How can I convert char[] array to char* [closed] - c

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I want to convert my char[] array to char* string in C programing language. Here is what I did. But I can't reach the solution . Please help me. Here is my code:
#include <stdio.h>
void fnk(char *chr){
printf("%s",chr);
}
int main(){
char charfoo[100];
int i=50;
gets(charfoo);
for(int j=0;j<i;j++){
fnk((char*) charfoo[j]);
}
}

Answer 1

I think you mean the following
fnk( &charfoo[i]);
Take into account that it would be better to write the loop the following way
for( int j=0; charfoo[j]; j++ ) {
fnk( &charfoo[j]);
}
Also function gets is unsafe and is not supported any more by the C Standard. Instead use function fgets For example
fgets( charfoo, sizeof( charfoo ), stdin );
In this case the loop can look like
for( int j=0; charfoo[j] != '\0' && charfoo[j] != '\n'; j++ ) {
fnk( &charfoo[j]);
}
If you want to output just one character in the function then the function should be defined like
void fnk(char chr){
printf("%c",chr);
}
and called like
fnk( charfoo[j]);

Answer 2

If we switch from gets() to fgets() as Vlad from Moscow suggests, then the return value from fgets() is of the same type as the argument to fnk() so we can just pass the result along:
#include <stdio.h>
void fnk(char *string) {
printf("%s", string);
}
int main() {
char charfoo[100];
fnk(fgets(charfoo, sizeof(charfoo), stdin));
return 0;
}
The next step in sophistication would be to save the result of fgets() to a char * pointer and test if it's NULL or not before passing it along (or not) to fnk().
If you want an actual char * copy of the array charfoo[] to exist in memory and not simply pass off char[] arrays as char * pointers, you could do the following which throws in lots of error checking too:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void fnk(char *string) {
printf("%s", string);
}
int main() {
char charfoo[100];
char *input = fgets(charfoo, sizeof(charfoo), stdin);
if (input != NULL) {
char *string = strdup(input);
if (string != NULL) {
fnk(string);
free(string);
}
}
return 0;
}
But in all cases, it seems like you're making things more difficult than they need to be.

Answer 3

Calling fnk with i would just be the same call 50 time -- assuming you want to use j instead the code would look like this
#include <stdio.h>
void fnk(char *chr){
printf("%s",chr);
}
int main(){
char charfoo[100];
int i=50;
gets(charfoo);
for(int j=0;j<i;j++){
fnk(charfoo + j);
}
}
be aware that gets could cause buffer overflow if the input is longer than the charfoo (i.e. 99 bytes + newline + null) -- but that is a different story