predicate in a pointer variable does not means nothing to the pointer.
The real difference is that the pointer (that is not const) is pointing to a const variable. and when you change the pointer it will be point to ANOTHER NEW const variable. That is why const has no effect on simple pointers.
Note: You can achieve different behaviours with pointers and const with more complex scenario. But with simple as it, it mostly has no effect.
Citing Malcolm McLean:
This is a confusing hangover from the days of early C. Early C didn't have const, so string literals were "char *". They remained char * to avoid breaking old code, but they became non-modifiable, so const char * in all but name.
Actually, string literals are not pointers, but arrays, this is why sizeof("hello world") works as a charm (yields 12, the terminating null character is included, in contrast to strlen...). Apart from this small detail, above statement is correct for good old C even in these days.
In C++, though, string literals have been arrays of constant characters (char const[]) right from the start:
C++ standard, 5.13.5.8:
Ordinary string literals and UTF-8 string literals are also referred to as narrow string literals. A narrow string literal has type “array of n const char”, where n is the size of the string as defined below, and has static storage duration.
(Emphasised by me.) In general, you are not allowed to assign pointer to const to pointer to non-const:
char const* s = "hello";
char ss = s;
This will fail to compile. Assigning string literals to pointer to non-const should normally fail, too, as the standard explicitly states in C.1.1, subclause 5.13.5:
Change: String literals made const.
The type of a string literal is changed from “array of char” to “array of const char”.
[...]char* p = "abc"; // valid in C, invalid in C++
Still, string literal assignement to pointer to non-const is commonly accepted by compilers (as an extension!), probably to retain compatibility to C. As this is, according to the standard, invalid, the compiler yields a warning, at least...
In C the definition of strchr is as follows
char * strchr(const char *s, int c);
This declaration guarantees the user that strchr will not modify the contents of 's' (unless the code uses explicit typecasting).
The issue as gleaned from web is that this definition casts way the "const" attribute when it returns a pointer and hence in 'C++' this is not "const correct".
The solution in C++ seems to be to have overloaded functions
const char * strchr ( const char * str, int character );
char * strchr ( char * str, int character );
While this is "const safe", I am unable to understand how the second declaration
char * strchr (char * str, int character)
promise its caller that it will not modify its parameter.
It doesn't make such a promise. Quite the contrary, it provides its caller with a way to modify the string through the returned pointer. That's why this version can only be invoked with a non-const pointer.
The C++ version is const-correct and C version is not because with C you can accidentally write this
const char str[] = "Hello world";
*strchr(str, 'o') = 'O'; // undefined behaviour
or this
*strchr("Hello world", 'o') = 'O'; // undefined behaviour
While the C++ version would fail at compile time
const char str[] = "Hello world";
*strchr(str, 'o') = 'O'; // error: assignment of read-only location
As you can see, given a signature you can't prove an arbitrary function won't modify the data pointed to (even worse, consider that a function could const_cast away the pointer and modify it anyway!). But that wasn't the point of this overloading, the point was to make accidental errors harder to make.
When I wrote the following code and executed it, the compiler said
deprecated conversion from string constant to char*
int main()
{
char *p;
p=new char[5];
p="how are you";
cout<< p;
return 0;
}
It means that I should have written const char *.
But when we pass arguments into main using char* argv[] we don't write const char* argv[].
Why?
Because ... argv[] isn't const. And it certainly isn't a (static) string literal since it's being created at runtime.
You're declaring a char * pointer then assigning a string literal to it, which is by definition constant; the actual data is in read-only memory.
int main(int argc, char **argv) {
// Yes, I know I'm not checking anything - just a demo
argv[1][0] = 'f';
std::cout << argv[1] << std::endl;
}
Input:
g++ -o test test.cc
./test hoo
Output:
foo
This is not a comment on why you'd want to change argv, but it certainly is possible.
Historical reasons. Changing the signature of main() would break too much existing code. And it is possible that some implementations allow you to change the parameters to main from your code. However code like this:
char * p = "helllo";
* p = 'x';
is always illegal, because you are not allowed to mess with string literals like that, so the pointer should be to a const char.
why is it required for char* to be constant while assigning it to a string
Because such literal strings (like "hi", "hello what's going on", etc), are stored in the read-only segment of your exe. As such, the pointers that point to them need to point to constant characters (eg, can't change them).
You are assigning a string constant (const char*) to a pointer to a non-constant string (char *p). This would allow you to modify the string constant, e.g. by doing p[0] = 'n'.
Anyway, why don't you use std::string instead ? (you seem to be using C++).
If you look at execution functions like execve, you will see that they actually don't accept const char* as parameters, but do indeed require char*, therefore you can't use a string constant to invoke main.
I have a class with a private char str[256];
and for it I have an explicit constructor:
explicit myClass(char *func)
{
strcpy(str,func);
}
I call it as:
myClass obj("example");
When I compile this I get the following warning:
deprecated conversion from string constant to 'char*'
Why is this happening?
