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.
Related
In C and C++ the rules are the same. In C,
[§6.4.4.4]/2 An integer character constant is a sequence of one or
more multibyte characters enclosed in single-quotes, as in 'x'.
In C++,
[§2.14.3]/1 A character literal is one or more characters enclosed
in single quotes, as in 'x', optionally preceded by one of the
letters u, U, or L, as in u'y', U'z', or L'x',
respectively.
The key phrase is "one or more". In contrast, a string literal can be empty, "", presumably because it consists of the null terminating character. In C, this leads to awkward initialization of a char. Either you leave it uninitialized, or use a useless value like 0 or '\0'.
char garbage;
char useless = 0;
char useless2 = '\0';
In C++, you have to use a string literal instead of a character literal if you want it to be empty.
(somecondition ? ' ' : '') // error
(somecondition ? " " : "") // necessary
What is the reason it is this way? I'm assuming C++'s reason is inherited from C.
The reason is that a character literal is defined as a character. There may be extensions that allow it to be more than one character, but it needs to be at least one character or it just doesn't make any sense. It would be the same as trying to do:
int i = ;
If you don't specify a value, what do you put there?
This is because an empty string still contains the the null character '\0' at the end, so there is still a value to bind to the variable name, whereas an empty character literal has no value.
String is a set of character terminated by a NULL character ( '\0' ).
So a Empty string will always have a NULL character in it at the end .
But in case of a character literal no value is there.
it needs at least one character.
This question already has answers here:
c++ compile error: ISO C++ forbids comparison between pointer and integer
(5 answers)
Closed 5 years ago.
Beginning programmer here...
I'm writing a very simply program for my computer science class and I ran into an issue that I'd like to know more about. Here is my code:
#include <iostream>
using namespace std;
int main(int argc, const char * argv[])
{
char courseLevel;
cout << "Will you be taking graduate or undergraduate level courses (enter 'U'"
" for undergraduate,'G' for graduate.";
cin >> courseLevel;
if (courseLevel == "U")
{
cout << "You selected undergraduate level courses.";
}
return 0;
}
I'm getting two error messages for my if statement:
1) Result of comparison against a string literal is unspecified (use strncmp instead).
2) Comparison between pointer and integer ('int' and 'const char*').
I seem to have resolved the issue by enclosing my U in single quotes, or the program at least works anyway. But, as I stated, I'd simply like to understand why I was getting the error so I can get a better understanding of what I'm doing.
You need to use single quotes instead.
In C, (and many other languages) a character constant is a single character1 contained in single quotes:
'U'
While a string literal is any number of characters contained in double quotes:
"U"
You declared courseLevel as a single character: char courseLevel; So you can only compare that to another single char.
When you do if (courseLevel == "U"), the left side is a char, while the right side is a const char* -- a pointer to the first char in that string literal. Your compiler is telling you this:
Comparison between pointer and integer ('int' and 'const char*')
So your options are:
if (courseLevel == 'U') // compare char to char
Or, for sake of example:
if (courseLevel == "U"[0]) // compare char to first char in string
Note for completeness: You can have mulit-character constants:
int a = 'abcd'; // 0x61626364 in GCC
But this is certainly not what you're looking for.
Rapptz is right, but I think some more elaboration should help...
courseLevel == "U"
In C and C++, double-quotes create string literals - which are arrays of characters finishing with a numerical-0 ASCII-NUL terminating sentinel character so programs can work out where the text ends. So, you basically are asking if a character is equal to an array of characters... they just can't be compared. Similar questions that are valid are:
does this character variable hold a specific character value: courseLevel == 'U'
does this character variable appear in a specific array: strchr(courseLevel, "U")
does this character variable match the first element in a specific array: courseLevel == "U"[0]
Of course, the first one of these is the one that makes intuitive sense in your program.
The reason why you get an error is because string literals in C and C++ end with a null terminated character \0 while single characters don't. So when you compare to a char to a string literal you're comparing the character literal to a char array {'U','\0'}.
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 .
I'm doing an assignment for school and I seem to have a problem with other characters being in the array. (Amongst other problems...)
