I understand that char pointers initialized to string literals are stored in read-only memory. But what about arrays of string literals?
In the array:
int main(void) {
char *str[] = {"hello", "world"};
}
Are "hello" and "world" stored as string literals in read-only memory? Or on the stack?
Technically, a string literal is a quoted string in source code. Colloquially, people use “string literal” to refer to the array of characters created for a string literal. Often we can overlook this informality, but, when asking about storage, we should be clear.
The array created for a string literal has static storage duration, meaning it exists (notionally, in the abstract computer the C standard uses as a model of computing) for the entire execution of the program. Because the behavior of attempting to modify the elements of this array is not defined by the C standard, the C implementation may treat them as constants and may place them in read-only memory. It is not required to do so by the C standard, but this is common practice in C implementations for general-purpose multi-user operating systems.
In the code you show, string literals are used as initializers for an array of pointers. In this use, the array of each string literal is converted to a pointer to its first element, and that address is used as the initial value for the corresponding element of the array of pointers.
The array of the string literal is the same as for any string literal; the C implementation may place it in read-only memory, and common practice is to do so.
Here is what the c17 standard says:
String literals [...] It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined. (6.4.5 p 7)
Like string literals, const-qualified compound literals can be placed into read-only memory and can even be shared. (6.5.2.5 p 13).
I am trying to understand the reason behind not being able to modify a string literal in C.
Why is the following illegal in C?
char* p = "abc";
*p = 'd';
From the C89 Rationale, 3.1.4 String literals:
String literals are specified to be unmodifiable. This specification allows implementations to share copies of strings with identical text, to place string literals in read-only memory, and perform certain optimizations. However, string literals do not have the type array of const char, in order to avoid the problems of pointer type checking, particularly with library functions, since assigning a pointer to const char to a plain pointer to char is not valid. Those members of the Committee who insisted that string literals should be modifiable were content to have this practice designated a common extension (see F.5.5).
I am trying to understanding the passing of string to a called function and modifying the elements of the array inside the called function.
void foo(char p[]){
p[0] = 'a';
printf("%s",p);
}
void main(){
char p[] = "jkahsdkjs";
p[0] = 'a';
printf("%s",p);
foo("fgfgf");
}
Above code returns an exception. I know that string in C is immutable, but would like to know what is there is difference between modifying in main and modifying the calling function. What happens in case of other date types?
I know that string in C is immutable
That's not true. The correct version is: modifying string literals in C are undefined behaviors.
In main(), you defined the string as:
char p[] = "jkahsdkjs";
which is a non-literal character array, so you can modify it. But what you passed to foo is "fgfgf", which is a string literal.
Change it to:
char str[] = "fgfgf";
foo(str);
would be fine.
In the first case:
char p[] = "jkahsdkjs";
p is an array that is initialized with a copy of the string literal. Since you don't specify the size it will determined by the length of the string literal plus the null terminating character. This is covered in the draft C99 standard section 6.7.8 Initialization paragraph 14:
An array of character type may be initialized by a character string literal, optionally
enclosed in braces. Successive characters of the character string literal (including the
terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.
in the second case:
foo("fgfgf");
you are attempting to modify a string literal which is undefined behavior, which means the behavior of program is unpredictable, and an exception is one possibility. From the C99 draft standard section 6.4.5 String literals paragraph 6 (emphasis mine):
It is unspecified whether these arrays are distinct provided their elements have the
appropriate values. If the program attempts to modify such an array, the behavior is
undefined.
The difference is in how you are initializing p[].
char p[] = "jkahsdkjs";
This initializas a writeable array called p, auto-sized to be large enough to contain your string and stored on the stack at runtime.
However, in the case of:
foo("fgfgf");
You are passing in a pointer to the actual string literal, which are usually enforced as read-only in most compilers.
What happens in case of other date types?
String literals are a very special case. Other data types, such as int, etc do not have an issue that is analogous to this, since they are stored strictly by value.
I know that string literal used in program gets storage in read only area for eg.
//global
const char *s="Hello World \n";
Here string literal "Hello World\n" gets storage in read only area of program .
