I am working on Microsoft Visual Studio environment. I came across a strange behavior
char *src ="123";
char *des ="abc";
printf("\nThe src string is %c", src[0]);
printf("\tThe dest string is %c",dest[0]);
des[0] = src[0];
printf("\nThe src string is %c", src[0]);
printf("\tThe dest string is %c",dest[0]);
The result is:
1 a
1 a
That means the des[0] is not being initialized. As src is pointing to the first element of the string. I guess by rules this should work.
This is undefined behavior:
des[0] = src[0];
Try this instead:
char des[] ="abc";
Since src and des are initialized with string literals, their type should actually be const char *, not char *; like this:
const char * src ="123";
const char * des ="abc";
There was never memory allocated for either of them, they just point to the predefined constants. Therefore, the statement des[0] = src[0] is undefined behavior; you're trying to change a constant there!
Any decent compiler should actually warn you about the implicit conversion from const char * to char *...
If using C++, consider using std::string instead of char *, and std::cout instead of printf.
Section 2.13.4 of ISO/IEC 14882 (Programming languages - C++) says:
A string literal is a sequence of characters (as defined in 2.13.2) surrounded by double quotes, optionally beginning with the letter L, as in "..." or L"...". A string literal that does not begin with L is an ordinary string literal, also referred to as a narrow string literal. An ordinary string literal has type “array of n const char” and static storage duration (3.7), where n is the size of the string as defined below, and is initialized with the given characters. ...
Whether all string literals are distinct (that is, are stored in nonoverlapping objects) is implementation defined. The effect of attempting to modify a string literal is undefined.
In C, string literals such as "123" are stored as arrays of char (const char in C++). These arrays are stored in memory such that they are available over the lifetime of the program. Attempting to modify the contents of a string literal results in undefined behavior; sometimes it will "work", sometimes it won't, depending on the compiler and the platform, so it's best to treat string literals as unwritable.
Remember that under most circumstances, an expression of type "N-element array of T" will be converted to an expression of type "pointer to T" whose value is the location of the first element in the array.
Thus, when you write
char *src = "123";
char *des = "abc";
the expressions "123" and "abc" are converted from "3-element array of char" to "pointer to char", and src will point to the '1' in "123", and des will point to the 'a' in "abc".
Again, attempting to modify the contents of a string literal results in undefined behavior, so when you write
des[0] = src[0];
the compiler is free to treat that statement any way it wants to, from ignoring it completely to doing exactly what you expect it to do to anything in between. That means that string literals, or a pointer to them, cannot be used as target parameters to calls like strcpy, strcat, memcpy, etc., nor should they be used as parameters to calls like strtok.
vinaygarg: That means the des[0] is not being initialized. As src is pointing to the first element of the string. I guess by rules this should work.
Firstly you must remember that *src and *dst are defined as pointers, nothing more, nothing less.
So you must then ask yourself what exactly "123" and "abc" are and why it cannot be altered? Well to cut a long story short, it is stored in application memory, which is read-only. Why? The strings must be stored with the program in order to be available to your code at run time, in theory you should get a compiler warning for assigning a non-const char* to a const char *. Why is it read-only? The memory for exe's and dll's need to be protected from being overwritten somehow, so it must be read-only to stop bugs and viruses from modifying executing code.
So how can you get this string into modifiable memory?
// Copying into an array.
const size_t BUFFER_SIZE = 256;
char buffer[BUFFER_SIZE];
strcpy(buffer, "abc");
strncpy(buffer, "abc", BUFFER_SIZE-1);
Related
A book on C programming says,
"There is another family of read-only objects that unfortunately are not protected by their type from being modified: string literals.
Takeaway - String literals are read-only.
If introduced today, the type of string literals would certainly be char const[], an array of const-qualified characters. Unfortunately, the const keyword was introduced to the C language much later than string literals, and therefore it remained as it is for backward
compatibility."
Question 1. How can strings be read only, like takeaway says, if they can be modified?
Question 2. "There is another family of read-only objects that unfortunately are not protected by their type from being modified: string literals."
What type is this referring to, which doesn't keep a string literal from being modified?
Question 3. If string literals were introduced today and had the type char const[], how'll they be an array i.e. I can't grasp as to how string literals will be an array of const qualified characters?
How can strings be read only, like takeaway says, if they can be modified?
Because Unfortunately, the const keyword was introduced to the C language much later than string literals, and therefore it remained as it is for backward compatibility.
String literals existed before protection (in the form of const keyword) existed in the language.
