I'm new with C++ and I started to wonder, what happens if you leave the null character out when defining a char array?
For example, if I define a char array with the null character:
char myarray[] = {'a', 'b', 'c', '\0'};
and then I define it without the null character:
char myarray[] = {'a', 'b', 'c'};
What is the importance of the null character in this scenario? Might the absence of null character in the example above cause some problems later on?...Do you recommend always including or excluding the null character when defining char arrays this way?
Thank you for any help :)
It means that anything that takes a char* as parameter, expecting it to be a null-terminated string, will invoke undefined behaviour, and fail in one way or another*.
Some examples are strlen, the std::string(const char*) constructor, the std::ostream operator<< specialization for char*...
* undefined behaviour means it could even work "correctly", but there is no guarantee this is reproducible.
char myarray[] = {'a', 'b', 'c'};
If you define with out nul character it is a valid character array not valid string.
1.You should not use this character array as an argument to the string functions like strlen(),strcpy(),etc..
2.You should not print this as we print string with %s in C.
3.You can print character by character.
4.You can compare character by character.
Further char myarray[] = {'a', 'b', 'c', '\0'}; is equal to "abc"
but char myarray[] = {'a', 'b', 'c'}; is not equal to "abc"
If you don't have the terminating null-character, you can still use your char array like you could with the null-char.
However, functions that expect null-terminated strings (like strlen), will not stop at the end, since they don't know where the end is. (That's what the null-char is for)
They will therefore continue to work in memory until they either go out of bounds and you get a segmentation fault or run until they find their null-char.
Basically, if you want your char array to be a string, append a null-char to denote the end.
what happens if you leave the null character out when defining a char array?
You get an array containing just the characters you specify.
What is the importance of the null character in this scenario?
In C, it's conventional to represent a string as a null-terminated character array. This convention is sometimes used in C++ to interoperate with C-style interfaces, or to work with string literals (which inherited their specification from C), or because the programmer thinks it's a good idea for some reason. If you're going to do this, then obviously you'll need to terminate all the arrays you want to interpret as strings.
The question seems to be about C++, although you've also tagged it C for some reason. In C++, you usually want to use std::string to manage strings for you. Life is too short for messing around with low-level arrays and pointers.
Might the absence of null character in the example above cause some problems later on?
If you pass a non-terminated array to a function expecting a terminated array, then it will stomp off the end of the array causing undefined behaviour.
Do you recommend always including or excluding the null character when defining char arrays this way?
I recommend understanding what the array is supposed to be used for, and include the terminator if it's supposed to be a C-style string.
What is the importance of the null character in this scenario?
High.
Might the absence of null character in the example above cause some problems later on?
Yes. C and C++ functions taking a char* that points to this C-string will require it to be null-terminated.
Do you recommend always including or excluding the null character when defining char arrays this way?
Neither. I recommend using std::string, since you said you are writing C++.
null character will be used by strlen like functions if you wish to use your array as a string. If I need a string because I want to use some text I write:
const char* mystr = "abc"; // it is already null terminated
writing:
char myarray[] = {'a', 'b', 'c', '\0'};
is to verbose
Related
I am new to c++ language,recently, as I was taught that:
we should put '\0' at the end of char array while doing initialization ,for example :
char x[6] = "hello"; //OK
However,if you do :
char x[5] = "hello";
Then this would raise the error :
initializer-string for array of chars is too long
Everything goes as I expect until the experssion below does not raise the compile error...:
char x[5] = {'h','e','l','l','o'};
This really confuses me , So I would like to ask two questions :
1.Why doesn't expression char x[5] = "hello"; raise error?
2.To my knowledge,the function strlen() would stop only if it finds '\0' to determine the lengh of char array,in this case,what would strlen(x) return?
Thanks!
The string literal "hello" has six characters, because there's an implied nul terminator. So
char x[] = "hello";
defines an array of six char. That's almost always what you want, because the C-style string functions (strlen, strcpy, strcat, etc.) operate on C-style strings, which are, by definition, nul terminated.
But that doesn't mean that every array of char will be nul terminated.
char x[] = { 'h', 'e', 'l', 'l', 'o' };
This defines an array of five char. Applying C-style string functions to this array will result in undefined behavior, because the array does not have a nul terminator.
You can do character-by-character initialization and create a valid C-style string by explicitly including the nul terminator:
char x[] = { 'h', 'e', 'l', 'l', 'o', '\0' };
This defines an array of six char that holds a C-style string (i.e., a nul terminated sequence of characters).
The key here is to separate in your mind the general notion of an array of char from the more specific notion of an array of char that holds a C-style string. The latter is almost always what you want to do, but that doesn't mean that there is never a use for the former. It's just that the former is uncommon.