In one function, I declared the array..
const int MAXSIZE = 100;
char inFix[MAXSIZE];
And this code is used to put chars into the array
//loop to store user input until enter is pressed
while((inputChar = static_cast<char>(cin.get()))!= '\n')
{
//if function to decide if the input should be stored or not
if(isOperator(inputChar) || isdigit(static_cast<int>(inputChar)) || inputChar == '(' || inputChar == ')')
{
inFix[a] = inputChar; //stores input
a++;
}
}
At the end of this, I append the null character to the array though I wasn't sure if I should do this:
inFix[MAXSIZE] = '\0';
Or if I should've used strcat.. either way... in my next function, I use strcat to append a parenthesis to the end.
But I've been having problems with the code, so I ran a for loop to print what is inside the infix array at the beginning of my next function, just to see...
And I get this annoying beeping sound, and a string of weird characters like hearts, and music signs... and.. a whole list of odd characters. What could be the problem? Thanks.
EDIT: by the way, I input 9*4, and I run the for loop after i append the parenthesis, so at the beginning of the output, I get:
9*4) and then the string of odd characters...
so I ran a for loop to print what is inside the infix array at the beginning of my next function, just to see...
And I get this annoying beeping sound, and a string of weird characters like hearts, and music signs... and.. a whole list of odd characters. What could be the problem?
The problem is that you're printing out array elements which you never initialized. The answer you have currently accepted advises you to initialize all those elements. Although this will not cause errors it is a mistake to let this answer prevent you from fully understanding the problem you encountered.
Reconsider your code where you inserted a null character at the end of the array:
inFix[MAXSIZE] = '\0';
You apparently know that a null character has something to do with marking the end of the string, but you've mistaken how to do that correctly. Everything from the beginning of the array until a null character will be treated as part of your string. If you copy three characters from the input 9*4 into your array then you should only want those three characters to be seen as part of your string. What you do not want is for everything in the array past those three characters, up to MAXSIZE to also be treated as part of your string. So you need to put the end-of-string marker, the '\0', right after the characters you care about.
(BTW, inFix[MAXSIZE] = '\0'; not only puts the end-of-string marker at the end of the array, it puts it outside the array, which you are not allowed to do. The program will behave unpredictably.)
inFix[0] = '9';
inFix[1] = '*';
inFix[2] = '4';
inFix[3] = '\0'; // <-- this is where you need to put the end-of-string marker, because this is the end of the characters you care about.
Putting the end-of-string marker at the end of the array effectively does this:
inFix[0] = '9';
inFix[1] = '*';
inFix[2] = '4';
inFix[3] = ???
inFix[4] = ???
.
.
.
inFix[98] = ???
inFix[99] = ??? // annoying bell noise? musical note symbol?
inFix[100] = '\0'; // Yes, Please!
The reason initializing the array to all zeros (which can also be done like this char inFix[MAXSIZE] = {};, empty braces instead of a 0) worked for you is because that means that no matter where you stop writing characters you care about, the next character will be a '\0'. That position, right after the characters you care about, is the only place it matters.
Since the loop that's copying the characters knows exactly where it stops it also knows exactly where to insert the end-of-string marker. It would be easy to just insert a single '\0' in the correct place.
Try appending the '\0' to position a in the array instead - that is, exactly after the last char you've read. Otherwise you're putting your characters in the array, then a random sequence of what was in the array before, and then after that is the \0 (or, in this case, it's one after the end of the array, which is even worse).
You say you null terminate string using inFix[MAXSIZE] = '\0' as you currently understand this should converted into inFix[MAXSIZE - 1] = '\0' but in this case you say that you have an string that ended in last possible character!! so how you should add something( for example a parenthesis ) to it?? because it is full and it can't accept any more characters.
So may be this is better to use inFix[a] = '\0' because you know that a point to end of string, and now if a < MAXSIZE then you have enough space to add more items to the string otherwise you may increase MAXSIZE (using strcat for example) or show some error to the user.
You declared the array with:
char inFix[MAXSIZE];
At this point in time, the content of the array is not initialized. Printing out these uninitialized characters will lead to the behavior you see (strange characters, sounds, etc.).
However, the contents of that array are not initialized. If you want to initialize this array, change this line to:
char inFix[MAXSIZE] = {0};
This initialization syntax causes the elements in the array to be initialized. The first elements are initialized to the values provided in the brackets, while any values not in the brackets will be initialized to 0. In the case above, all of the array elements will be initialized to 0 (which happens to correspond to the NULL character \0).
Since all the characters are initialized to NULL in this case, you don't need to append the null character after your input.
That said, a better solution would be not to use a C-Style array at all, but a std::vector<char>, or simply a std::string, as pointed out above.
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'} = "||"