empty char array, the ways I check it are the same?-CodePudding

I have a char array and I want to check if it is null.

if(_char_array[0] != '\0') {...}

is the same as

if(_char_array != '\0') {...}

thanks

CodePudding user response：

This if statement

if(_char_array[0] != '\0') {...}

checks whether a character array contains a non-empty string.

In fact there are compared two integers: the first character of the array _char_array promoted to the type int and the integer character constant '\0' that has the type int.

This if statement

if(_char_array != '\0') {...}

does not make a sense because any array occupies a memory extent. So converted to a pointer to its first element it can not be a null pointer.

That is in this statement there are compared a pointer to the first element of the array due to the implicit conversion of the array to a pointer to its first element and a null pointer constant.

If _char_array is initially declared as a pointer then the above if statement checks whether the pointer is not a null pointer.

CodePudding user response：

An array name is not a pointer, but an identifier for a variable of type array, which is implicitly converted to a pointer to the element type. These two are not the same at all, the array will always have value.

CodePudding user response：

In the first if statement you were checking whether the first element of _char_array is 0 or not, whereas in the second if statement you were checking address of _char_array to be 0 or not.

If the address is 0 then there would be 2 cases:

CASE 1

char *_char_array = NULL;

CASE 2

char *_char_array = malloc(<any_size>); // but malloc failed and returned `NULL`

NOTE: Statically allocated arrays never have an address of 0.