If I write:
char *aPtr = "blue"; //would be better const char *aPtr = "blue"
aPtr[0]='A';
I have a warning. The code above can work but isn't standard, it has a undefined behavior because it's read-only memory with a pointer at string litteral. The question is:
Why is it like this?
with this code rather:
char a[]="blue";
char *aPtr=a;
aPtr[0]='A';
is ok. I want to understand under the hood what happens
The first is a pointer to a read-only value created by the compiler and placed in a read-only section of the program. You cannot modify the characters at that address because they are read-only.
The second creates an array and copies each element from the initializer (see this answer for more details on that). You can modify the contents of the array, because it's a simple variable.
The first one works the way it does because doing anything else would require dynamically-allocating a new variable, and would require garbage collection to free it. That is not how C and C++ work.
The primary reason that string literals can't be modified (without undefined behavior) is to support string literal merging.
Long ago, when memory was much tighter than today, compiler authors noticed that many programs had the same string literals repeated many times--especially things like mode strings being passed to fopen (e.g., f = fopen("filename", "r");) and simple format strings being passed to printf (e.g., printf("%d\n", a);).
To save memory, they'd avoid allocating separate memory for each instance of these strings. Instead, they'd allocate one piece of memory, and point all the pointers at it.
In a few cases, they got even trickier than that, to merge literals that were't even entirely identical. For example consider code like this:
printf("%s\t%d\n", a);
/* ... */
printf("%d\n", b);
In this case, the string literals aren't entirely identical, but the second one is identical part of the end of the first. In this case, they'd still allocate one piece of memory. One pointer would point to the beginning of the memory, and the other to the position of the %d in that same block of memory.
With a possibility (but no requirement for) string literal merging, it's essentially impossible to say what behavior you'll get when you modify a string literal. If string literals are merged, modifying one string literal might modify others that are identical, or end identically. If string literals are not merged, modifying one will have no effect on any other.
MMUs added another dimension: they allowed memory to be marked as read-only, so attempting to modify a string literal would result in a signal of some sort--but only if the system had an MMU (which was often optional at one time) and also depending on whether the compiler/linker decided to put the string literals in memory they'd marked constant or not.
Since they couldn't define what the behavior would be when you modified a string literal, they decided that modifying a string literal would produce undefined behavior.
The second case is entirely different. Here you've defined an array of char. It's clear that if you define two separate arrays, they're still separate, regardless of content, so modifying one can't possibly affect the other. The behavior is clear and always has been, so doing so gives defined behavior. The fact that the array in question might be initialized from a string literal doesn't change that.
Related
If I have an embedded null terminator [aside: is that UB?], is it well-defined for me to access the values after it?
#include <stdio.h>
const char foo[] = "abc\0def";
int main() {
printf("%s", foo+4);
return sizeof(foo);
}
For the record, it prints what you might expect:
def
An embedded null is NOT Undefined Behavior. It could be a logic error, if you work with functions that expect Strings to be null-terminated. But there's nothing wrong or evil or undefined about accessing the full breadth of an array you've successfully allocated, regardless of its contents.
One thing to observe though: if you attempt to store this data in a std::string (which is how you should handle all strings, TBH), how you store the string can be important.
std::string str1 = foo; //contents of str1 becomes "abc".
std::string str2 = std::string(foo, sizeof(foo)); //contents of str2 becomes "abc\0def"
[dcl.init.string] states
An array of narrow character type (3.9.1), char16_t array, char32_t array, or wchar_t array can be initialized by a narrow string literal, char16_t string literal, char32_t string literal, or wide string literal, respectively, or by an appropriately-typed string literal enclosed in braces (2.14.5). Successive characters of the value of the string literal initialize the elements of the array.
emphasis mine
So the embedded null it not a problem, it just becomes a element of the array. Since the array is sized to container all the characters and escape sequeneces we know there are elements after that embedded null and it is safe to access those.
Really the only issue with the embedded null is that any C function is going to stop when it hits that null and won't full process the string. You might consider using a std::string instead which doesn't have those issues.
Accessing a C string beyound the terminating null character per se never is undefined behaviour. Still, we can yield undefined behaviour this way, but for a totally different reason:
If the terminating null character happens to reside at the last position in the char array reserved for the string, then we access this underlying array out of its bounds if we access the string beyond its end. And this out-of-bounds-access is what really yields the undefined behaviour...
Edit:
[aside: is that UB?]
UB, undefined behaviour, is behaviour that cannot be defined, because there is no meaningful behaviour for. Relying on undefined behaviour can result in anything, including getting the expected results, but can fail miserably any other time (e. g. on another platform, after switching compiler version, after simply recompiling, even after just restarting one and the same program). Thus a program relying on undefined behaviour is considered not to be well defined.
