This question already has answers here:
Why does writing to a string literal in this C program segfault?
(6 answers)
What's wrong with my strcpy? [closed]
(4 answers)
Closed 9 years ago.
Yesterday I had a little trouble with a homemade "strcpy" function. It works now though but I'm a little confused!
char* a = "Hello, World!"; //Works
char b[] = "Hello, World!"; //Works also
strcpy(a, "Hello!"); //Segmentation fault
strcpy(b, "Haha!!"); //Works..
Where is the difference? Why does the char pointer cause a "Segmentation fault"?
Why does this even work? :
char* a = "Haha"; //works
a = "LOL"; //works..
char* a = "Hello, World!";
gives you a pointer to a string literal. A string literal may exist in read only memory so its contents cannot be changed.
char* a = "Haha"; //works
a = "LOL"; //works..
changes the pointer a to point to a different string literal. It doesn't attempt to modify the contents of either string literal so is safe/correct.
char b[] = "Hello, World!"
declares an array on the stack and initialises it with the contents of a string literal. Stack memory is writeable so its perfectly safe to change the contents of this memory.
In your first example since you are trying to write to a read only memory pointed by a,you will get the segmentation fault.If you want to use pointers then allocate the memory on heap,use and delete it after its use.
Where as b is an array of chars initialized with "Hello, World!".
In the second example you are making the pointer to point to different string literal which should be fine.
Related
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)
What is the difference between char s[] and char *s?
(14 answers)
Closed 3 years ago.
My question comes in three part.
char str[] = "hello!";
char *str1 = "heello!";
puts(str);
str1[1] = 1
puts(str1);
printf("%s\n", str1);
printf("%p\n", str1);
Ouput of code:
hello!
h1ello!
heello!
0x10735ef9b
1) I understand that when you're declaring a char * type, you're declaring a pointer that points to char type. And usually for pointer, since it contains the address of the variable its pointing to, when you print the pointer itself, you'll get the address stored in the pointer and only when you dereference the pointer, then you'll get the content stored in the variable the pointer is pointing to.
However for a char * type, why is it that it doesn't function like a regular pointer? You can print the content of the variable which the 'char *' pointer is pointing to without dereferencing it and also use it like a string variable(rather than a pointer) by changing the value of the str1[1] as shown above.
2) I was taught that you shouldn't declare a pointer variable as such in the topic of pointer:
char *str1 = "heello!";
because the pointer variable would be pointing to whatever address happens to be in the memory at that time, thus running a risk of changing the value in a memory location that you don't mean to and resulting in segfault.
Hence you should do the following:
char *str1;
char c;
str1 = &c;
*addr = "hello!"
By doing the above, you'll make sure that your pointer is pointing to the right location before dereferencing and writing to the location.
However in the notes on string, it says to declare and initialise as such:
char *str1 = "heello!";
I'm quite confused as to what I should follow now when declaring a char * pointer. Please help.
3) It says in my notes that the location of string is stored in a read only region of the memory called the text region.
Hence doing this:
char *str1 = "hello!";
str[1] = '.';
Will crash your program.
But when you do this:
char str2[7] = "hello!";
str2[1] = '.';
This is okay and will successfully change the second element in the string from 'e' to '.'
The explanation to this was that str1 points to a read only region in the memory, while str2 contains a copy of the string on the stack.
I do not understand how the above 2 different ways of declaring the string variable will make such a big difference when executing the code. Please help.
Thank you!
String literals like "heello!" are really read-only arrays of characters (including the string null-terminator).
With
char *str1 = "heello!";
you make str1 point to the first character of such an array.
And when you do str1[1] = 1 you attempt to modify the read-only array, which is forbidden and leads to undefined behavior (which can but doesn't have to lead to crashes).
That's why you either should use your own arrays for strings, or use const char * to point to literal strings.
This question already has answers here:
What is the difference between char s[] and char *s?
(14 answers)
Closed 8 years ago.
I have a very simple c question regarding memcopy that I have been able to fix but not understand why. The following code crashes with an "access violation" at runtuime:
char *src = "HELLO WORLD!";
char * dest = "hello world!";
memcpy(dest, src, strlen(src)+1);
dest would appear to be big enough and as I understand it memcpy should just blindly copy the bytes so I don't understand the problem.
Changing dest to an array e.g char dest[13]; fixes things, so is it a requirement that dest must be uninitialized memory and creating an array does this for you whereas the pointer declaration does not?
Cheers
When you do:
char *dest = "hello world!";
"hello world!" is kept in the read-only parts of the memory and pointer named dest is pointed to it, making any writing operation on this memory illegal. So if you try to use it as a destination in your memcpy; it's an access violation.
