why in the below code int *p = 22 will give compile time error and ptr will print the value successfully .
int main()
{
/*taking a character pointer and assigning a string to it*/
char *ptr = "Stackoverflow" ; //correct
/*taking a int pointer and assigning a string to it*/
int *p = 22 ; //incorrect
printf("%s",ptr); // correct and print
printf("%d",p); //incorrect and give compile time error.
return 0;
}
If you have a character array as for example
char s[] = "Stackoverflow";
then the array designator used in expressions it is converted to pointer to its first element. So you may write
char *ptr = s;
The pointer ptr now points to the first character of the array s.
String literals in C are also represented like character arrays and stored in memory like character arrays with the static storage duration.
So for example the string literal "Stackoverflow" has the type char[14] (including the terminating zero).
So then you write
char *ptr = "Stackoverflow";
then this statement in fact the same if there would be
static char unnamed[] = "Stackoverflow";
char *ptr = unnamed;
As for this statement
int *p = 22 ;
then the integer literal is not converted to a pointer to itself. It represents literally the number 22 and nothing more.
So the compiler issues a message because if you want that the pointer indeed contained the integer value 22 then you have to use a casting.
The expression
22 == 22
always yields true.
While this expression
"Stackoverflow" == "Stackoverflow"
is not necessary yields true because depending on compiler options the compiler can place coincidental string literals in different memory areas. And in this expression it is the pointers to the first characters of the string literals that are compared.
Take into account that if you are going to output an integer object pointed to by a pointer you need to use dereferencing. So in any case instead of
printf("%d",p);
you should write
printf("%d", *p);
Or if you want to output the value stored in a pointer you have to use another format specifier
printf("%p", p);
Because the string is actually an array of char's and when assigning an array to a variable, the arrays will behave as a pointer to the first element of the array. This can be confusing. More on this here:
when does a array act as a pointer in c?
The integer on the other hand is just an integer, not an array of integers.
This line assigns the memory address to be 22. This is incorrect. While this line will compile with only a warning, it will crash your program at runtime when you try to access memory address 22 later in the code.
int *p = 22 ; //incorrect
This will give you the result you're looking for:
int *p = NULL;
int val = 22;
p = &val;
Also
printf("%d",p); //incorrect and give compile time error.
printf("%d",*p); //you must derefence the pointer
%s takes a pointer to char, where it is a null terminated string, while %d takes an int, not an int*.
