Printing pointers of an array

TIMBERings · 10-10-2009, 09:45 PM

I am brand new to C, so I've been trying to do some research. This seems to be right by everything I read, but I could use a hand.

Code:

char *signedtwobit(char *x, char *y);

int main(void) {
	char x[]={0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
	char y[]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1};
	char *p, *q;
	p=x;
	q=y;
	
	char *s;
	s=signedtwobit(p,q);
	int i=0;
	for (i=0;i<=32;i++) {
		printf("|%d|, ", *s);
		s++; 
	}
	
	
}

char *signedtwobit(char *x, char *y) {
	char s[33];
	int i;
	
	for (i=31; i>=0; --i) {
		s[i]=(int)x+(int)y;
		if (s[i]==2) {
			s[i]=0;
			x++;
			*x++;
			--x;
		x=x+1;
		y=y+1;
	}
}
	
	char *p=&s[0];
	return p;
	
}

This is what I get.

Code:

|-64|, |-12|, |111|, |5|, |-72|, |16|, |-122|, |4|, |8|, |96|, |-125|, |4|, |8|, |8|, |-50|, |-104|, |-65|, |96|, |27|, |-12|, |-73|, |-64|, |116|, |5|, |-72|, |-64|, |-122|, |4|, |8|, |24|, |-50|, |-104|, |-65|,

I should get.

Code:

|0|, |1|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|, |0|,

Something is wrong in the signedtwobit function. Because instead of running through the function, I assigned a pointer to the x array and got the correct string.

I would love any type of help.

tabstop · 10-10-2009, 09:50 PM

You are returning a pointer to a local variable. Since the local variable goes away when the function is done, you are left with a pointer to nothing in particular.

Also (int)x casts the pointer to an int, so you might get something in the 1 billion range. It certainly won't be the value of one of the array elements.

TIMBERings · 10-10-2009, 10:34 PM

So instead I should pass the char array instead of the pointer?

tabstop · 10-10-2009, 10:38 PM

You are passing the array. You had it right up top: *s to get at the innards of the array. It hasn't changed just because you've gone twelve lines down in your program.

TIMBERings · 10-10-2009, 10:41 PM

sorry, i meant return the array instead of the pointer.

tabstop · 10-10-2009, 10:53 PM

You can't return an array.

If you want this array to be available outside the function, you must manage the memory yourself (say hello to malloc, your new friend).

TIMBERings · 10-10-2009, 11:26 PM

One more quick question. I'm wondering why I can't do this.

Code:

	char x[]={0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
	char y[]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1};
	char *p, *q;
	p=x;
	q=y;
	signedtwobit(p,q);

//says it needs an expression before ']'	
	x[]={0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
	y[]={1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
	p=x;
	q=y;
	signedtwobit(p,q);

	x[]={1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
	y[]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1};
	p=x;
	q=y;
	signedtwobit(p,q);


//Says it expects an expression before '{'
	x={1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
	y={1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
	p=x;
	q=y;
	signedtwobit(p,q);

	x={0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
	y={1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
	p=x;
	q=y;
	signedtwobit(p,q);

tabstop · 10-10-2009, 11:35 PM

You can never assign to an array. You can only assign to individual slots in an array. (That thing you do at first is "initializing" and can only happen when the memory is first allocated, i.e., when it is declared.)

10-10-2009, 09:50 PM	#2
tabstop and the Hat of Guessing Join Date: Nov 2007 Posts: 8,740	You are returning a pointer to a local variable. Since the local variable goes away when the function is done, you are left with a pointer to nothing in particular. Also (int)x casts the pointer to an int, so you might get something in the 1 billion range. It certainly won't be the value of one of the array elements.
tabstop is offline

10-10-2009, 10:34 PM	#3
TIMBERings Registered User Join Date: Oct 2009 Posts: 14	So instead I should pass the char array instead of the pointer?
TIMBERings is offline

10-10-2009, 10:38 PM	#4
tabstop and the Hat of Guessing Join Date: Nov 2007 Posts: 8,740	You are passing the array. You had it right up top: *s to get at the innards of the array. It hasn't changed just because you've gone twelve lines down in your program.
tabstop is offline

10-10-2009, 10:41 PM	#5
TIMBERings Registered User Join Date: Oct 2009 Posts: 14	sorry, i meant return the array instead of the pointer.
TIMBERings is offline

10-10-2009, 10:53 PM	#6
tabstop and the Hat of Guessing Join Date: Nov 2007 Posts: 8,740	You can't return an array. If you want this array to be available outside the function, you must manage the memory yourself (say hello to malloc, your new friend).
tabstop is offline

10-10-2009, 11:35 PM	#8
tabstop and the Hat of Guessing Join Date: Nov 2007 Posts: 8,740	You can never assign to an array. You can only assign to individual slots in an array. (That thing you do at first is "initializing" and can only happen when the memory is first allocated, i.e., when it is declared.)
tabstop is offline

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