When I construct a string like this:
char string[1] = {'a'};
printf("%s", string)
it returns a a4.
Why is there a four at the end? How can I get rid of it?
I choose this method because I need to make a string from character indexes, such as char array[4] = {string[i],string[j],string[k]};.
Your string should end with terminating char '\0'
You can do it by:
char string[2] = {'a','\0'};
Or:
char string[] = "a";
"strings" in C are essentially arrays of characters ending with the \0 character (null terminated).
So if you want an array of characters, what you did is fine, but it is not a "string". Dont try to print it as such.
If you would also like to print it or treat it as a "string", then increase it's length by 1, and add a '\0' char at the end.
The conversion specification %s is used to output strings that is a sequence of characters terminated by a zero character.
The array declared this way
char string[1] = {'a'};
does not contain a string.
So to output its elements you need to specify the exact number of characters you are going to output. For example
printf("%*.*s", 1, 1, string);
Otherwise reserve one more element in the array for the terminating zero and use the conversion specification %s. For example
char string[2] = {'a'};
printf( "%s", string );
Related
In an introductory course of C, I have learned that while storing the strings are stored with null character \0 at the end of it. But what if I wanted to print a string, say printf("hello") although I've found that that it doesn't end with \0 by following statement
printf("%d", printf("hello"));
Output: 5
but this seem to be inconsistent, as far I know that variable like strings get stored in main memory and I guess while printing something it might also be stored in main memory, then why the difference?
The null byte marks the end of a string. It isn't counted in the length of the string and isn't printed when a string is printed with printf. Basically, the null byte tells functions that do string manipulation when to stop.
Where you will see a difference is if you create a char array initialized with a string. Using the sizeof operator will reflect the size of the array including the null byte. For example:
char str[] = "hello";
printf("len=%zu\n", strlen(str)); // prints 5
printf("size=%zu\n", sizeof(str)); // prints 6
printf returns the number of the characters printed. '\0' is not printed - it just signals that the are no more chars in this string. It is not counted towards the string length as well
int main()
{
char string[] = "hello";
printf("szieof(string) = %zu, strlen(string) = %zu\n", sizeof(string), strlen(string));
}
https://godbolt.org/z/wYn33e
sizeof(string) = 6, strlen(string) = 5
Your assumption is wrong. Your string indeed ends with a \0.
It contains of 5 characters h, e, l, l, o and the 0 character.
What the "inner" print() call outputs is the number of characters that were printed, and that's 5.
In C all literal strings are really arrays of characters, which include the null-terminator.
However, the null terminator is not counted in the length of a string (literal or not), and it's not printed. Printing stops when the null terminator is found.
All answers are really good but I would like to add another example to complete all these
#include <stdio.h>
int main()
{
char a_char_array[12] = "Hello world";
printf("%s", a_char_array);
printf("\n");
a_char_array[4] = 0; //0 is ASCII for null terminator
printf("%s", a_char_array);
printf("\n");
return 0;
}
For those don't want to try this on online gdb, the output is:
Hello world
Hell
https://linux.die.net/man/3/printf
Is this helpful to understand what escape terminator does? It's not a boundary for a char array or a string. It's the character that will say to the guy that parses -STOP, (print) parse until here.
PS: And if you parse and print it as a char array
for(i=0; i<12; i++)
{
printf("%c", a_char_array[i]);
}
printf("\n");
you get:
Hell world
where, the whitespace after double l, is the null terminator, however, parsing a char array, will just the char value of every byte. If you do another parse and print the int value of each byte ("%d%,char_array[i]), you'll see that (you get the ASCII code- int representation) the whitespace has a value of 0.
In C function printf() returns the number of character printed, \0 is a null terminator which is used to indicate the end of string in c language and there is no built in string type as of c++, however your array size needs to be a least greater than the number of char you want to store.
Here is the ref: cpp ref printf()
But what if I wanted to print a string, say printf("hello") although
I've found that that it doesn't end with \0 by following statement
printf("%d", printf("hello"));
Output: 5
You are wrong. This statement does not confirm that the string literal "hello" does not end with the terminating zero character '\0'. This statement confirmed that the function printf outputs elements of a string until the terminating zero character is encountered.
