Solution for Exercise 02_05-modify.c

02_05-modify.c

#include <stdio.h>

int main()
{
	char string[] = "become";
	char *s;

	/* initialize the pointer! */
	s = string;   /* no ampersand needed */

	printf("%s to ",string);
	s += 3;       /* 4th letter */
	*s = 'a';
	printf("%s\n",string);

	return 0;
}

Output

become to became

Notes

* Array string[] contains the string "become".

* Pointer s is the same data type as string[], char.

* When pointer s is assigned to string[], no & operator is required; this is due to the special relationship between arrays and strings.

* The fourth letter of string[] is at offset 3. To modify this position, the value of pointer s is increased by three: s += 3; This expression adds three to the address stored in s, which references the location of the 4th letter, 'o'.

* Then the dereferencing operator (*) is used to alter the character at address s: *s = 'a'; Letter 'a' is assigned to the memory location held in pointer variable s. At this point, variable s no longer references the start of the string.

* The result is then output.

Exercise 03_02-scanf2.c →