Arguments passed to puts function in C - c

Question

I have only recently started learning C. I was going through the concept of arrays and pointers, when I came across a stumbling block in my understanding of it.
Consider this code -
#include<stdio.h>
int main()
{
char string[]="Hello";
char *ptr;
ptr=string;
puts(*ptr);
return(0);
}
It compiles, but runs into segmentation fault on execution.
The warning that I get is:
type error in argument 1 to `puts'; found 'char' expected 'pointer to char'
Now *ptr does return a character "H" and my initial impression was that it would just accept a char as an input.
Later, I came to understand that puts() expects a pointer to a character array as it's input, but my question is when I pass something like this - puts("H"), isn't that the same thing as puts(*ptr), given that *ptr does contain the character "H".

Answer 1

"H" is a string literal that consists of 2 bytes 'H' and '\0'. Whenever you have "H" in your code, a pointer to the memory region with 2 bytes is meant. *ptr simply returns a single char variable.

Answer 2

By doing puts(*str), you're dereferencing the str variable. This would then try and use the 'H' character as a memory address (since that's what str) points to, then segfault since it will be an invalid pointer (since it will probably fall outside your process' memory). This is because the puts function accepts a pointer as an argument.
What you really want is puts(str).
As an aside, the latter example puts("h") populates the string table with "h" at compile time and replaces the definition there with an implicit pointer.

Answer 3

The puts() function takes a pointer to a string and what you are doing is specifying a single character.
Take a look at this Lesson 9: C Strings.
So rather than doing
#include<stdio.h>
int main()
{
char string[]="Hello";
char *ptr;
ptr=string; // store address of first character of the char array into char pointer variable ptr
// ptr=string is same as ptr=&string[0] since string is an array and an
// array variable name is treated like a constant pointer to the first
// element of the array in C.
puts(*ptr); // get character pointed to by pointer ptr and pass to function puts
// *ptr is the same as ptr[0] or string[0] since ptr = &string[0].
return(0);
}
You should instead be doing
#include<stdio.h>
int main()
{
char string[]="Hello";
char *ptr;
ptr=string; // store address of first character of the char array into char pointer variable ptr
puts(ptr); // pass pointer to the string rather than first character of string.
return(0);
}

Answer 4

When ever you enter string in gets or want to display it using puts you had to actually pass the location of the pointer or the string
for example
char name[] = "Something";
if you want to print that
you have to write printf("%s",name); --> name actually stores the address of the string "something"
and by using puts if you want to display
puts(name) ----> same as here address is put in the arguments

Answer 5

No.
'H' is the character literal.
"H" is, in effect, a character array with two elements, those being 'H' and the terminating '\0' null byte.

Answer 6

puts is waiting as input a string pointer so it's waiting a memory address. but in your example you provided the content of the memory which is *ptr. the *ptr is the content of the memory with address ptr which is h
ptr is memory address
*ptr is the content of this memory
the input parameter of puts is an address type but you have provided a char type (content of the address)
the puts start the printing character by character starting by the address you give it as input until the memory which contain 0 and then it stop printing