Ergo, they are not protected, because the tools to protect it did not exist. A classic example of undefined behavior is writing to a string literal, see Undefined, unspecified and implementation-defined behavior .
What type is this referring to, which doesn't keep a string literal from being modified?
It has the type char[N] - "array of N chars" - where N is the number of characters in the string plus one for zero terminating character.
See https://en.cppreference.com/w/c/language/string_literal .
char *str = "string literal";
// ^^^ - implicit decay from `char [N]` -> `char *`
str[0] = 'a'; // compiler will be fine, but it's invalid code
// or super shorter form:
"string literal"[0] = 'a'; // invalid code
If string literals were introduced today and had the type char const[], how'll they be an array i.e. I can't grasp as to how string literals will be an array of const qualified characters?
The type would be const char[N] - an array of N constant chars, which means you can't modify the characters.
// assuming string literal has type const char [N]
const char *str = "string literal";
// ^^^ - implicit decay from `const char [N]` -> `const char *`
str[0] = 'a'; // compile time error
// or super shorter form:
"string literal"[0] = 'a'; // with `const char [N]` would be a compiler time error
With gcc compiler use -Wwrite-strings to protect against mistakes like writing to string literals.
Question 1. How can strings be read only, like takeaway says, if they can be modified?
The text does not say they can be modified. It says they are not protected from being modified. That is a slight error; properly, it should say they are not protected from attempts to modify them: The rules of the C standard do not prevent you from writing code that attempts to modify a string literal, and they do not define the results when a program executes such an attempt. In some circumstances, attempting to modify a string literal may result in a signal, usually ending program execution by default. In other circumstances, the attempt may succeed, and the string literal will be modified. In other circumstances, nothing will happen; there will be neither a signal nor a change to the string literal. It is also possible other behaviors may occur.
Question 2. "There is another family of read-only objects that unfortunately are not protected by their type from being modified: string literals."
What type is this referring to, which doesn't keep a string literal from being modified?
Technically, a string literal is a piece of source code that has a character sequence inside quotes, optionally with an encoding prefix. During compilation or program execution, an array is generated with the contents of the character sequence and a terminating null character. For string literals without a prefix, the type of that array is char []. (If there is a prefix, the type may also be wchar_t [], char16_t [], or char32_t, depending on the prefix.)
Colloquially, we often refer to this array as the string literal, even though the array is the thing that results from a string literal (an array in memory) not the actual string literal (in the source code).
The type char [] does not contain const, so it does not offer the protections that const char [] does. (Those protections are fairly mild.)
Question 3. If string literals were introduced today and had the type char const[], how'll they be an array i.e. I can't grasp as to how string literals will be an array of const qualified characters?
Your confusion here is unclear. When a string literal appears in source code, the compiler arranges for its contents to be in the memory of the running program. Those contents are in memory as an array of characters. If the rules of C were different, the type of that array would be const char [] instead of char [].
This question already has answers here:
Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"?
(19 answers)
Closed 3 years ago.
I am trying to change value of character array components using a pointer. But I am not able to do so. Is there a fundamental difference between declaring arrays using the two different methods i.e. char A[] and char *A?
I tried accessing arrays using A[0] and it worked. But I am not able to change values of the array components.
{
char *A = "ab";
printf("%c\n", A[0]); //Works. I am able to access A[0]
A[0] = 'c'; //Segmentation fault. I am not able to edit A[0]
printf("%c\n", A[0]);
}
Expected output:
a
c
Actual output:
a
Segmentation fault
The difference is that char A[] defines an array and char * does not.
The most important thing to remember is that arrays are not pointers.
In this declaration:
char *A = "ab";
the string literal "ab" creates an anonymous array object of type char[3] (2 plus 1 for the terminating '\0'). The declaration creates a pointer called A and initializes it to point to the initial character of that array.
The array object created by a string literal has static storage duration (meaning that it exists through the entire execution of your program) and does not allow you to modify it. (Strictly speaking an attempt to modify it has undefined behavior.) It really should be const char[3] rather than char[3], but for historical reasons it's not defined as const. You should use a pointer to const to refer to it:
const char *A = "ab";
so that the compiler will catch any attempts to modify the array.
In this declaration:
char A[] = "ab";
the string literal does the same thing, but the array object A is initialized with a copy of the contents of that array. The array A is modifiable because you didn't define it with const -- and because it's an array object you created, rather than one implicitly created by a string literal, you can modify it.
An array indexing expression, like A[0] actually requires a pointer as one if its operands (and an integer as the other). Very often that pointer will be the result of an array expression "decaying" to a pointer, but it can also be just a pointer -- as long as that pointer points to an element of an array object.
