Why can an int return type function return a string?

Question

This is the function I'm talking about:

static int getInput()
{
    printf("Enter your name: ");
    return fgets(input, sizeof(input), stdin) != NULL;
}

How does this work? Why can an int return type use fgets like this?

Answer 1

"Why can an int return type function return a string?"

In fact, It can´t.

Let´s take a closer look at this statement:

return fgets(input, sizeof(input), stdin) != NULL;

fgets() returns a pointer to char. This pointer is checked for a null pointer. The validation of this boolean expression either evaluates to 1 if the returned pointer is a non-null pointer (fgets() actually returned a pointer to input) or 0 if it is a null pointer - means an error occurred at consuming input by the use of fgets().

This int value (0 or 1) is then returned from the function getInput(); not a pointer or even a string by value itself (which is impossible regarding the C syntax).

---

Side notes:

1. input seems to be a global char array. Avoid global arrays. It can confuse you and readers. Pass a pointer to the buffer by reference and also pass the size of the buffer by value instead.

   static char* getInput (char* buf, size_t len)
   {
       if ( len > INT_MAX || len < 2 )
       {
           return NULL;
       }
   
       printf("Enter your name: ");
       return fgets(buf, len, stdin);
   }
   

   and call it like:

   char a[20];
   if ( getInput(a, sizeof(a)) == NULL )
   {
       fputs("Error occured at consuming input!", stderr);
   }
   

2. For the meaning of the static qualifer in this example, take a look at:

   What is a "static" function in C?