When you are using a string literal as in the statement above then the compiler
creates a character array with the static storage duration that contains elements of the string literal.
So in fact this expression
printf("hello")
is processed by the compiler something like the following
static char string_literal_hello[] = { 'h', 'e', 'l', 'l', 'o', '\0' };
printf( string_literal_hello );
Th action of the function printf in this you can imagine the following way
int printf( const char *string_literal )
{
int result = 0;
for ( ; *string_literal != '\0'; ++string_literal )
{
putchar( *string_literal );
++result;
}
return result;
}
To get the number of characters stored in the string literal "hello" you can run the following program
#include <stdio.h>
int main(void)
{
char literal[] = "hello";
printf( "The size of the literal \"%s\" is %zu\n", literal, sizeof( literal ) );
return 0;
}
The program output is
The size of the literal "hello" is 6
You have to clear your concept first..
As it will be cleared when you deal with array, The print command you are using its just counting the characters that are placed within paranthesis. Its necessary in array string that it will end with \0
A string is a vector of characters. Contains the sequence of characters that form the
string, followed by the special ending character
string: '\ 0'
Example:
char str[10] = {'H', 'e', 'l', 'l', 'o', '\0'};
Example: the following character vector is not one string because it doesn't end with '\ 0'
char str[2] = {'h', 'e'};
So I just read an example of how to create an array of characters which represent a string.
The null-character \0 is put at the end of the array to mark the end of the array. Is this necessary?
If I created a char array:
char line[100];
and put the word:
"hello\n"
in it, the chars would be placed at the first six indexes line[0] - line[6], so the rest of the array would be filled with null characters anyway?
This books says, that it is a convention that, for example the string constant "hello\n" is put in a character array and terminated with \0.
Maybe I don't understand this topic to its full extent and would be glad for enlightenment.
The \0 character does not mark the "end of the array". The \0 character marks the end of the string stored in a char array, if (and only if) that char array is intended to store a string.
A char array is just a char array. It stores independent integer values (char is just a small integer type). A char array does not have to end in \0. \0 has no special meaning in a char array. It is just a zero value.
But sometimes char arrays are used to store strings. A string is a sequence of characters terminated by \0. So, if you want to use your char array as a string you have to terminate your string with a \0.
So, the answer to the question about \0 being "necessary" depends on what you are storing in your char array. If you are storing a string, then you will have to terminate it with a \0. If you are storing something that is not a string, then \0 has no special meaning at all.
'\0' is not required if you are using it as character array. But if you use character array as string, you need to put '\0'. There is no separate string type in C.
There are multiple ways to declare character array.
Ex:
char str1[] = "my string";
char str2[64] = "my string";
char str3[] = {'m', 'y', ' ', 's', 't', 'r', 'i', 'n', 'g', '\0'};
char str4[64] = {'m', 'y', ' ', 's', 't', 'r', 'i', 'n', 'g' };
All these arrays have the same string "my string". In str1, str2, and str4, the '\0' character is added automatically, but in str3, you need to explicitly add the '\0' character.
(When the size of an array is explicitly declared, and there are fewer items in the initializer list than the size of the array, the rest of the array is initialized with however many zeros it takes to fill it -- see C char array initialization and The N_ELEMENTS macro .).
When/Why is '\0' necessary to mark end of an (char) array?
The terminating zero is necessary if a character array contains a string. This allows to find the point where a string ends.
As for your example that as I think looks the following way
char line[100] = "hello\n";
then for starters the string literal has 7 characters. It is a string and includes the terminating zero. This string literal has type char[7]. You can imagine it like
char no_name[] = { 'h', 'e', 'l', 'l', 'o', '\n', '\0' };
When a string literal is used to initialize a character array then all its characters are used as initializers. So relative to the example the seven characters of the string literal are used to initialize first 7 elements of the array. All other elements of the array that were not initialized by the characters of the string literal will be initialized implicitly by zeroes.
If you want to determine how long is the string stored in a character array you can use the standard C function strlen declared in the header <string.h>. It returns the number of characters in an array before the terminating zero.
