Indefinite-length array as member of struct: how?

[Bill Svirsky]

In article <669 at kl-cs.UUCP> pc at cs.keele.ac.uk (Phil Cornes) writes:
+With a structure declaration:
+
+	struct node1 {
+		struct node1 *next1;
+		char string1[1];
+	} *nodeptr1;
+
+and a block of code:
+
+	nodeptr1=(struct node1 *)malloc(sizeof(struct node1)+strlen(data));
+	(void)strcpy(nodeptr1-+string1,data);

[stuff deleted]

+On the other hand, with a structure declared as:
+
+	struct node2 {
+		struct node2 *next2;
+		char *string2;
+	} nodeptr2;
+
+and a block of code like:
+
+	nodeptr2=(struct node2 *)malloc(sizeof(struct node2)+strlen(data)+1);
+	nodeptr2-+string2 = (char *)nodeptr2+sizeof(struct node2);
+	(void)strcpy(nodeptr2-+string2,data);

[more stuff deleted]

+                                              My own preference in
+this case is to use the second of these solutions for two reasons:
+
+1. The first solution relies upon a lot more knowledge of the way that C
+   operates internally, in order to be confident that it will work.
+
+2. The first solution also suffers from the problem that it relies upon
+   accessing the string1 array outside its declared subscript range,
+   which must be classed as a 'tacky' practice at best.

One more advantage to the 2nd solution is that it is easily expandable
and very flexible.  For instance, given:

	struct node {
		struct node *next;
		char *string1;
		char *string2;
	} nodeptr;

use a block of code like:

	nodeptr=(struct node *)malloc(sizeof(struct node)+
			strlen(data1)+strlen(data2)+2);

	nodeptr->string1 = (char *)nodeptr+sizeof(struct node);
	nodeptr->string2 = nodeptr->string1+strlen(data1)+1);

	(void)strcpy(nodeptr->string1,data1);
	(void)strcpy(nodeptr->string2,data2);

-- 
Bill Svirsky, Citicorp+TTI, 3100 Ocean Park Blvd., Santa Monica, CA 90405
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