This program works in C:
#include <stdio.h>
int main(void) {
char a[10] = "Hello";
char *b = a;
printf("%s",b);
}
There are two things I would expect to be different. One is that we in the second line in the main write: "char *b = &a", then the program is like this:
#include <stdio.h>
int main(void) {
char a[10] = "Hello";
char *b = &a;
printf("%s",b);
}
But this does not work. Why is that? Isn’t this the correct way to initialize a pointer with an adress?
The second problem I have is in the last line we should have: printf("%s",*b) so the program is like this:
#include <stdio.h>
int main(void) {
char a[10] = "Hello";
char *b = a;
printf("%s",*b);
}
But this gives a segmentation fault. Why does this not work? Aren’t we supposed to write "*" in front of a pointer to get its value?
>Solution :
There is a special rule in C. When you write
char *b = a;
you get the same effect as if you had written
char *b = &a[0];
That is, you automatically get a pointer to the array’s first element. This happens any time you try to take the "value" of an array.
Aren’t we supposed to write "*" in front of a pointer to get its value?
Yes, and if you wanted to get the single character pointed to by b, you would therefore need the *. This code
printf("first char: %c\n", *b);
would print the first character of the string. But when you write
printf("whole string: %s\n", b);
you get the whole string. %s prints multiple characters, and it expects a pointer. Down inside printf, when you use %s, it loops over and prints all the characters in the string.