C Code: Pass string from one function to another function - c

Question

The main problem is: as soon as I send a string from one function to another, this second function doesn't really get the string as a parameter.
In detailled:
I have a function void myfunc() contains a word. This word should be send to another function, so it can count the length of it. That's what I've written so far:
void myfunc(int (*countlength)(char ch)){
char word[10] = "Hello\n";
int result = countlength(&word);
printf("Length of word: %d\n", result);
}
The word is being send to this function countlength(char* word):
int countlength(char* word) {
int length = strlen(word);
return length;
}
However the function countlength() can't count it's length and I don't know why...
The thing is, it works when the word is in the main function. Does anybody know why my Code doesn't work?

Answer 1

Two mistakes:
void myfunc(int (*countlength)(char ch)){
should be
void myfunc(int (*countlength)(char* ch)){
instead, as the function accepts char pointers.
Secondly,
int result = countlength(&word);
should be
int result = countlength(word);
as word is already a char*.

Answer 2

What you're passing to the function doesn't match what it's expecting.
&word has type char (*)[10], i.e. a pointer to an array of size 10. The function expects a char *, so just pass word. Arrays are converted to a pointer to their first element when passed to a function, so the types will match.

Answer 3

This parameter declaration of a function pointer
int (*countlength)(char ch)
does not correspond to the function declaration used as an argument for this parameter
int countlength(char* word)
So you need to declare the parameter like
int (*countlength)(char *ch)
In fact the identifier ch is redundant. You could just write
int (*countlength)(char *)
That is the declaration of the function myfunc will look like
void myfunc(int (*countlength)(char *));
You declared a character array within the function like
char word[10] = "Hello\n";
So the expression used as an argument in this call
countlength(&word)
has the type char ( * )[10] instead of the expected type char *.
There is no need to use the address of operator. The array designator used as an argument in this call
countlength( word )
is implicitly converted to a pointer to the first element of the array and has the type char *.
This function
int countlength(char* word) {
int length = strlen(word);
return length;
}
does not change its argument. So it should be declared at least like
int countlength( const char* word) {
int length = strlen(word);
return length;
}
The used standard C string function strlen has the return type size_t. In general an object of the type int can be not enough large to store possible lengths of strings.
So the function should be declared like
size_t countlength( const char* word) {
return strlen(word);
}
Thus returning to the function myfunc it should look like
void myfunc( size_t ( *countlength )( const char * ) )
{
char word[10] = "Hello\n";
size_t result = countlength( word );
printf( "Length of word: %zu\n", result );
}

Answer 4

Well; if you use the code like this, it's working just fine
When you declare an array, its name has a type of a pointer so here word has a type of char* and it's the pointer of the array's first element
#include <stdio.h>
#include <string.h>
int countlength(char* word) {
int length = strlen(word);
return length;
}
void myfunc(){
char word[10] = "Hello\n";
int result = countlength(word);
printf("Length of word: %d\n", result);
}
main(){
myfunc();
}
Length of word: 6