I can't understand what the '\0' in the two different place mean in the following code:
string x = "hhhdef\n";
cout << x << endl;
x[3]='\0';
cout << x << endl;
cout<<"hhh\0defef\n"<<endl;
Result:
hhhdef
hhhef
hhh
Can anyone give me some pointers?
C++ std::strings are "counted" strings - i.e., their length is stored as an integer, and they can contain any character. When you replace the third character with a \0 nothing special happens - it's printed as if it was any other character (in particular, your console simply ignores it).
In the last line, instead, you are printing a C string, whose end is determined by the first \0 that is found. In such a case, cout goes on printing characters until it finds a \0, which, in your case, is after the third h.
C++ has two string types:
The built-in C-style null-terminated strings which are really just byte arrays and the C++ standard library std::string class which is not null terminated.
Printing a null-terminated string prints everything up until the first null character. Printing a std::string prints the whole string, regardless of null characters in its middle.
\0 is the NULL character, you can find it in your ASCII table, it has the value 0.
It is used to determinate the end of C-style strings.
However, C++ class std::string stores its size as an integer, and thus does not rely on it.
You're representing strings in two different ways here, which is why the behaviour differs.
The second one is easier to explain; it's a C-style raw char array. In a C-style string, '\0' denotes the null terminator; it's used to mark the end of the string. So any functions that process/display strings will stop as soon as they hit it (which is why your last string is truncated).
The first example is creating a fully-formed C++ std::string object. These don't assign any special meaning to '\0' (they don't have null terminators).
The \0 is treated as NULL Character. It is used to mark the end of the string in C.
In C, string is a pointer pointing to array of characters with \0 at the end. So following will be valid representation of strings in C.
char *c =”Hello”; // it is actually Hello\0
char c[] = {‘Y’,’o’,’\0′};
The applications of ‘\0’ lies in determining the end of string .For eg : finding the length of string.
The \0 is basically a null terminator which is used in C to terminate the end of string character , in simple words its value is null in characters basically gives the compiler indication that this is the end of the String Character
Let me give you example -
As we write printf("Hello World"); /* Hello World\0
here we can clearly see \0 is acting as null ,tough printinting the String in comments would give the same output .
Related
I'm little confused. What is the logically difference between these codes?
#include <iostream>
using namespace std;
int main(){
char a[5]="ABCD"; // this
cout << a;
return 0;
}
Second is
char a[5]={"ABCD"}; // this
Third is
char a[5]={'A','B','C','D'}; // this
char a[5]={"ABCD"};
char a[5]={'A','B','C','D','\0'};
In both cases, the array of characters a is declared with a size of 5 elements of type char: the 4 characters that compose the word "ABCD", plus a final null character ('\0'), which specifies the end of the sequence and that, in the second case, when using double quotes (") it is appended automatically.Attention adding null character separating via commas. A series of characters enclosed in double quotes ("") is called a string constant. The C compiler can automatically add a null character '\0' at the end of a string constant to indicate the end of the string.
Source:This link can help you better
The first two are assignment of a char[5] source to a char[5] array with different syntax only. (the 5 being the four letters plus a null terminator)
The last one will also do the same, but it doesn't explicitly specify a null terminator. Since you are assigning to a char[5], the last one will still zero-fill the remaining space, effectively adding a null terminator and acting the same, but the last one will not throw a compiler error if you assign to a char[4]; it will just leave you with an unterminated array of characters.
In my code, I have a string variable named ChannelPacket.
when I print Channelpacket in gdb, it gives following string :
"\020\000B\001\237\246&\b\000\016\000\002\064\001\000\000\005\000\021\002\000\000\006\000\f\001\001\000\000sZK"
But if i print Channelpacket.c_str(), it gives just "\020 output.
Please help me.
c_str() returns a pointer to char that's understood to be terminated by a NUL character ('\0').
Since your string contains an embedded '\0', it's seen as the end of the string when viewed as a pointer to char.
When viewed as an actual std::string, the string's length is known, so the whole thing is written out, regardless of the embedded NUL characters.
The second byte is a zero, which means the end of the string. If you want to output the raw bytes, rather than treating them as a null-terminated string, you can't use cout << Channelpacket.c_str() - use cout << Channelpacket instead.
I've always had a question about null-terminated strings in C++/C. For example, if you have a character array like so:
char a[10];
And then you wanted to read in characters like so:
for(int i = 0; i < 10; i++)
{
cin >> a[i];
}
And lets in input the following word: questioner
as the input.
Now my question is what happens to the '\0'? If I were to reverse the string, and make it print out
renoitseuq
Where does the null-terminating character go? I thought that good programming practice was to always leave one extra character for the zero-terminating character. But in this example, everything was printed correctly, so why care about the null-terminating character? Just curious. Thanks for your thoughts!
There are cases where you're given a null-terminator, and cases where you have to ask for one yourself.
const char* x = "bla";
is a null-terminated C-style string. It actually has 4 characters - the 3 + the null terminator.
Your string isn't null-terminated. In fact, treating it as a null-terminated string leads to undefined behavior. If you were to cout << it, you'd be attempting to read beyond the memory you're allowed to access, because the runtime will keep looking for a null-terminator and spit out characters until it reaches one. In your case, you were lucky there was one right at the end, but that's not a guarantee.
char a[10]; is just like any other array - un-initialized values, 10 characters - not 11 just because it's a char array. You wouldn't expect int b[10] to contain 10 values for you to play with and an extra 0 at the end just because, would you?
Well, reading that back, I don't see why you'd expect that from a C-string as well - it's not all intuitive.
