pointers #2 - comp.lang.c

Which is the correct way, to initialise a pointer to an element of an array. I am sorry if my terminology is not correct /* declare pointer */ int *ptr; /* define and initialise array */ int array[10] = {,0,1,2,3,4,5,6,7,8,9}; /* assign value to pointer */ ptr = &array[9]; or ptr = (array + 9); I have found both work but to avoid future problems is one better than the other. And if so why

darklight wrote: > Which is the correct way, to initialise a pointer to an element of > an array. I am sorry if my terminology is not correct > > /* declare pointer */ > int *ptr; > /* define and initialise array */ > int array[10] = {,0,1,2,3,4,5,6,7,8,9}; > /* assign value to pointer */ > ptr = &array[9]; or ptr = (array + 9); > > I have found both work but to avoid future problems is one > better than the other. And if so why The two expressions are equivalent. Each points ptr to the last element of the array. In a value context, the following pairs of expressions are equivalent: (arrayname) == (&arrayname[0]) (arrayname + n) == (&arrayname[n]) (*(arrayname + n)) == (arrayname[n]) Which you choose is entirely a matter of taste. I should perhaps add that the external parentheses are there purely as an attempted defence against nitpicking! :-) -- Richard Heathfield : binary@eton.powernet.co.uk "Usenet is a strange place." - Dennis M Ritchie, 29 July 1999. C FAQ: http://www.eskimo.com/~scs/C-faq/top.html K&R answers, C books, etc: http://users.powernet.co.uk/eton

darklight wrote: > Which is the correct way, to initialise a pointer to an element of > an array. I am sorry if my terminology is not correct > > /* declare pointer */ > int *ptr; > /* define and initialise array */ > int array[10] = {,0,1,2,3,4,5,6,7,8,9}; > /* assign value to pointer */ > ptr = &array[9]; or ptr = (array + 9); > > I have found both work but to avoid future problems is one > better than the other. And if so why > Either way will be valid. Generally you will find the [] operator the most accepted method. To me, it it easier to understand the operations. By the way the () operator is not required in the above assignment, but you may add them if it gives you more clarity. It could be, ptr = array + 9; However, in dereferencing, you will need the () operator to yield the value of the last element of the array named array. int i = *(array + 9;) will assign to i the value 9. int i = *array + 9; will assign to i the value 18. -- Al Bowers Tampa, Fl USA mailto: xabowers@myrapidsys.com (remove the x to send email) http://www.geocities.com/abowers822/

darklight <nglen702@netscapel.net> wrote: # Which is the correct way, to initialise a pointer to an element of # an array. I am sorry if my terminology is not correct # # /* declare pointer */ # int *ptr; # /* define and initialise array */ # int array[10] = {,0,1,2,3,4,5,6,7,8,9}; # /* assign value to pointer */ # ptr = &array[9]; or ptr = (array + 9); # # I have found both work but to avoid future problems is one # better than the other. And if so why &(array[9]) = &(*(array+9)) = (array+9) The notations are equivalent, and you can expect a compiler to generate the exact same code for both. Which is better is style question for you. Which do you think looks better? Which do you more easily understand? That is the correct one. -- Derk Gwen http://derkgwen.250free.com/html/index.html Don't say anything. Especially you.

darklight <nglen702@netscapel.net> wrote: > Which is the correct way, to initialise a pointer to an element of > an array. I am sorry if my terminology is not correct > > /* declare pointer */ > int *ptr; > /* define and initialise array */ > int array[10] = {,0,1,2,3,4,5,6,7,8,9}; > /* assign value to pointer */ > ptr = &array[9]; or ptr = (array + 9); > > I have found both work but to avoid future problems is one > better than the other. And if so why Either one works equally well and the decision on which to use is generally based on personal preference and the context. -- == Eric Gorr ========= http://www.ericgorr.net ========= ICQ:9293199 === "Therefore the considerations of the intelligent always include both benefit and harm." - Sun Tzu == Insults, like violence, are the last refuge of the incompetent... ===

Hi ptr=&array[9] 0r (array+9) both refer same operation. The basic difference is that by accessing with pointers the operation will be quite faster while comparing array[9]. Hence good C programmers will prefer pointers. Hope this information will be useful to u. If u find anymore idea like this please let me know. Regards, Anand.