The relationship between arrays and pointers in C is complicated, and there's a lot of misinformation out there. I recommend reading section 6 of the comp.lang.c FAQ.
You can use either an array name or a pointer to refer to elements of an array object. You ran into a problem with an array object that's read-only. For example:
#include <stdio.h>
int main(void) {
char array_object[] = "ab"; /* array_object is writable */
char *ptr = array_object; /* or &array_object[0] */
printf("array_object[0] = '%c'\n", array_object[0]);
printf("ptr[0] = '%c'\n", ptr[0]);
}
Output:
array_object[0] = 'a'
ptr[0] = 'a'
String literals like "ab" are supposed to be immutable, like any other literal (you can't alter the value of a numeric literal like 1 or 3.1419, for example). Unlike numeric literals, however, string literals require some kind of storage to be materialized. Some implementations (such as the one you're using, apparently) store string literals in read-only memory, so attempting to change the contents of the literal will lead to a segfault.
The language definition leaves the behavior undefined - it may work as expected, it may crash outright, or it may do something else.
String literals are not meant to be overwritten, think of them as read-only. It is undefined behavior to overwrite the string and your computer chose to crash the program as a result. You can use an array instead to modify the string.
char A[3] = "ab";
A[0] = 'c';
Is there a fundamental difference between declaring arrays using the two different methods i.e. char A[] and char *A?
Yes, because the second one is not an array but a pointer.
The type of "ab" is char /*readonly*/ [3]. It is an array with immutable content. So when you want a pointer to that string literal, you should use a pointer to char const:
char const *foo = "ab";
That keeps you from altering the literal by accident. If you however want to use the string literal to initialize an array:
char foo[] = "ab"; // the size of the array is determined by the initializer
// here: 3 - the characters 'a', 'b' and '\0'
The elements of that array can then be modified.
Array-indexing btw is nothing more but syntactic sugar:
foo[bar]; /* is the same as */ *(foo + bar);
That's why one can do funny things like
"Hello!"[2]; /* 'l' but also */ 2["Hello!"]; // 'l'
One does usually associate 'unmodifiable' with the term literal
char* str = "Hello World!";
*str = 'B'; // Bus Error!
However when using compound literals, I quickly discovered they are completely modifiable (and looking at the generated machine code, you see they are pushed on the stack):
char* str = (char[]){"Hello World"};
*str = 'B'; // A-Okay!
I'm compiling with clang-703.0.29. Shouldn't those two examples generate the exact same machine code? Is a compound literal really a literal, if it's modifiable?
EDIT: An even shorter example would be:
"Hello World"[0] = 'B'; // Bus Error!
(char[]){"Hello World"}[0] = 'B'; // Okay!
A compound literal is an lvalue and values of its elements are modifiable. In case of
char* str = (char[]){"Hello World"};
*str = 'B'; // A-Okay!
you are modifying a compound literal which is legal.
C11-§6.5.2.5/4:
If the type name specifies an array of unknown size, the size is determined by the initializer list as specified in 6.7.9, and the type of the compound literal is that of the completed array type. Otherwise (when the type name specifies an object type), the type
of the compound literal is that specified by the type name. In either case, the result is an lvalue.
As it can be seen that the type of compound literal is a complete array type and is lvalue, therefore it is modifiable unlike string literals
Standard also mention that
§6.5.2.5/7:
String literals, and compound literals with const-qualified types, need not designate distinct objects.101
Further it says:
11 EXAMPLE 4 A read-only compound literal can be specified through constructions like:
(const float []){1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6}
12 EXAMPLE 5 The following three expressions have different meanings:
"/tmp/fileXXXXXX"
(char []){"/tmp/fileXXXXXX"}
(const char []){"/tmp/fileXXXXXX"}
The first always has static storage duration and has type array of char, but need not be modifiable; the last two have automatic storage duration when they occur within the body of a function, and the first of these
two is modifiable.
13 EXAMPLE 6 Like string literals, const-qualified compound literals can be placed into read-only memory and can even be shared. For example,
(const char []){"abc"} == "abc"
might yield 1 if the literals’ storage is shared.
The compound literal syntax is a short hand expression equivalent to a local declaration with an initializer followed by a reference to the unnamed object thus declared:
char *str = (char[]){ "Hello World" };
is equivalent to:
char __unnamed__[] = { "Hello world" };
char *str = __unnamed__;
The __unnamed__ has automatic storage and is defined as modifiable, it can be modified via the pointer str initialized to point to it.