This is an error message you see whenever you have a situation like the following:
char* pointer_to_nonconst = "string literal";
Why? Well, C and C++ differ in the type of the string literal. In C the type is array of char and in C++ it is constant array of char. In any case, you are not allowed to change the characters of the string literal, so the const in C++ is not really a restriction but more of a type safety thing. A conversion from const char* to char* is generally not possible without an explicit cast for safety reasons. But for backwards compatibility with C the language C++ still allows assigning a string literal to a char* and gives you a warning about this conversion being deprecated.
So, somewhere you are missing one or more consts in your program for const correctness. But the code you showed to us is not the problem as it does not do this kind of deprecated conversion. The warning must have come from some other place.
The warning:
deprecated conversion from string constant to 'char*'
is given because you are doing somewhere (not in the code you posted) something like:
void foo(char* str);
foo("hello");
The problem is that you are trying to convert a string literal (with type const char[]) to char*.
You can convert a const char[] to const char* because the array decays to the pointer, but what you are doing is making a mutable a constant.
This conversion is probably allowed for C compatibility and just gives you the warning mentioned.
As answer no. 2 by fnieto - Fernando Nieto clearly and correctly describes that this warning is given because somewhere in your code you are doing (not in the code you posted) something like:
void foo(char* str);
foo("hello");
However, if you want to keep your code warning-free as well then just make respective change in your code:
void foo(char* str);
foo((char *)"hello");
That is, simply cast the string constant to (char *).
There are 3 solutions:
Solution 1:
const char *x = "foo bar";
Solution 2:
char *x = (char *)"foo bar";
Solution 3:
char* x = (char*) malloc(strlen("foo bar")+1); // +1 for the terminator
strcpy(x,"foo bar");
Arrays also can be used instead of pointers because an array is already a constant pointer.
Update: See the comments for security concerns regarding solution 3.
A reason for this problem (which is even harder to detect than the issue with char* str = "some string" - which others have explained) is when you are using constexpr.
constexpr char* str = "some string";
It seems that it would behave similar to const char* str, and so would not cause a warning, as it occurs before char*, but it instead behaves as char* const str.
Details
Constant pointer, and pointer to a constant. The difference between const char* str, and char* const str can be explained as follows.
const char* str : Declare str to be a pointer to a const char. This means that the data to which this pointer is pointing to it constant. The pointer can be modified, but any attempt to modify the data would throw a compilation error.
str++ ; : VALID. We are modifying the pointer, and not the data being pointed to.
*str = 'a'; : INVALID. We are trying to modify the data being pointed to.
char* const str : Declare str to be a const pointer to char. This means that point is now constant, but the data being pointed too is not. The pointer cannot be modified but we can modify the data using the pointer.
str++ ; : INVALID. We are trying to modify the pointer variable, which is a constant.
*str = 'a'; : VALID. We are trying to modify the data being pointed to. In our case this will not cause a compilation error, but will cause a runtime error, as the string will most probably will go into a read only section of the compiled binary. This statement would make sense if we had dynamically allocated memory, eg. char* const str = new char[5];.
const char* const str : Declare str to be a const pointer to a const char. In this case we can neither modify the pointer, nor the data being pointed to.
str++ ; : INVALID. We are trying to modify the pointer variable, which is a constant.
*str = 'a'; : INVALID. We are trying to modify the data pointed by this pointer, which is also constant.
In my case the issue was that I was expecting constexpr char* str to behave as const char* str, and not char* const str, since visually it seems closer to the former.
Also, the warning generated for constexpr char* str = "some string" is slightly different from char* str = "some string".
Compiler warning for constexpr char* str = "some string": ISO C++11 does not allow conversion from string literal to 'char *const'
Compiler warning for char* str = "some string": ISO C++11 does not allow conversion from string literal to 'char *'.
Tip
You can use C gibberish ↔ English converter to convert C declarations to easily understandable English statements, and vice versa. This is a C only tool, and thus wont support things (like constexpr) which are exclusive to C++.
In fact a string constant literal is neither a const char * nor a char* but a char[]. Its quite strange but written down in the c++ specifications; If you modify it the behavior is undefined because the compiler may store it in the code segment.
Maybe you can try this:
void foo(const char* str)
{
// Do something
}
foo("Hello")
It works for me
I solve this problem by adding this macro in the beginning of the code, somewhere. Or add it in <iostream>, hehe.
#define C_TEXT( text ) ((char*)std::string( text ).c_str())
I also got the same problem. And what I simple did is just adding const char* instead of char*. And the problem solved. As others have mentioned above it is a compatible error. C treats strings as char arrays while C++ treat them as const char arrays.
For what its worth, I find this simple wrapper class to be helpful for converting C++ strings to char *:
class StringWrapper {
std::vector<char> vec;
public:
StringWrapper(const std::string &str) : vec(str.begin(), str.end()) {
}
char *getChars() {
return &vec[0];
}
};
The following illustrates the solution, assign your string to a variable pointer to a constant array of char (a string is a constant pointer to a constant array of char - plus length info):
#include <iostream>
void Swap(const char * & left, const char * & right) {
const char *const temp = left;
left = right;
right = temp;
}
int main() {
const char * x = "Hello"; // These works because you are making a variable
const char * y = "World"; // pointer to a constant string
std::cout << "x = " << x << ", y = " << y << '\n';
Swap(x, y);
std::cout << "x = " << x << ", y = " << y << '\n';
}