Now suppose I declare some literal in body of some function like
func1(char *name)
{
const char *s="Hello World\n";
}
As local variables to function are stored on activation record of that function, is this the
same case for string literals also? Again assume I call func1 from some function func2 as
func2()
{
//code
char *s="Mary\n";
//call1
func1(s);
//call2
func1("Charles");
//code
}
Here above,in 1st call of func1 from func2, starting address of 's' is passed i.e. address of s[0], while in 2nd call I am not sure what does actually happens. Where does string literal "Charles" get storage. Whether some temperory is created by compiler and it's address is passed or something else happens?
I found literals get storage in "read-only-data" section from
String literals: Where do they go?
but I am unclear about whether that happens only for global literals or for literals local to some function also. Any insight will be appreciable. Thank you.
A C string literal represents an array object of type char[len+1], where len is the length, plus 1 for the terminating '\0'. This array object has static storage duration, meaning that it exists for the entire execution of the program. This applies regardless of where the string literal appears.
The literal itself is an expression type char[len+1]. (In most but not all contexts, it will be implicitly converted to a char* value pointing to the first character.)
Compilers may optimize this by, for example, storing identical string literals just once, or by not storing them at all if they're never referenced.
If you write this:
const char *s="Hello World\n";
inside a function, the literal's meaning is as I described above. The pointer object s is initialized to point to the first character of the array object.
For historical reasons, string literals are not const in C, but attempting to modify the corresponding array object has undefined behavior. Declaring the pointer const, as you've done here, is not required, but it's an excellent idea.
Where string literals (or rather, the character arrays they are compiled to) are located in memory is an implementation detail in the compiler, so if you're thinking about what the C standard guarantees, they could be in a number of places, and string literals used in different ways in the program could end up in different places.
But in practice most compilers will treat all string literals the same, and they will probably all end up in a read-only segment. So string literals used as function arguments, or used inside functions, will be stored in the same place as the "global" ones.
Wouldn't the pointer returned by the following function be inaccessible?
char *foo(int rc)
{
switch (rc)
{
case 1:
return("one");
case 2:
return("two");
default:
return("whatever");
}
}
So the lifetime of a local variable in C/C++ is practically only within the function, right? Which means, after char* foo(int) terminates, the pointer it returns no longer means anything, right?
I'm a bit confused about the lifetime of a local variable. What is a good clarification?
Yes, the lifetime of a local variable is within the scope({,}) in which it is created.
Local variables have automatic or local storage. Automatic because they are automatically destroyed once the scope within which they are created ends.
However, What you have here is a string literal, which is allocated in an implementation-defined read-only memory. String literals are different from local variables and they remain alive throughout the program lifetime. They have static duration [Ref 1] lifetime.
A word of caution!
However, note that any attempt to modify the contents of a string literal is an undefined behavior (UB). User programs are not allowed to modify the contents of a string literal.
Hence, it is always encouraged to use a const while declaring a string literal.
const char*p = "string";
instead of,
char*p = "string";
In fact, in C++ it is deprecated to declare a string literal without the const though not in C. However, declaring a string literal with a const gives you the advantage that compilers would usually give you a warning in case you attempt to modify the string literal in the second case.
Sample program:
#include<string.h>
int main()
{
char *str1 = "string Literal";
const char *str2 = "string Literal";
char source[]="Sample string";
strcpy(str1,source); // No warning or error just Undefined Behavior
strcpy(str2,source); // Compiler issues a warning
return 0;
}
Output:
cc1: warnings being treated as errors
prog.c: In function ‘main’:
prog.c:9: error: passing argument 1 of ‘strcpy’ discards qualifiers from pointer target type
Notice the compiler warns for the second case, but not for the first.
To answer the question being asked by a couple of users here:
What is the deal with integral literals?
In other words, is the following code valid?
int *foo()
{
return &(2);
}
The answer is, no this code is not valid. It is ill-formed and will give a compiler error.
Something like:
prog.c:3: error: lvalue required as unary ‘&’ operand
String literals are l-values, i.e: You can take the address of a string literal, but cannot change its contents.
