Pointers in C with chars, cant call the value - c

Question

This might just be the dumbest question, but i havent found any explanation when it comes to char, just when it comes to int, which i can manage.
However, i got this huge problem, i've pointed the adress of my pointer to a char and from that char i want to print out the value.
This is how my code looks:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void function(char *pn) {
char text[64];
char nei;
text[64] = &pn;
printf("%s\n", *text[64]);
}
int main()
{
function("Hello");
getchar();
return(0);
}
I can't figure out whats wrong with it to be honest. Why cant I put a * in front of text[64], since thats how you print the value, correct?

Answer 1

'*' has meaning when applied to a pointer. Applied to a char, I'm not sure what it would do. So, no, putting a * in front of text[64] is not how you print the value.
In your code text is a 64 byte long array of char. Valid indexes into it are 0 to 63.
The expression text[64] refers to memory you don't have access to, or may be a reference into one of your other automatic variables. text[64] is beyond the end of the array.
void function(char *pn) {
char text[64];
char nei;
text[64] = &pn; /* <- this won't work */
strcpy(text, pn); /* <- probably will work, may overflow text if string at pn is too long */
printf("%s\n", *text[64]); /* <- this won't work */
printf("%s\n", pn); /* <- probably will work */
printf("%s\n", text); /* after strcpy above, will likely work */
}

Answer 2

You cannot use both * and [ ], and also text[64] is out of bounds as its said in comment. You can use function strcpy to copy one string to another.

Answer 3

char text[64]; declares text as an array of 64 characters.
printf("%s\n", *text[64]) makes no sense, because:
text[64] points to the character after the end of the array; i.e. text[63] is the last character in the array.
*text[64] attempts to dereference the 64th item in the text array. But each item is not a pointer, but a character.
Perhaps you meant &text[n] where n is a number between 0 and 63, which would be a pointer to a character within the array, and hence a string. In C a string (what %s expects) is a pointer to a character, and the string is that character and all following up until a NUL.
This bit of your question is unclear:
Why cant I put a * in front of text[64],
(answered above)
since thats how you print the value, correct
I have no idea what that bit means. As per the above, %s expects a char *, i.e. a pointer to a character.

Answer 4

If you are trying to copy the contents of the pointer *pn to the character array text then you better use strcpy() function.
strcpy(text,pn);
will let you copy the contents.
You trying to print out the contents of the array. you can't use the * and [] together.
In case of a array
for example
char text[64]
as you have mentioned in your code
printf("%x",text);
will print the start of the allocated memory where the array lies.
when you do text[0] the start of the allocated memory address will be added with 0 and the contents of that memory will given back.
This is how an array work
Now if you are to use a dereferencing operator * then *text will give the 0th element of the array
*(text + 1)
will give you the second element of the array and so and so forth.
one more thing you can't use
printf("%s",text);
as the value that you are passing to the function is not a string but a character array as it does not have '\0' at its end.

Answer 5

Why cant I put a * in front of text[64]
You can! but not the way you have tried to do it.
For example, in a declaration, char *text[64]; would be used to create an array of strings, each 64 bytes in length. And would require allocation of memory using malloc()orcalloc()`, or an initializer such as:
char *text[64]={"this is a string1","this is string2"};
Which creates an array of 2 strings, each with 64 bytes (although only partially used), which could be used like this:
printf("%s\n", text[0]);
printf("%s\n", text[1]);
since that's how you print the value, correct? (referring to:printf("%s\n", *text[64]);)
No.
The argument char *pn refers to a single string.
If you want to copy this argument to the local variable:
First, change the line:
text[64] = &pn;
to
if(!pn) return; //test and leave if pn is null pointer
strcpy(text, pn);
Then just use:
printf("%s\n", text);
to print the line