Consider the following example
#include <stdio.h>
#include <string.h>
int main(void)
{
char line[100] = "hello\n";
printf( "The size of the array is %zu"
"\nand the length of the stored string \n%s is %zu\n",
sizeof( line ), line, strlen( line ) );
return 0;
}
Its output is
The size of the array is 100
and the length of the stored string
hello
is 6
In C you may use a string literal to initialize a character array excluding the terminating zero of the string literal. For example
char line[6] = "hello\n";
In this case you may not say that the array contains a string because the sequence of symbols stored in the array does not have the terminating zero.
You need the null character to mark the end of the string. C does not store any internal information about the length of the character array or the length of a string, and so the null character/byte \0 marks where it ends.
This is only required for strings, however – you can have any ordinary array of characters that does not represent a string.
For example, try this piece of code:
#include <stdio.h>
int main(void) {
char string[1];
string[0] = 'a';
printf("%s", string);
}
Note that the character array is completely filled with data. Thus, there is no null byte to mark the end. Now, printf will keep printing until it hits a null byte – this will be somewhere past the end of the array, so you will print out a lot of junk in addition to just "a".
Now, try this:
#include <stdio.h>
int main(void) {
char string[2];
string[0] = 'a';
string[1] = '\0';
printf("%s", string);
}
It will only print "a", because the end of the string is explicitly marked.
The length of a C string (an array containing the characters and terminated with a '\0' character) is found by searching for the (first) NUL byte. \0 is zero character. In C it is mostly used to indicate the termination of a character string.
I make an example to you:
let's say you've written a word into a file:
word = malloc(sizeof(cahr) * 6);
word = "Hello";
fwrite(word, sizeof(char), 6, fp);
where in word we allocate space for the 5 character of "Hello" plus one more for its terminating '\0'. The fp is the file.
An now, we write another word after the last one:
word2 = malloc(sizeof(cahr) * 7);
word2 = "world!";
fwrite(word2, sizeof(char), 7, fp);
So now, let's read the two words:
char buff = malloc(sizeof(char)*1000); // See that we can store as much space as we want, it won't change the final result
/* 13 = (5 chacater from 'Hello')+(1 character of the \0)+(6 characters from 'world!')+(1 character from the \0) */
fread(buff, sizeof(char), 13, fp); // We read the words 'Hello\0' and 'world!\0'
printf("the content of buff is: %s", buff); // This would print 'Hello world!'
This last is due to the ending \0 character, so C knows there are two separated strings into buffer. If we had not put that \0 character at the end of both words, and repeat the same example, the output would be "Helloworld!"
This can be used for many string methods and functions!.
I want to assign the first two values from the hash array to the salt array.
char hash[] = {"HAodcdZseTJTc"};
char salt[] = {hash[0], hash[1]};
printf("%s", salt);
However, when I attempt this, the first two values are assigned and then all thirteen values are also assigned to the salt array. So my output here is not:
HA
but instead:
HAHAodcdZseTJTC
salt is not null-terminated. Try:
char salt[] = {hash[0], hash[1], '\0'};
Since you are adding just two characters to the salt array and you are not adding the '\0' terminator.
Passing a non nul terminated array as a parameter to printf() with a "%s" specifier, causes undefined behavior, in your case it prints hash in my case
HA#
was printed.
Strings in c use a special convetion to know where they end, a non printable special character '\0' is appended at the end of a sequence of non-'\0' bytes, and that's how a c string is built.
For example, if you were to compute the length of a string you would do something like
size_t stringlength(const char *string)
{
size_t length;
for (length = 0 ; string[length] != '\0' ; ++length);
return length;
}
there are of course better ways of doing it, but I just want to illustrate what the significance of the terminating '\0' is.
Now that you know this, you should notice that
char string[] = {'A', 'B', 'C'};
is an array of char but it's not a string, for it to be a string, it needs a terminating '\0', so
char string[] = {'A', 'B', 'C', '\0'};
would actually be a string.
Notice that then, when you allocate space to store n characters, you need to allocate n + 1 bytes, to make room for the '\0'.
