Why do I get a from the console with the following code?
char array1[] = "Hello World";
char ch = array1;
printf(" %s" , ch);
(We are instructed not to do this with a pointer)
1 Answer
Because you need to make ch a pointer to a character. The compiler should have given you a warning that you were assigning a character pointer to a character variable, and it should have warned you that you were trying to printf("%s") on a non-character-pointer variable -- if it didn't, you should turn up your compiler's warning level!
Just adding that one little, easily-forgotten star makes it work correctly:
#include <stdio.h>
int main()
{
char array1[] = "Hello World\n";
char *ch = array1; printf(" %s" , ch);
return 0;
}
I also took the liberty of adding a newline on the end of your string and putting it in a proper main function, for the sake of completeness.
I just noticed that you said you "aren't allowed to do this with pointers" -- pretty much the only other way to do it is to print out character-by-character until you hit a null terminator:
#include <stdio.h>
int main()
{
char array1[] = "Hello World\n";
int i;
for (i = 0 ; array1[i] != '\0' ; i++)
{
printf("%c", array1[i]);
}
return 0;
}
output:
$ ./a.out Hello World
A more robust approach would be to also put a limit on how many characters you will print out, but just checking for \0 is probably fine enough for what seems to be a homework assignment.
3 Comments
user133466
Mark, I've entered the above code, but Im trying to check the 4th element in this string with printf("%s", s[3]); but the project freezes on me =(
Robert Groves
@metashockwave s[3] is a character refrence so you need to change the %s to %c to print just that character.
Aubin
why not printf( "%s", array1 );?
*from pointer notation?