Initialize a string in C to empty string

Question

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

string[0] = \"\"; 

but it wrote

\"warning: assignment makes inte         


        

Answer 1

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.

Answer 2

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

Answer 3

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));

Answer 4

To achieve this you can use:

strcpy(string, "");

Answer 5

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.

Answer 6

Assuming your array called 'string' already exists, try

string[0] = '\0';

\0 is the explicit NUL terminator, required to mark the end of string.