In the case of char *str = "Hello World!"; the object pointed to by str is not supposed to be modified. In fact attempting to modify it has undefined behavior.
The C Standard could have defined such string literals as having type const char[] instead of char[], but this would generate many warnings and errors in legacy code.
Yet it is advisable to pass a flag to the compiler to make such string literals implicitly const and make the whole project const correct, ie: defining all pointer arguments that are not used to modify their object as const. For gcc and clang, the command line option is -Wwrite-strings. I also strongly advise to enable many more warnings and make them fatal with -Wall -W -Werror.
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 was going through some documentation which states that
First Case
char * p_var="Sack";
will create a constant string literal.
And hence code like
p_var[1]="u";
will fail because of that property.
Second Case
Also mentioned is that this is possible only for character literals and not for other data types through pointers. So code like
float *p="3.14";
will fail, resulting in a compiler error.
But when i try it out i don't get compiler errors ,accessing it though gives me 0.000000f(using gcc on Ubuntu).
So regarding the above, i have three queries:
Why are string literals created in First Case read-only?
Why are only string literals allowed to be created and not other constants like float through pointers?
3. Why is Second Case not giving me compiler errors?
Update
Please discard the 3rd question and second case. I tested it by adding quotes.
Thanks
The premise is wrong: pointers don’t create any string literals, neither read-only nor writeable.
What does create a read-only string literal is the literal itself: "foo" is a read-only string literal. And if you assign it to a pointer, then that pointer points to a read-only memory location.
With that, let’s turn to your questions:
Why are string literals created in First Case read-only?
The real question is: why not? In most cases, you won’t want to change the value of a string literal later on so the default assumption makes sense. Furthermore, you can create writeable strings in C via other means.
Why are only string literals allowed to be created and not other constants like float?
Again, wrong assumption. You can create other constants:
float f = 1.23f;
Here, the 1.23f literal is read-only. You can also assign it to a constant variable:
const float f = 1.23f;
Why is Second Case not giving me compiler errors?
Because the compiler cannot check in general whether your pointer points to read-only memory or to writeable memory. Consider this:
char* p = "Hello";
char str[] = "world"; // `str` is a writeable string!
p = &str[0];
p[1] = 'x';
Here, p[1] = 'x' is entirely legal – if we hadn’t re-assigned p beforehand, it would have been illegal. Checking this cannot be generally done at compile-time.
Regarding your question:
Why are string literals created in First Case read-only?
char *p_var="Sack";
Well, the p_var is assigned with the starting address of the memory allocated to the string "Sack". p_var content is not read-only, since you haven't put the const keyword anywhere in C constructs. Although manipulating the p_var contents like strcpy or strcat may cause undefined behavior.
Quote C ISO 9899:
The declaration
char s[] = "abc", t[3] = "abc";
defines plain char array objects s and t whose elements are initialized with character string literals.
This declaration is identical to:
char s[] = { 'a', 'b', 'c', '\0' },
t[] = { 'a', 'b', 'c' };
The contents of the arrays are modifiable. On the other hand, the declaration:
char *p = "abc";
defines p with type pointer to char and initializes it to point to an object with type array of char with length 4 whose elements are initialized with a character string literal. If an attempt is made to use p to modify the contents of the array, the behavior is undefined.
An explanation of why it could be read-only per your platform and compiler:
Commonly string literals will be put in "read-only-data" section which gets mapped into the process space as read-only (which is why you seem to not being allowed to change it).
But some platforms do allow, the data segment to be writable.
Why are only string literals allowed to be created and not other constants like float? and the third question.
To create a float constant you should use:
const float f=1.5f;
Now, when you are doing:
float *p="3.14";
you are basically assigning the string literal's address to a float pointer.
Try compiling with -Wall -Werror -Wextra. You will find out what is happening.
It works because, in practice, there's no difference between a char * and a float * under the hood.
Its as if you are writing this:
float *p=(float*) "3.14";
This is a well-defined behaviour, unless the memory alignment requirements of float and char differ, in which case it results in undefined behaviour (Reference: C99, 6.3.2.3 p7).
Efficiency
they are
It's a string mind the quotes
float *p="3.14";
This is also a string literal !
Why are string literals created in First Case read-only?
No, both "sack" and "3.14" are string literals and both are read-only.
Why are only string literals allowed to be created and not other constants like float?
If you want to create a float const then do:
const float p=3.14;
Why is Second Case not giving me compiler errors?
You are making the pointer p point to a string literal. When you dereference p, it expects to read a float value. So there's nothing wrong as far as the compiler can see.