Example: Dereferencing a pointer pointing to an object that has already been deleted (a "dangling pointer"), or close to the question: accessing an array out of bounds (could result in trying to access memory not asigned to the current process or even not existing, but could read or (bad!!!) overwrite memory of a totally different object that happens to be located at the given address (it does not even have to be the same object every time your program runs, not even within one single program run).
Undefined behaviour is not to be mixed up with unspecified behaviour (or synonymously, implementation defined behaviour): In this case, the behaviour for a given input is well defined, but it is left to the compiler vendor to define the behaviour within some given reasonable limitations.
Example: right shift of negative integers - it can occur with or without sign extension (so can be an arithmetic or a logical shift). Which one applies is not specified by the standard, though, but using right shift on negative integers is well defined.
Why can't we add two C-strings in C++? This is what I know, please correct me, and add to it.
Is it because the '+' operator is not overloaded to do the operation. The compiler essentially interprets the variable name as a pointer. Since, we can not add two pointers, so we can't add two string variables like this: str = str + "str"?
First, realize that concatenating strings as std::string::operator+() and std::string::operator+=() do, is a conceptually different operation from addition.
The things you're asking about "adding" are C-style strings, which come from C, so C++ supports them. There is no operator overloading in C.
A C-style string is just a pointer to an array of characters in memory, which is (hopefully) terminated by a NULL character '\0'. A char * may point to a C-string, or it may not. You have to manage the memory yourself. The memory may be allocated statically, and there may be only enough room for the characters you put there to start with, such as when you define a string char myString[] = "blah";. There are no built-in, automatic memory reallocation mechanisms, so even if such a concatenation function were defined, it couldn't guarantee that there would be room for whatever you want to append to a string in the target buffer.
By contrast, in C++, std::string is a class that dynamically allocates memory as needed. std::string objects are always std::strings; they don't point to other things. Concatenation operators are defined (+, +=), and they handle memory (re)allocation for you.
...So now you know the two main types of strings used in C++, so you can do some further research on "C-style strings" or "C strings"1; and std::string, which is a "C++ string". Maybe start at Wikipedia: String (C++).
1 FYI, "C-String" is also the name of a type of revealing clothing, so you may get some NSFW search results mixed in using that term.
I am having a mental block and I know I should know this but I need a little help.
If I declare a string variable like this:
string word = "Hello";
How do I find the memory address of "Hello"?
Edit: This is what I am trying to do...
Write a function that takes one argument, the address of a string, and prints that string once. (Note: you will need to use a pointer to complete this part.)
However, if a second argument, type int, is provided and is nonzero, the function should print the string a number of times equal to the number of times that function has been called at that point. (Note that the number of times the string is printed is not equal to the value of the second argument; it is equal to the number of times the function has been called so far.)
Use either:
std::string::data() if your data isn't null-terminated c-string like.
or
std::string::c_str() if you want the data and be guaranteed to get the null-termination.
Note that the pointer returned by either of these calls doesn't have to be the underlying data the std::string object is manipulating.
Take the address of the first character is the usual way to do it. &word[0]. However, needing to do this if you're not operating with legacy code is usually a sign that you're doing something wrong.
I guess you want a pointer to a plain old C-string? Then use word.c_str(). Note that this is not guaranteed to point to the internal storage of the string, it's just a (constant) C-string version you can work with.
You can use the c_str() function to get a pointer to the C string (const char *); however note that the pointer is invalidated whenever you modify the string; you have to invoke c_str() again as the old string may have been deallocated.
OK, so, I know this question is old but I feel like the obvious answer here is actually:
std::string your_string{"Hello"};
//Take the address of the beginning
auto start_address = &(*your_string.begin())
//Take the address at the end
auto end_address = &(*your_string.end())
In essence this will accomplish the same thing as using:
auto start_address = your_string.c_str();
auto end_address = your_string.c_str() + strlen(your_string.c_str());
However I would prefer the first approach (taking the address of the dereferenced iterator) because:
a) Guaranteed to work with begin/end compatible containers which might not have the c_str method. So for example, if you decided you wanted a QString (the string QT uses) or AWS::String or std::vector to hold your characters, the c_str() approach wouldn't work but the one above would.
b) Possibly not as costly as c_str()... which generally speaking should be implemented similarly to the call I made in the second line of code to get the address but isn't guaranteed to be implemented that way (E.g. if the string you are using is not null terminated it might require the reallocation and mutation of your whole string in order to add the null terminator... which will suddenly make it thread unsafe and very costly, this is not the case for std::string but might be for other types of strings)
c) It communicates intent better, in the end what you want is an address and the first bunch of code expresses exactly that.