As an aside, in C you're allowed to elide the nul terminator:
char x[5] = "hello";
this is legal C, and it creates an array of 5 char, with no nul terminator. In C++ that's not legal.
Why doesn't expression char x[5] = "hello"; raise an error?
This is not true. The appearance of an error is expected in this case.
To my knowledge, the function strlen() would stop only if it finds '\0' to determine the length of the char array, in this case, what would strlen(x) return?
If you can run the code somehow, the program will undergo an undefined-behavior. That is, you will not get what you would expect. The strlen() will only stop counting when it finds a null-terminator, i.e. it may go outside the initialized part of the char array and access the uninitialized ones – it's where the UB is invoked.
This question already has answers here:
What is a null-terminated string?
(7 answers)
Closed 2 years ago.
There is a saying when we declare char variable.
We should declare like this -> char ArrayName[Maximum_C-String_Size+1];
For example:
char arr[4+1] = {'a', 'b', 'c', 'd'}
but
arr[4] = {'a', 'b', 'c', 'd'} is also work
why need to add 1?
thanks!
There is no need to do this, unless you are defining something that will be used as a null-terminated string.
// these two definitions are equivalent
char a[5] = { 'a', 'b', 'c', 'd' };
char b[5] = { 'a', 'b', 'c', 'd', '\0' };
If you only want an array with 4 char values in it, and you won't be using that with anything that expects to find a string terminator, then you don't need to add an extra element.
If you’re storing a C-style string in an array, then you need an extra element for the string terminator.
Unlike C++, C does not have a unique string data type. In C, a string is simply a sequence of character values including a zero-valued terminator. The string "foo" is represented as the sequence {'f','o','o',0}. That terminator is how the various string handling functions know where the string ends. The terminator is not a printable character and is not counted towards the length of the string (strlen("foo") returns 3, not 4), however you need to set aside space to store it. So, if you need to store a string that’s N characters long, then the array in which it is stored needs to be at least N+1 elements wide to account for the terminator.
However, if you’re storing a sequence that’s not meant to be treated as a string (you don’t intend to print it or manipulate it with the string library functions), then you don’t need to set aside the extra element.
I'm new to programming and very new to C++, and I recently came across strings.
Why do we need a null terminator at the end of a character list?
I've read answers like since we might not use all the spaces of an array therefore we need the null terminator for the program to know where the string ends e.g. char[100] = "John"
but why can't the program just loop through the array to check how many spaces are filled and hence decide the length?
And if only four characters are filled in the array for the word "John", what are the others spaces filled with?
The other characters in the array char john[100] = "John" would be filled with zeros, which are all null-terminators. In general, when you initialize an array and don't provide enough elements to fill it up, the remaining elements are default-initialized:
int foo[3] {5}; // this is {5, 0, 0}
int bar[3] {}; // this is {0, 0, 0}
char john[5] = "John"; // this is {'J', 'o', 'h', 'n', 0}
char peter[5] = "Peter"; // ERROR, initializer string too long
// (one null-terminator is mandatory)
Also see cppreference on Array initialization. To find the length of such a string, we just loop through the characters until we find 0 and exit.
The motivation behind null-terminating strings in C++ is to ensure compatibility with C-libraries, which use null-terminated strings. Also see What's the rationale for null terminated strings?
Containers like std::string don't require the string to be null-terminated and can even store a string containing null-characters. This is because they store the size of the string separately. However, the characters of a std::string are often null-terminated anyways so that std::string::c_str() doesn't require a modification of the underlying array.
C++-only libraries will rarely -if ever- pass C-strings between functions.
The existance of a null terminator is a design decision. The purpose it serves is marking the end of the string. There are other ways to do this, for example in Pascal the first element of a string is it's size so no null terminator is needed.
In the example you give only the first 5 elements of the array will be initialized, the rest are zero initialized. Notice how I said 5 elements and not just four. The fifth element is the null terminator.
Sure the program can loop through the string to find out it's length but how will it know when to stop looping?
The nul terminator is what tells you what spaces are filled. Everything up to and including the nul terminator has been filled. Everything after it has not.
There is no general notion of which elements of an array have been filled. An array holds some number of elements; its size is determined when it is created. All of its elements have some value initially; there's no way, in general, to determine which ones have been assigned a value and which ones have not from looking at the values of the elements.
Strings are arrays of char and a coding convention that the "end" of the string is marked by a nul character. Most of the string manipulation functions rely on this convention.
A string literal, such as "John", is an array of char. "John" has 5 elements in the array: 'J', 'o', 'h', 'n', '\0'. The function strcpy, for example, copies characters until it sees that nul terminator:
char result[100]; // no meaningful values here
strcpy(result, "John");
After the call to strcpy, the first five elements of result are 'J', 'o', 'h', 'n', and '\0'. The rest of the array elements have no meaningful values.
