[c] Initialize a string in C to empty string

I want to initialize string in C to empty string. I tried:

string[0] = ""; 

but it wrote

"warning: assignment makes integer from pointer without a cast"

How should I do it then?

calloc allocates the requested memory and returns a pointer to it. It also sets allocated memory to zero.

In case you are planning to use your string as empty string all the time:

char *string = NULL;
string = (char*)calloc(1, sizeof(char));

In case you are planning to store some value in your string later:

char *string = NULL;
int numberOfChars = 50; // you can use as many as you need
string = (char*)calloc(numberOfChars + 1, sizeof(char));

It's a bit late but I think your issue may be that you've created a zero-length array, rather than an array of length 1.

A string is a series of characters followed by a string terminator ('\0'). An empty string ("") consists of no characters followed by a single string terminator character - i.e. one character in total.

So I would try the following:

string[1] = ""

Note that this behaviour is not the emulated by strlen, which does not count the terminator as part of the string length.

string[0] = "";

"warning: assignment makes integer from pointer without a cast

Ok, let's dive into the expression ...

0 an int: represents the number of chars (assuming string is (or decayed into) a char*) to advance from the beginning of the object string

string[0]: the char object located at the beginning of the object string

"": string literal: an object of type char[1]

=: assignment operator: tries to assign a value of type char[1] to an object of type char. char[1] (decayed to char*) and char are not assignment compatible, but the compiler trusts you (the programmer) and goes ahead with the assignment anyway by casting the type char* (what char[1] decayed to) to an int --- and you get the warning as a bonus. You have a really nice compiler :-)

To achieve this you can use:

strcpy(string, "");

Assigning string literals to char array is allowed only during declaration:

char string[] = "";

This declares string as a char array of size 1 and initializes it with \0.

Try this too:

char str1[] = ""; 
char str2[5] = ""; 
printf("%d, %d\n", sizeof(str1), sizeof(str2)); //prints 1, 5

In addition to Will Dean's version, the following are common for whole buffer initialization:

char s[10] = {'\0'};

or

char s[10];
memset(s, '\0', sizeof(s));

or

char s[10];
strncpy(s, "", sizeof(s));

You want to set the first character of the string to zero, like this:

char myString[10];
myString[0] = '\0';

(Or myString[0] = 0;)

Or, actually, on initialisation, you can do:

char myString[10] = "";

But that's not a general way to set a string to zero length once it's been defined.

I think Amarghosh answered correctly. If you want to Initialize an empty string(without knowing the size) the best way is:

//this will create an empty string without no memory allocation. 
char str[]="";// it is look like {0}

But if you want initialize a string with a fixed memory allocation you can do:

// this is better if you know your string size.    
char str[5]=""; // it is look like {0, 0, 0, 0, 0}