However, any other literals (int, float, char, etc.) are r-values (the C standard uses the term the value of an expression for these) and their address cannot be taken at all.
[Ref 1]C99 standard 6.4.5/5 "String Literals - Semantics":
In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals. The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence. For character string literals, the array elements have type char, and are initialized with the individual bytes of the multibyte character sequence; for wide string literals, the array elements have type wchar_t, and are initialized with the sequence of wide characters...
It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.
It's valid. String literals have static storage duration, so the pointer is not dangling.
For C, that is mandated in section 6.4.5, paragraph 6:
In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals. The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence.
And for C++ in section 2.14.5, paragraphs 8-11:
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 (3.7).
9 A string literal that begins with u, such as u"asdf", is a char16_t string literal. A char16_t string literal has type “array of n const char16_t”, where n is the size of the string as defined below; it has static storage duration and is initialized with the given characters. A single c-char may produce more than one char16_t character in the form of surrogate pairs.
10 A string literal that begins with U, such as U"asdf", is a char32_t string literal. A char32_t string literal has type “array of n const char32_t”, where n is the size of the string as defined below; it has static storage duration and is initialized with the given characters.
11 A string literal that begins with L, such as L"asdf", is a wide string literal. A wide string literal has type “array of n const wchar_t”, where n is the size of the string as defined below; it has static storage duration and is initialized with the given characters.
String literals are valid for the whole program (and are not allocated not the stack), so it will be valid.
Also, string literals are read-only, so (for good style) maybe you should change foo to const char *foo(int)
Yes, it is valid code, see case 1 below. You can safely return C strings from a function in at least these ways:
const char* to a string literal. It can't be modified and must not be freed by caller. It is rarely useful for the purpose of returning a default value, because of the freeing problem described below. It might make sense if you actually need to pass a function pointer somewhere, so you need a function returning a string..
char* or const char* to a static char buffer. It must not be freed by the caller. It can be modified (either by the caller if not const, or by the function returning it), but a function returning this can't (easily) have multiple buffers, so it is not (easily) threadsafe, and the caller may need to copy the returned value before calling the function again.
char* to a buffer allocated with malloc. It can be modified, but it must usually be explicitly freed by the caller and has the heap allocation overhead. strdup is of this type.
const char* or char* to a buffer, which was passed as an argument to the function (the returned pointer does not need to point to the first element of argument buffer). It leaves responsibility of buffer/memory management to the caller. Many standard string functions are of this type.
One problem is, mixing these in one function can get complicated. The caller needs to know how it should handle the returned pointer, how long it is valid, and if caller should free it, and there's no (nice) way of determining that at runtime. So you can't, for example, have a function, which sometimes returns a pointer to a heap-allocated buffer which caller needs to free, and sometimes a pointer to a default value from string literal, which caller must not free.
Good question. In general, you would be right, but your example is the exception. The compiler statically allocates global memory for a string literal. Therefore, the address returned by your function is valid.
That this is so is a rather convenient feature of C, isn't it? It allows a function to return a precomposed message without forcing the programmer to worry about the memory in which the message is stored.
See also #asaelr's correct observation re const.
Local variables are only valid within the scope they're declared, however you don't declare any local variables in that function.
It's perfectly valid to return a pointer to a string literal from a function, as a string literal exists throughout the entire execution of the program, just as a static or a global variable would.
If you're worrying about what you're doing might be invalid undefined, you should turn up your compiler warnings to see if there is in fact anything you're doing wrong.
str will never be a dangling pointer, because it points to a static address where string literals resides.
It will be mostly read-only and global to the program when it will be loaded.
Even if you try to free or modify, it will throw a segmentation fault on platforms with memory protection.
A local variable is allocated on the stack. After the function finishes, the variable goes out of scope and is no longer accessible in the code. However, if you have a global (or simply - not yet out of scope) pointer that you assigned to point to that variable, it will point to the place in the stack where that variable was. It could be a value used by another function, or a meaningless value.
In the above example shown by you, you are actually returning the allocated pointers to whatever function that calls the above. So it would not become a local pointer. And moreover, for the pointers that are needed to be returned, memory is allocated in the global segment.