So I'd say this approach is better for:
Clarity, compatibility and efficiency.
Edit:
Note that with either approach, if you change your initial string after taking the address the address will be invalidated (since the string might be rellocated). The compiler will not warn you against this and it could cause some very nasty bugs :/
Declare a pointer to the variable and then view it how you would.
Any idea why I get "Maya is not Maya" as a result of this code?
if ("Maya" == "Maya")
printf("Maya is Maya \n");
else
printf("Maya is not Maya \n");
Because you are actually comparing two pointers - use e.g. one of the following instead:
if (std::string("Maya") == "Maya") { /* ... */ }
if (std::strcmp("Maya", "Maya") == 0) { /* ... */ }
This is because C++03, §2.13.4 says:
An ordinary string literal has type “array of n const char”
... and in your case a conversion to pointer applies.
See also this question on why you can't provide an overload for == for this case.
You are not comparing strings, you are comparing pointer address equality.
To be more explicit -
"foo baz bar" implicitly defines an anonymous const char[m]. It is implementation-defined as to whether identical anonymous const char[m] will point to the same location in memory(a concept referred to as interning).
The function you want - in C - is strmp(char*, char*), which returns 0 on equality.
Or, in C++, what you might do is
#include <string>
std::string s1 = "foo"
std::string s2 = "bar"
and then compare s1 vs. s2 with the == operator, which is defined in an intuitive fashion for strings.
The output of your program is implementation-defined.
A string literal has the type const char[N] (that is, it's an array). Whether or not each string literal in your program is represented by a unique array is implementation-defined. (§2.13.4/2)
When you do the comparison, the arrays decay into pointers (to the first element), and you do a pointer comparison. If the compiler decides to store both string literals as the same array, the pointers compare true; if they each have their own storage, they compare false.
To compare string's, use std::strcmp(), like this:
if (std::strcmp("Maya", "Maya") == 0) // same
Typically you'd use the standard string class, std::string. It defines operator==. You'd need to make one of your literals a std::string to use that operator:
if (std::string("Maya") == "Maya") // same
What you are doing is comparing the address of one string with the address of another. Depending on the compiler and its settings, sometimes the identical literal strings will have the same address, and sometimes they won't (as apparently you found).
Any idea why i get "Maya is not Maya" as a result
Because in C, and thus in C++, string literals are of type const char[], which is implicitly converted to const char*, a pointer to the first character, when you try to compare them. And pointer comparison is address comparison.
Whether the two string literals compare equal or not depends whether your compiler (using your current settings) pools string literals. It is allowed to do that, but it doesn't need to. .
To compare the strings in C, use strcmp() from the <string.h> header. (It's std::strcmp() from <cstring>in C++.)
To do so in C++, the easiest is to turn one of them into a std::string (from the <string> header), which comes with all comparison operators, including ==:
#include <string>
// ...
if (std::string("Maya") == "Maya")
std::cout << "Maya is Maya\n";
else
std::cout << "Maya is not Maya\n";
C and C++ do this comparison via pointer comparison; looks like your compiler is creating separate resource instances for the strings "Maya" and "Maya" (probably due to having an optimization turned off).
My compiler says they are the same ;-)
even worse, my compiler is certainly broken. This very basic equation:
printf("23 - 523 = %d\n","23"-"523");
produces:
23 - 523 = 1
Indeed, "because your compiler, in this instance, isn't using string pooling," is the technically correct, yet not particularly helpful answer :)
This is one of the many reasons the std::string class in the Standard Template Library now exists to replace this earlier kind of string when you want to do anything useful with strings in C++, and is a problem pretty much everyone who's ever learned C or C++ stumbles over fairly early on in their studies.
Let me explain.
Basically, back in the days of C, all strings worked like this. A string is just a bunch of characters in memory. A string you embed in your C source code gets translated into a bunch of bytes representing that string in the running machine code when your program executes.
The crucial part here is that a good old-fashioned C-style "string" is an array of characters in memory. That block of memory is often referred to by means of a pointer -- the address of the start of the block of memory. Generally, when you're referring to a "string" in C, you're referring to that block of memory, or a pointer to it. C doesn't have a string type per se; strings are just a bunch of chars in a row.
When you write this in your code:
"wibble"
Then the compiler provides a block of memory that contains the bytes representing the characters 'w', 'i', 'b', 'b', 'l', 'e', and '\0' in that order (the compiler adds a zero byte at the end, a "null terminator". In C a standard string is a null-terminated string: a block of characters starting at a given memory address and continuing until the next zero byte.)
And when you start comparing expressions like that, what happens is this:
if ("Maya" == "Maya")
At the point of this comparison, the compiler -- in your case, specifically; see my explanation of string pooling at the end -- has created two separate blocks of memory, to hold two different sets of characters that are both set to 'M', 'a', 'y', 'a', '\0'.