I would be remiss if I didn't mention that this style of string comes from C, and is often referred to as C-style strings. C++ supports all of the C string stuff, but it also has a more sophisticated notion of a string, std::string, which is completely different. In general, you should be using C++-style strings and not C-style strings.
I'm a bit baffled that this is allowed:
char num[6] = "a";
What is happening here? Am I assigning a pointer to the array or copying the literal values into the array (and therefore I'm able to modify them later)?
Why can I assign a string literal less than the array itself? What is happening here?
This is well defined. When initialize character arrays with string literal,
If the size of the array is specified and it is larger than the number
of characters in the string literal, the remaining characters are
zero-initialized.
So,
char num[6] = "a";
// equivalent to char num[6] = {'a', '\0', '\0', '\0', '\0', '\0'};
Am I assigning a pointer to the array or copying the literal values into the array (and therefore I'm able to modify them later)?
The value will be copied, i.e. the elements of the array will be initialized by the chars of the string literal (including '\0').
String literals can be used to initialize character arrays. If an array is initialized like char str[] = "foo";, str will contain a copy of the string "foo".
Successive characters of the string literal (which includes the implicit terminating null character) initialize the elements of the array.
char num[6] = "a";
is equivalent to
char num[6] = {'a', '\0', '\0', '\0', '\0', '\0'};
Why can I assign a string literal less than the array itself?
This is allowed by the language. It is often useful to be able to add more characters to the array later, which wouldn't be possible if the existing characters filled the entire array.
Am I assigning a pointer to the array
No. You cannot assign a pointer to an array, so that is not happening.
or copying the literal values into the array
That is exactly what is happening.
and therefore I'm able to modify them later
You are able to modify the array, indeed.
Just use char num[6] = {"a"};. It works.
This kind of declaration is a special syntax sugar thing. It's equivalent to
char num[6] = {'a', 0}
The array is always modifiable. Its contents after such a declaration would be a character representing 'a', a zero (NUL terminator) and the remainder of the array will also be zeroed (zero initialization).
That is one type of declaration whcih is equivalent to
char num[6] = {'a','\0'};
You declared c-string with length of max. 5 normal chars, at the end must me \0 to end c - string.
With declaration you can use
char num[6] = "a";
then you need to assign value:
With strcpy(dest,src)
strcpy(num,"test");
Char by char
num[0]='t';
num[1]='e';
num[2]='s';
num[3]='t';
num[4]='\0';
i am learning c/c++ recently.but i don't understand the difference between
int a(chat* str,int len)
{
cout<<str<<len;
}
and
int a(char* str)
{
cout<<str<<strlen(str);
}
When you pass char* without a length, how would you know how many elements to process? char* means a pointer to a character. When you pass a pointer, you have no idea (and cannot find out) how much memory (if any) was allocated for the pointer.
That's why C-strings use are null-terminated (they end with a '\0' character), so you can detect their length by iterating the pointer. Hence, if you want to use a pointer without giving the length of its allocated memory, you need to obey some conventions. But in general, e.g. when passing a buffer, you shouldn't expect any end-signalling character, so in this case you need to pass the length, otherwise may end up reading/writing out of bounds.
For your particular example, you're fine with passing only a pointer provided you use your function only on C-strings, since strlen(str) uses this convention of counting until encountering a '\0'.
Buffer overflows are one the most messy and nightmarish programming errors, which can result in serious security issues. That's why you should try (whenever possible) to use std::string from the C++ standard library instead of C-style char* strings.
A C-String should always contain a termination character, we call it null character. It's technically 0 (not the number 0, but ASCII 0)
When we create a char* and initialize it with some text, it automatically adds the '\0' to the end.
char* c = "Hello";
This will create an array of char with six elements. Yes, six elements.
c = {'H', 'e', 'l', 'l', 'o', '\0'}
When you print c, it will search till it finds that '\0'. What if someone replaces it.
c[5] = '!';
Then the system can't determine the end of the text. Then it will keep on reading the memory (which does not belong to that variable, or may be even the program) until it hits a null char.
That is the main reason to pass the size (or number or chars to read) to a function.
On the other hand, if you need to read some data from a stream, you can use a buffer. In that case, you should specify how many bytes to read, in that way you will not cause buffer overflows.
Above answers are to the point. So I'm going to discuss other perspective behind of practise of passing length along with char *.
As others said, not always, the string pointed by char * end up with \0. Only when the string ends with \0 strlen() would actually work. There are certain use-cases for example binary coding, where data is represented as string. In such case, char * would not end with \0. Besides, there can be certain use-cases to read / write only up to certain length / size. In such case, it is always necessary to test whether the input length is within the range of length of total string. So as a common case, length has been passed explicitly, which can be used in any way as desired by the caller.