Thread: What does & mean?

Hi,

May I ask a newbie question again? Why does the int a need to have a prefix of '&'? In java, I use readline() to read the inputstream only.

Thanks for help.

Code:

#include <stdio.h>


void main() {
	int a, b;
    printf("Enter the number to be reversed.\n");
	scanf("%d", &a);
	do {
		b = a % 10;
		printf("%d", b);
		a = a / 10;
	}while(a != 0);
	printf("\n");
}

The & is the "address-of" operator. Whenever you call a function if you just pass the value of a ( no & ) then if the function modifys a it will not be changed when you return from the function. If you pass a pointer to a you can safely change the value and it will stay changed when you return.

Although I do not catch what did you mean, may I ask the following question to clarify the pointer issue?

Code:

void main() {
	char *pw="west", *pe="east", *ps="South", *pn="North";
	enum location {east=1, west=2, South=3, North=4};
	enum location direction;
	direction = west;

	if(direction == west)
		printf("Cannot go %s\n", pw);
}

Why the pw needs to add the prefix of '*'? Is it because pw points to a string object?

If I printf("Cannot go %s\n", &pw), it returns 'D'. Why does it return 'D'?

Thanks a lot

The pw needs a * in front of it to signify its a pointer. The type is only char, you cannot fit "west" into a single character so you need a series ( or array ) or characters. The char * will sometimes put it in read-only memory so you may want to use an array if you plan on changing it later.

Code:

char buffer[] = "West";

Whenever you are printing using the %s and &pw you will get strange results. %s expects a string and the address of pw is a memory location not a string.