You should use:
char dest[] = "hello world!";
A string literal, "...", when used to initialise a char pointer, has type const char *, i.e. it's read-only.
dest should therefore have type const char *, at which point your code won't compile.
If you turn all your compiler warnings on then you'll get a warning for the "char * dest = ..." line.
char dest [] = "hello world!" and char* dest = "hello world!" is not the same thing. char dest[]="hello world!" is an array with 13 elements (the string is null-terminated). char* dest ="hello world!" is a pointer to an constant, null terminated string (also 13 characters).
This question already has answers here:
Segmentation fault with strcpy() [duplicate]
(7 answers)
Closed 9 years ago.
int main()
{
char *s="Hello";
*s="World";
printf("%s\n",s);
}
Why does the above program result in a segmentation fault?
int main()
{
char *s="Hello"; // makes 's' point to a constant
*s="World"; // modifies what 's' points to
printf("%s\n",s);
}
The first line of code makes s point to a constant. The second line tries to modify what s points to. So you are trying to modify a constant, which you can't do because a constant is ... well ... constant.
because *s is a char not a char*(string)
char *s="Hello";
declares a pointer to a string literal "Hello". This may exist in read-only memory so the line
*s="World";
is results in undefined behaviour. A crash is a valid (and useful) form of undefined behaviour.
Either of the following alternatives would work
const char* s = "Hello";
s="World";
printf("%s\n",s);
char s[16] = "Hello";
strcpy(s, "World";)
printf("%s\n",s);
s points to static (global) memory when it is created. You cannot reassign to that memory at run-time, hence, the crash.
*s is the first char of the string, so assigning string to character makes error.
If you want to assing string use s = "world"
int main()
{
char *s="Hello";
s="World";
printf("%s\n",s);
}
now try it will work.
char *s="hello"; Here s is in readonly location. So we can assign another string, but cannot rewrite new string.
s = "hello"; //work
strcpy(s, "hello"); //segmentation fault
There are two problems here.
The statement
*s = "World";
dereferences s, which gives you the first character of the string "Hello", or 'H'. So you're trying to assign a pointer value (the address of the string "World") to a single char object (the first character of the "Hello" string literal).
But...
On some systems (such as yours, apparently), string literals are stored in a read-only data segment, and attempting to modify read-only memory will lead to a runtime error on some systems. Hence the crash.
To change s to point to the "World" string literal, simply drop the dereference:
s = "World";
*s is the same as s[0]. s[0] has room to store a single character; in this case a 'W'.
There's not room to store the location of "World" in that character.
That's why you're getting a segmentation fault.
This question already has answers here:
What is the difference between char s[] and char *s?
(14 answers)
Closed 9 years ago.
In my GCC 32-bit compiler, the following code gives the output
char *str1="United";
printf("%s",str1);
output:
United
then should I consider char *str1="United"; the same as char str1[]="United"?
The two are not the same: char *str1="United" gives you a pointer to a literal string, which must not be modified (else it's undefined behavior). The type of a literal string should really be const but it's non-const for historical reasons. On the other hand, char str1[]="United" gives you a modifiable local string.
char* str = "United"; is read-only. You wouldn't be able to reach inside the string and change parts of it:
*str = 'A';
Will most likely give you a segmentation fault.
On the other hand char str1[] = "United"; is an array and so it can be modified as long as you don't exceed the space allocated for it (arrays cannot be resized). For example this is perfectly legal:
char str[] = "United";
str[0] = 'A';
printf("%s\n", str);
This will print Anited.
See the comp.lang.c.faq, question 1.32. It basically boils down to the fact that declaring the string in array form (char str[] = "foo") is identical to char str[] = {'f','o','o'}, which is identical to char str[] = {102, 111, 111}; that is, it is a normal array on the stack. But when you use a string literal in any other context it becomes "an unnamed, static array of characters, [which] may be stored in read-only memory, and which therefore cannot necessarily be modified." (And trying to modify it results in undefined behavior wherever it happens to be stored, so don't).
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is the difference between char a[] = “string”; and char *p = “string”;
char *str = "Hello";
printf("%c",++*str);
This gives segmentation fault on linux with gcc.
The moment the first statement is changes to as
char str[10] = "Hello";
It works. What may be the reason?
It is undefined behaviour to attempt to modify a string literal.
The compiler is free to place it in read-only memory (as it probably does in your case). Attempting to modify read-only memory is what's probably triggering the segfault.
This statement char *str = "Hello"; stores the string "Hello" in RO-section and assigns the address of the area of RO-section(in which "Hello"is stored) to str. The data stored in RO-section cannot be modified thus you are getting a segfault.
char str[10] = "Hello";
is also wrong. You should instead write
char str[10];
strncpy(str,"Hello",sizeof(str));