When the compiler sees a string in quotes like this, "under the hood" it builds an array of characters, and the string itself, "Maya", acts as the name of the array of characters. Because the names of arrays are effectively pointers, pointing at the first character of the array, the type of the expression "Maya" is pointer to char.
When you compare these two expressions using "==", what you're actually comparing is the pointers, the memory addresses of the beginning of these two different blocks of memory. Which is why the comparison is false, in your particular case, with your particular compiler.
If you want to compare two good old-fashioned C strings, you should use the strcmp() function. This will examine the contents of the memory pointed two by both "strings" (which, as I've explained, are just pointers to a block of memory) and go through the bytes, comparing them one-by-one, and tell you whether they're really the same.
Now, as I've said, this is the kind of slightly surprising result that's been biting C beginners on the arse since the days of yore. And that's one of the reasons the language evolved over time. Now, in C++, there is a std::string class, that will hold strings, and will work as you expect. The "==" operator for std::string will actually compare the contents of two std::strings.
By default, though, C++ is designed to be backwards-compatible with C, i.e. a C program will generally compile and work under a C++ compiler the same way it does in a C compiler, and that means that old-fashioned strings, "things like this in your code", will still end up as pointers to bits of memory that will give non-obvious results to the beginner when you start comparing them.
Oh, and that "string pooling" I mentioned at the beginning? That's where some more complexity might creep in. A smart compiler, to be efficient with its memory, may well spot that in your case, the strings are the same and can't be changed, and therefore only allocate one block of memory, with both of your names, "Maya", pointing at it. At which point, comparing the "strings" -- the pointers -- will tell you that they are, in fact, equal. But more by luck than design!
This "string pooling" behaviour will change from compiler to compiler, and often will differ between debug and release modes of the same compiler, as the release mode often includes optimisations like this, which will make the output code more compact (it only has to have one block of memory with "Maya" in, not two, so it's saved five -- remember that null terminator! -- bytes in the object code.) And that's the kind of behaviour that can drive a person insane if they don't know what's going on :)
If nothing else, this answer might give you a lot of search terms for the thousands of articles that are out there on the web already, trying to explain this. It's a bit painful, and everyone goes through it. If you can get your head around pointers, you'll be a much better C or C++ programmer in the long run, whether you choose to use std::string instead or not!
If I call a function like
myObj.setType("fluid");
many times in a program, how many copies of the literal "fluid" are saved in memory? Can the compiler recognize that this literal is already defined and just reference it again?
This has nothing to do with C++(the language). Instead, it is an "optimization" that a compiler can do. So, the answer yes and no, depending on the compiler/platform you are using.
#David This is from the latest draft of the language:
§ 2.14.6 (page 28)
Whether all string literals are
distinct (that is, are stored in
non overlapping objects) is
implementation defined. The effect of
attempting to modify a string literal
is undefined.
The emphasis is mine.
In other words, string literals in C++ are immutable because modifying a string literal is undefined behavior. So, the compiler is free, to eliminate redundant copies.
BTW, I am talking about C++ only ;)
Yes, it can. Of course, it depends on the compiler. For VC++, it's even configurable:
http://msdn.microsoft.com/en-us/library/s0s0asdt(VS.80).aspx
Yes it can, but there's no guarantee that it will. Define a constant if you want to be sure.
This is a compiler implementation issue. Many compilers that I have used have an option to share or merge duplicate string literals. Allowing duplicate string literals speeds up the compilation process but produces larger executables.
I believe that in C/C++ there is no specified handling for that case, but in most cases would use multiple definitions of that string.
2.13.4/2: "whether all string literals are distinct (that is, are stored in nonoverlapping objects) is implementation-defined".
This permits the optimisation you're asking about.
As an aside, there may be a slight ambiguity, at least locally within that section of the standard. The definition of string literal doesn't quite make clear to me whether the following code uses one string literal twice, or two string literals once each:
const char *a = "";
const char *b = "";
But the next paragraph says "In translation phase 6 adjacent narrow string literals are concatenated". Unless it means to say that something can be adjacent to itself, I think the intention is pretty clear that this code uses two string literals, which are concatenated in phase 6. So it's not one string literal twice:
const char *c = "a" "a";
Still, if you did read that "a" and "a" are the same string literal, then the standard requires the optimisation you're talking about. But I don't think they are the same literal, I think they're different literals that happen to consist of the same characters. This is perhaps made clear elsewhere in the standard, for instance in the general information on grammar and parsing.
Whether it's made clear or not, many compiler-writers have interpreted the standard the way I think it is, so I might as well be right ;-)