In the case of printf() it will try to consume all the bytes that the passed pointer points at, until one of them is '\0', there it would stop iterating through the bytes.
That also explains the Undefined Behavior thing, because clearly printf() would be reading out of bounds, and anything could happen, it depends on what is actually there at the memory address that does not belong the the passed data but is off bounds.
There are many functions in the standard library that expect strings, i.e. _sequences of non nul bytes, followed by a nul byte.
Consider this code:
char name[]="123";
char name1[]="1234";
And this result
The size of name (char[]):4
The size of name1 (char[]):5
Why the size of char[] is always plus one?
Note the difference between sizeof and strlen. The first is an operator that gives the size of the whole data item. The second is a function that returns the length of the string, which will be less than its sizeof (unless you've managed to get string overflow), depending how much of its allocated space is actually used.
In your example
char name[]="123";
sizeof(name) is 4, because of the terminating '\0', and strlen(name) is 3.
But in this example:
char str[20] = "abc";
sizeof(str) is 20, and strlen(str) is 3.
As Michael pointed out in the comments the strings are terminated by a zero. So in memory the first string will look like this
"123\0"
where \0 is a single char and has the ASCII value 0. Then the above string has size 4.
If you had not this terminating character, how would one know, where the string (or char[] for that matter) ends? Well, indeed one other way is to store the length somewhere. Some languages do that. C doesn't.
In C, strings are stored as arrays of chars. With a recognised terminating character ('\0' or just 0) you can pass a pointer to the string, with no need for any further meta-data. When processing a string, you read chars from the memory pointed at by the pointer until you hit the terminating value.
As your array initialisation is using a string literal:
char name[]="123";
is equivalent to:
char name[]={'1','2','3',0};
If you want your array to be of size 3 (without the terminating character as you are not storing a string, you will want to use:
char name[]={'1','2','3'};
or
char name[3]="123";
(thanks alk)
which will do as you were expecting.
Because there is a null character that is attached to the end of string in C.
Like here in your case
name[0] = '1'
name[1] = '2'
name[2] = '3'
name[3] = '\0'
name1[0] = '1'
name1[1] = '2'
name1[2] = '3'
name1[3] = '4'
name1[4] = '\0'
A String in C (and in, probably, every programming language - behind the scenes) is an array of characters which is terminated by \0 with the ASCII value of 0.
When assigning: char arr[] = "1234";, you assign a string literal, which is, by default, null-terminated (\0 is also called null) as you can see here.
To avoid a null (assuming you want just an array of chars and not a string), you can declare it the following way char arr[] = {'1', '2', '3', '4'}; and the program will behave as you wish (sizeof(arr) would be 4).
name = {'1','2','3','\0'};
name1 = {'1','2','3','4','\0'};
So
sizeof(name) = 4;
sizeof(name1) = 5;
sizeof returns the size of the object and in this case the object is an array and it is defined that your array is 4 bytes long in first case and 5 bytes in second case.
In C, string literals have a null terminating character added to them.
Your strings,
char name[]="123";
char name1[]="1234";
look more like:
char name[]="123\0";
char name1[]="1234\0";
Hence, the size is always plus one. Keep in mind when reading strings from files or from whatever source, the variable where you store your string, should always have extra space for the null terminating character.
For example if you are expected to read string, whose maximum size is 100, your buffer variable, should have size of 101.
Every string is terminated with the char nullbyte '\0' which add 1 to your length.
I need to convert a string to a char array in C; how can I do this?
Or at least, how can I extract single chars from a string incrementally?
In C, a string is actually stored as an array of characters, so the 'string pointer' is pointing to the first character. For instance,
char myString[] = "This is some text";
You can access any character as a simple char by using myString as an array, thus:
char myChar = myString[6];
printf("%c\n", myChar); // Prints s
Hope this helps!
David
In C, there's no (real, distinct type of) strings. Every C "string" is an array of chars, zero terminated.
Therefore, to extract a character c at index i from string your_string, just use
char c = your_string[i];
Index is base 0 (first character is your_string[0], second is your_string[1]...).
In this simple way
char str [10] = "IAmCute";
printf ("%c",str[4]);