orville@weyrich.UUCP (Orville R. Weyrich) (05/04/91)
I am studying for a C language exam, and a study guide I am using states that the declaration char **ch; is equivalent to char *ch; I am having difficulty understanding this. My interpretation of the declaration char **ch is that we have a pointer to a pointer to a char. I think that this would be equivalent to the declaration char *ch[] rather than what the study guide says. Am I missing something obvious? Thanks, orville. -------------------------------------- ****************************** Orville R. Weyrich, Jr. Certified Systems Professional Internet: orville%weyrich@uunet.uu.net Weyrich Computer Consulting Voice: (602) 391-0821 POB 5782, Scottsdale, AZ 85261 Fax: (602) 391-0023 (Yes! I'm available) -------------------------------------- ******************************
toma@swsrv1.cirr.com (Tom Armistead) (05/05/91)
In article <1991May4.062007.3264@weyrich.UUCP> orville@weyrich.UUCP (Orville R. Weyrich) writes: >I am studying for a C language exam, and a study guide I am using states >that the declaration char **ch; is equivalent to char *ch; > >I am having difficulty understanding this. My interpretation of the >declaration char **ch is that we have a pointer to a pointer to a char. >I think that this would be equivalent to the declaration char *ch[] >rather than what the study guide says. > >Am I missing something obvious? > >Thanks, > >orville. > >-------------------------------------- ****************************** >Orville R. Weyrich, Jr. Certified Systems Professional >Internet: orville%weyrich@uunet.uu.net Weyrich Computer Consulting >Voice: (602) 391-0821 POB 5782, Scottsdale, AZ 85261 >Fax: (602) 391-0023 (Yes! I'm available) >-------------------------------------- ****************************** Nope, your not missing anything. Your 'study guide' is wrong! Tom -- Tom Armistead - Software Services - 2918 Dukeswood Dr. - Garland, Tx 75040 =========================================================================== toma@swsrv1.cirr.com {egsner,letni,ozdaltx,void}!swsrv1!toma
lwb@pensoft.uucp (Lance Bledsoe) (05/07/91)
In article <1991May4.062007.3264@weyrich.UUCP> orville@weyrich.UUCP (Orville R. Weyrich) writes: >I am studying for a C language exam, and a study guide I am using states >that the declaration char **ch; is equivalent to char *ch; > >I am having difficulty understanding this. My interpretation of the >declaration char **ch is that we have a pointer to a pointer to a char. >I think that this would be equivalent to the declaration char *ch[] >rather than what the study guide says. You are correct. -- Lance Bledsoe Off: (512) 343-1111 Pencom Software, Inc. Fax (512) 343-9650 8716 Loop 360 N. Suite 300 UUCP: cs.utexas.edu!pensoft!lwb Austin, Texas 78759 UUNET: uunet!uudell!pensoft!lwb
gwyn@smoke.brl.mil (Doug Gwyn) (05/07/91)
In article <1991May4.062007.3264@weyrich.UUCP> orville@weyrich.UUCP (Orville R. Weyrich) writes: >I am studying for a C language exam, and a study guide I am using states >that the declaration char **ch; is equivalent to char *ch; No, they're not at all equivalent. They might not even have the same size. >I am having difficulty understanding this. My interpretation of the >declaration char **ch is that we have a pointer to a pointer to a char. Correct. >I think that this would be equivalent to the declaration char *ch[] >rather than what the study guide says. The only place where char**ch and char*ch[] are equivalent is as a declaration of a function parameter.
robc@cup.portal.com (Rob X Cowan) (05/07/91)
char **ch and char *ch are not equivalent, but if the reference isn't just a typo, perhaps the author meant that internally a pointer is a pointer is a.. and that a pointer to pointer to a char is much like a pointer to a char at the machine level. Pure conjecture. -Rob robc@cup.portal.com