We have a statement like-
printf("Hello World");
Does printf append a null character after 'd' in the given string?
When you write
printf("xyz");
you are actually passing a string consisting of three characters and a null-terminator to printf.
printf("xyz");
// is equivalent to
static const char xyz[] = { 'x', 'y', 'z', 0 };
printf(xyz);
both printfs have the same effect: they write the characters x, y and z to the console. The null-terminator is not written.
Try this:
#include <stdio.h>
int main()
{
char string0[] = {'a', 'b', 'c'};
char string1[] = {'a', 'b', 'c','\0'};
printf("%s", string0);
puts("");
printf("%s", string1);
return 0;
}
If you are lucky enough, you will see something like:
abc$#r4%&^3
abc
printf() does not append a '\0' to the string. It doesn't make any change to the string to be output, because its task is to "print", rather than "modify". Instead, it is the null characters that tell printf() where is the end of the strings.
When a string is defined e.g. "hello world" it is null terminated by default. printf does nothing related to null terminate expect its own print processing. It only accepts char* as input. In your example "Hello World" is temp string and null terminated already before passed to printf. If the passed string is not null terminated then the behavior is undefined.
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'};
I want to print char array in C like Arrays.ToString(array); of Java does. It prints what I want but puts some characters at the end. I guess it's because of the special character \0.
I declared a char array char letters[] = {'g','y','u','c','n','e'};
And tried to print: printf("\n [%s]:", letters);
The output is: [gyucneÇ_=]
Here is the Java code:
char[] letters= {'g','y','u','c','n','e'};
System.out.print( Arrays.toString(letters) );
The output is:
[g, y, u, c, n, e]
I wanted to have the output of Java code. I wonder if I want it to contain commas too, do I have to print the characters one by one or can I print it at once ?
And of course my priority is to remove the special character that is printed at the end of C code.
Print each letter on its own. You do not have a string. You cannot call most functions from <string.h> or printf() or a bunch of others that expect a string.
char letters[] = {'g', 'y', 'u', 'c', 'n', 'e'}; // ATTENTION: letters is not a string!
for (int i = 0; i < sizeof letters; i++) {
putchar(letters[i]);
}
putchar('\n'); // end with a newline
I declared a char array: char letters[] = {'g','y','u','c','n','e'};
But that is not a C string (since it is not NUL terminated ! ). You should have coded instead:
const char letters[] = {'g','y','u','c','n','e',(char)0};
(or use '\0' instead of (char)0 ....) or better yet:
const char letters[] = "gyucne";
and both are exactly equivalent.
Then you can code something like printf("letters are %s\n", letters); since your letters are now a C string.
NB. Please read also http://utf8everywhere.org/ & How to debug small programs - both are practically very relevant for your case. See also at least some C reference site.
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 am trying to understand outputs of printing character arrays and it is giving me variable outputs on ideone.com(C++ 4.3.2) and on my machine (Dev c++, MinGW compiler)
1)
#include<stdio.h>
main()
{
char a[] = {'s','t','a','c','k','o'};
printf("%s ",a);
}
It prints "stacko" on my machine BUT doesn't print anything on ideone
2)
#include<stdio.h>
main()
{
char a[] = {'s','t','a','c','k','o','v','e'};
printf("%s ",a);
}
on ideone : it prints "stackove" only the first time then prints nothing the subsequent times when i run this program
on my dev-c : it prints "stackove.;||w"
what should be the IDEAL OUTPUT when I try to print this kind of character array without any '\0' at the end , it seems to give variable outputs everywhere . please help !
%s conversion specifier expects a string. A string is a character array containing a terminating null character '\0' which marks the end of the string. Therefore, your program as such invokes undefined behaviour because printf overruns the array accessing memory out of the bound of the array looking for the terminating null byte which is not there.
What you need is
#include <stdio.h>
int main(void)
{
char a[] = {'s', 't', 'a', 'c', 'k', 'o', 'v', 'e', '\0'};
// ^
// include the terminating null character to make the array a string
// output a newline to immediately print on the screen as
// stdout is line-buffered by default
printf("%s\n", a);
return 0;
}
You can also initialize your array with a string literal as
#include <stdio.h>
int main(void)
{
char a[] = "stackove"; // initialize with the string literal
// equivalent to
// char a[] = {'s', 't', 'a', 'c', 'k', 'o', 'v', 'e', '\0'};
// output a newline to immediately print on the screen as
// stdout is line-buffered by default
printf("%s\n", a);
return 0;
}
I'm trying to print out an element in my array:
#include <stdio.h>
int main(void) {
char greeting[6] = {'H', 'e', 'l', 'l', 'o', '\0'};
printf("%s", greeting[0]);
return 0;
}
I expect it to print out H, but instead it crashed and the Windows dialog popped up:
"program.exe has stopped working"
What did I do wrong?
Try printf("%s", &greeting[1]); it should print "ello":
1) "greeting" is a character array containing the string "Hello\0".
2) You can call printf("%s\n", greeting); with no problem.
3) "greeting[0]" is the first character in the array.
"&greeting[0]" is a pointer to the first character in the array.
printf("%s", s) expects s to be a pointer, not a character.
4) Alternatively, you might want to print just a character.
In this case, try printf("%c", greeting[0]);
5) Use "%s" to print a string, use "c%" to print a character. Use "%d" or "0x%02x"to print the ASCII representation of the character.
'Hope that helps
You should write
printf("%c", greeting[1]);
to write out a single character ('H') instead of trying to print a string. Your program crashes because %s expects a char* parameter to be passed, but greeting[1] is of type char.
You should do:
printf("%c \n", greeting[1]);
The format specifier to print a char is c. So the format string to be used is %c.