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From: grogers@convex.com (Geoffrey Rogers)
Newsgroups: comp.lang.c
Subject: Re: Help With Array of Pointers to chars
Message-ID: <1991Apr03.015905.23203@convex.com>
Date: 3 Apr 91 01:59:05 GMT
References: <1991Apr2.193849.17442@daffy.cs.wisc.edu>
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In article <1991Apr2.193849.17442@daffy.cs.wisc.edu> jansa@cat27.cs.wisc.edu (Dean Jansa) writes:
>What is the fastest way, ( read efficent use of memory and quick ) to
>transfer strings from a struct i.e:
>
>	struct something 
>		  {
>		   string[10];
>		   string2[10];
>		   .
>                  };
>into a array of pointers to chars:
>       char *myarray[10];
>
>I need to malloc each pointer to char to be able to hold 10 chars in this
>example then do a strcpy.   Any easier ways out there??

How about:

	char *myarray[10], *p;
	struct something x;
	int i;

	p = myarray = (char *) malloc(sizeof(x));
	for ( p += 10, i = 1; i < 10; ++i, p += 10 )
		myarray[i] = p;

	memcpy((void *) myarray[0], (void *) &x, sizeof(x));

If you have copy the strings into the array of pointers. Otherwise do:

	myarray[0] = x.string;
	myarray[1] = x.string2;
	  "              "
	  "              "
	  "              "
	myarray[9] = x.string9;

If you know you are not going to overwrite x.

For very short loops it is almost best to unroll the loop yourself, because
not all compilers will do this.

+------------------------------------+---------------------------------+
| Geoffrey C. Rogers   		     | "Whose brain did you get?"      |
| grogers@convex.com                 | "Abbie Normal!"                 |
| {sun,uunet,uiucdcs}!convex!grogers |                                 |
+------------------------------------+---------------------------------+