You are reading 10 chars, not a string. I assume that you also output 10 chars in reverse, so the 0-char plays no role, coz you dont use the array as string, but as an array of single chars...
char a[10] is ten characters, any of which can be a '\0'.
If you put "questioner" in there none of them are.
To get that you'd need a[11] and fill it with "questioner" and then '\0'.
If you were reversing it, you'd get the position of the first '\0' in a[?], reverse up to that and then add a null terminator.
This is a classic banana skin in C, unfortunately it still manages to get under your foot at the most inopportune of moments, even if you are all too familiar with it.
First, I'd like to say that I'm new to C / C++, I'm originally a PHP developer so I am bred to abuse variables any way I like 'em.
C is a strict country, compilers don't like me here very much, I am used to breaking the rules to get things done.
Anyway, this is my simple piece of code:
char IP[15] = "192.168.2.1";
char separator[2] = "||";
puts( separator );
Output:
||192.168.2.1
But if I change the definition of separator to:
char separator[3] = "||";
I get the desired output:
||
So why did I need to give the man extra space, so he doesn't sleep with the man before him?
That's because you get a not null-terminated string when separator length is forced to 2.
Always remember to allocate an extra character for the null terminator. For a string of length N you need N+1 characters.
Once you violate this requirement any code that expects null-terminated strings (puts() function included) will run into undefined behavior.
Your best bet is to not force any specific length:
char separator[] = "||";
will allocate an array of exactly the right size.
Strings in C are NUL-terminated. This means that a string of two characters requires three bytes (two for the characters and the third for the zero byte that denotes the end of the string).
In your example it is possible to omit the size of the array and the compiler will allocate the correct amount of storage:
char IP[] = "192.168.2.1";
char separator[] = "||";
Lastly, if you are coding in C++ rather than C, you're better off using std::string.
If you're using C++ anyway, I'd recommend using the std::string class instead of C strings - much easier and less error-prone IMHO, especially for people with a scripting language background.
There is a hidden nul character '\0' at the end of each string. You have to leave space for that.
If you do
char seperator[] = "||";
you will get a string of size 3, not size 2.
Because in C strings are nul terminated (their end is marked with a 0 byte). If you declare separator to be an array of two characters, and give them both non-zero values, then there is no terminator! Therefore when you puts the array pretty much anything could be tacked on the end (whatever happens to sit in memory past the end of the array - in this case, it appears that it's the IP array).
Edit: this following is incorrect. See comments below.
When you make the array length 3, the extra byte happens to have 0 in it, which terminates the string. However, you probably can't rely on that behavior - if the value is uninitialized it could really contain anything.
In C strings are ended with a special '\0' character, so your separator literal "||" is actually one character longer. puts function just prints every character until it encounters '\0' - in your case one after the IP string.
In C, strings include a (invisible) null byte at the end. You need to account for that null byte.
char ip[15] = "1.2.3.4";
in the code above, ip has enough space for 15 characters. 14 "regular characters" and the null byte. It's too short: should be char ip[16] = "1.2.3.4";
ip[0] == '1';
ip[1] == '.';
/* ... */
ip[6] == '4';
ip[7] == '\0';
Since no one pointed it out so far: If you declare your variable like this, the strings will be automagically null-terminated, and you don't have to mess around with the array sizes:
const char* IP = "192.168.2.1";
const char* seperator = "||";
Note however, that I assume you don't intend to change these strings.
But as already mentioned, the safe way in C++ would be using the std::string class.
A C "String" always ends in NULL, but you just do not give it to the string if you write
char separator[2] = "||". And puts expects this \0 at the ned in the first case it writes till it finds a \0 and here you can see where it is found at the end of the IP address. Interesting enoiugh you can even see how the local variables are layed out on the stack.
The line: char seperator[2] = "||"; should get you undefined behaviour since the length of that character array (which includes the null at the end) will be 3.
Also, what compiler have you compiled the above code with? I compiled with g++ and it flagged the above line as an error.
String in C\C++ are null terminated, i.e. have a hidden zero at the end.
So your separator string would be:
{'|', '|', '\0'} = "||"
Why is it that you can insert a '\0' char in a std::basic_string and the .length() method is unaffected but if you call char_traits<char>::length(str.c_str()) you get the length of the string up until the first '\0' character?
e.g.
string str("abcdefgh");
cout << str.length(); // 8
str[4] = '\0';
cout << str.length(); // 8
cout << char_traits<char>::length(str.c_str()); // 4
Great question!
The reason is that a C-style string is defined as a sequence of bytes that ends with a null byte. When you use .c_str() to get a C-style string out of a C++ std::string, then you're getting back the sequence the C++ string stores with a null byte after it. When you pass this into strlen, it will scan across the bytes until it hits a null byte, then report how many characters it found before that. If the string contains a null byte, then strlen will report a value that's smaller than the whole length of the string, since it will stop before hitting the real end of the string.
An important detail is that strlen and char_traits<char>::length are NOT the same function. However, the C++ ISO spec for char_traits<charT>::length (§21.1.1) says that char_traits<charT>::length(s) returns the smallest i such that char_traits<charT>::eq(s[i], charT()) is true. For char_traits<char>, the eq function just returns if the two characters are equal by doing a == comparison, and constructing a character by writing char() produces a null byte, and so this is equal to saying "where is the first null byte in the string?" It's essentially how strlen works, though the two are technically different functions.
A C++ std::string, however, it a more general notion of "an arbitrary sequence of characters." The particulars of its implementation are hidden from the outside world, though it's probably represented either by a start and stop pointer or by a pointer and a length. Because this representation does not depend on what characters are being stored, asking the std::string for its length tells you how many characters are there, regardless of what those characters actually are.
Hope this helps!