Anand wrote: > > Hi > > ptr=&array[9] 0r (array+9) both refer same operation. > The basic difference is that by accessing with pointers > the operation will be quite faster while comparing array[9]. Unless it isn't faster. The compiler has enough information to know that (&array[9]) and (array+9) are the same. I'm unfamiliar with the address operator as being translated into a time consuming machine operation. -- pete

On 12 Jan 2004 02:53:40 -0800, anandattek@yahoo.co.in (Anand) wrote: > Hi > > ptr=&array[9] 0r (array+9) both refer same operation. The basic difference > is that by accessing with pointers the operation will be quite faster while > comparing array[9]. Hence good C programmers will prefer pointers. Hope this > information will be useful to u. If u find anymore idea like this please let me > know. My compilers generate the same code for p = &array[9] and p = array+9 What about your compilers? Regards Horst

Anand wrote: > Hi > > ptr=&array[9] 0r (array+9) both refer same operation. The basic > difference > is that by accessing with pointers the operation will be quite faster > while comparing array[9]. Get A Better Compiler. Yours Is Broken. > Hence good C programmers will prefer pointers. Better C programmers know that the compiler knows perfectly well that the two expressions are equivalent. > Hope this information will be useful to u. Let us hope that u, whoever he is, never reads it, since it's fundamentally flawed. -- Richard Heathfield : binary@eton.powernet.co.uk "Usenet is a strange place." - Dennis M Ritchie, 29 July 1999. C FAQ: http://www.eskimo.com/~scs/C-faq/top.html K&R answers, C books, etc: http://users.powernet.co.uk/eton

Eric <egDfAusenetE5fz@verizon.net> spoke thus: > darklight <nglen702@netscapel.net> wrote: >> /* declare pointer */ >> int *ptr; >> /* define and initialise array */ >> int array[10] = {0,1,2,3,4,5,6,7,8,9}; ^ extra comma removed >> /* assign value to pointer */ >> ptr = &array[9]; or ptr = (array + 9); > Either one works equally well and the decision on which to use is > generally based on personal preference and the context. But the equivalence breaks down for ptr=array+10; /* Legal */ ptr=&array[10]; /* Illegal (?), since array[10] is a dereference */ , right? -- Christopher Benson-Manica | I *should* know what I'm talking about - if I ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

Christopher Benson-Manica wrote: > Eric <egDfAusenetE5fz@verizon.net> spoke thus: >> darklight <nglen702@netscapel.net> wrote: > >>> /* declare pointer */ >>> int *ptr; >>> /* define and initialise array */ >>> int array[10] = {0,1,2,3,4,5,6,7,8,9}; > ^ extra comma removed >>> /* assign value to pointer */ >>> ptr = &array[9]; or ptr = (array + 9); > >> Either one works equally well and the decision on which to use is >> generally based on personal preference and the context. > > But the equivalence breaks down for > > ptr=array+10; /* Legal */ > ptr=&array[10]; /* Illegal (?), since array[10] is a dereference */ > > , right? Wrong, at least in C99. [6.5.3.2] 3 The unary & operator returns the address of its operand. [...] Similarly, if the operand is the result of a [] operator, neither the & operator nor the unary * that is implied by the [] is evaluated and the result is as if the & operator were removed and the [] operator were changed to a + operator. Jeremy.

Jeremy Yallop <jeremy@jdyallop.freeserve.co.uk> spoke thus: > 3 The unary & operator returns the address of its operand. [...] > Similarly, if the operand is the result of a [] operator, neither > the & operator nor the unary * that is implied by the [] is > evaluated and the result is as if the & operator were removed and > the [] operator were changed to a + operator. Thank you, I am yet again abashed yet wiser :) -- Christopher Benson-Manica | I *should* know what I'm talking about - if I ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

In article <cdf5b6.0401120253.4d2b99ac@posting.google.com>, anandattek@yahoo.co.in (Anand) wrote: > Hi > > ptr=&array[9] 0r (array+9) both refer same operation. You should have just stopped there. > The basic difference > is that by accessing with pointers the operation will be quite faster while > comparing array[9]. Says who? I have *never* seen a compiler that generates more efficent code for (&array[9]) than it does for *(array+9).