What does while(*pointer) means in C?

Question

When I recently look at some passage about C pointers, I found something interesting. What it said is, a code like this:

char var[10];
char *pointer = &v         


        

Answer 1

Yes, you can go for it.

Please note that *pointer is the value at the memory location the pointer point to(or hold the address of).

Your *pointer is now pointing to the individual characters of the character array var.

So, while(*pointer) is shorthand usage of the equivalent

while(*pointer!='\0').

Suppose, your string is initialized to 9 characters say "123456789" and situated at an address say addr(memory location).

Now because of the statement:

char *pointer=&var;

pointer will point to first element of string "1234567890".

When you write the *pointer it will retrieve the value stored at the memory location addr which is 1.

Now, the statement:

while(*pointer)

will be equivalent to

while(49)

because ASCII Value of 1 is 49, and condition is evaluated to true.

This will continue till \0 character is reached after incrementing pointer for nine times.

Now, the statement:

while(*pointer)

will be equivalent to

while(0)

because ASCII value of \0 is 0. Thus, condition is evaluated to false and loop stops.

Summary:

• In while(condition), condition must be non-zero to continue loop execution. If condition evaluates to zero then loop stops executing.

• while(*pointer) will work till the value at memory location being pointed to is a non-zero ASCII value.

• Also you can use:

  if(*ptr){     //instead of if(*ptr!='\0')
    //do somthing
  }
  
  if(!*ptr){     //instead of if(*ptr=='\0')
    //do somthing
  }