Passing INT Array by reference

[Jeff K]

Can you pass an int array by reference to a function and modify
selective elements?

Here is my code:

#include <stdio.h>

#define COLUMNSIZE 30
#define ASIZE 5
int calcfldpos(int *row, int *column, int *numArray)
{
int i=0;
printf("\tbefor e-->++numArray=%d\ n",*numArray );
*numArray+=4;
printf("\t after-->++numArray=%d\ n",*numArray );
return(1);
}
int updintArray(int *numArray[5])
{
int i=0;
printf("\nupdin tArray\n");
for ( i=0; i<ASIZE; i++ )
{
printf("numArra y[%d]=%d\n",i,numArr ay[i]);
*numArray[i]+=1;
printf("numArra y[%d]=%d\n\n",i,numA rray[i]);
}
return(0);
}
int main(int argc,char** argv)
{
int row=10;
int column=0;
int i=0;
int numArray[ASIZE];

numArray[0]=0; numArray[1]=1; numArray[2]=2; numArray[3]=3;
numArray[4]=4;

for ( i=0; i<ASIZE; i++ )
{
printf("numArra y[%d]=%d\n",i,numArr ay[i]);
}

printf("before->calcfldpos:\tn umArray[2] = %d\n",numArray[2]);
calcfldpos(&row ,&column,&numAr ray[2]);
printf(" after->calcfldpos:\tn umArray[2] = %d\n",numArray[2]);
for ( i=0; i<ASIZE; i++ )
{
printf("numArra y[%d]=%d\n",i,numArr ay[i]);
}
printf("Now updintArray\n") ;
updintArray(&nu mArray);
return(0);
}

I'm getting a core dumb on this statement:
*numArray[i]+=1;

I do not know how the pointer are messed up?
Any comment would be appreciated.

[Barry Schwarz]

Re: Passing INT Array by reference

On 14 Jul 2005 13:59:51 -0700, "Jeff K" <jfkearnssr@gma il.com> wrote:
[color=blue]
>Can you pass an int array by reference to a function and modify
>selective elements?[/color]

C passes by value exclusively. The technique used for arrays does
allow the called function to updated the arrays directly.
[color=blue]
>
>Here is my code:
>[/color]

snip
[color=blue]
>int updintArray(int *numArray[5])
>{
> int i=0;
> printf("\nupdin tArray\n");
> for ( i=0; i<ASIZE; i++ )
> {
> printf("numArra y[%d]=%d\n",i,numArr ay[i]);
> *numArray[i]+=1;
> printf("numArra y[%d]=%d\n\n",i,numA rray[i]);
> }
> return(0);
>}
>int main(int argc,char** argv)
>{
> int row=10;
> int column=0;
> int i=0;
> int numArray[ASIZE];
>[/color]

snip
[color=blue]
> for ( i=0; i<ASIZE; i++ )
> {
> printf("numArra y[%d]=%d\n",i,numArr ay[i]);
> }
> printf("Now updintArray\n") ;[/color]

Please learn to indent consistently. It will save you (and us) a lot
of aggravation.
[color=blue]
> updintArray(&nu mArray);
> return(0);
>}
>
>I'm getting a core dumb on this statement:
> *numArray[i]+=1;[/color]

How did you get past the compilation stage. updint is expecting an
array of 5 pointer to int (int *[5]). You pass it a pointer to an
array of 5 int (int (*)[5]). These types are not compatible an should
result in some diagnostic.[color=blue]
>
>I do not know how the pointer are messed up?
>Any comment would be appreciated.[/color]



<<Remove the del for email>>

[Suman]

Re: Passing INT Array by reference



Barry Schwarz wrote:[color=blue]
> On 14 Jul 2005 13:59:51 -0700, "Jeff K" <jfkearnssr@gma il.com> wrote:
>[color=green]
> >Can you pass an int array by reference to a function and modify
> >selective elements?[/color]
>
> C passes by value exclusively. The technique used for arrays does
> allow the called function to updated the arrays directly.
>[color=green]
> >
> >Here is my code:
> >[/color]
>
> snip
>[color=green]
> >int updintArray(int *numArray[5])
> >{
> > int i=0;
> > printf("\nupdin tArray\n");
> > for ( i=0; i<ASIZE; i++ )
> > {
> > printf("numArra y[%d]=%d\n",i,numArr ay[i]);
> > *numArray[i]+=1;
> > printf("numArra y[%d]=%d\n\n",i,numA rray[i]);
> > }
> > return(0);
> >}
> >int main(int argc,char** argv)
> >{
> > int row=10;
> > int column=0;
> > int i=0;
> > int numArray[ASIZE];
> >[/color]
>
> snip
>[color=green]
> > for ( i=0; i<ASIZE; i++ )
> > {
> > printf("numArra y[%d]=%d\n",i,numArr ay[i]);
> > }
> > printf("Now updintArray\n") ;[/color]
>
> Please learn to indent consistently. It will save you (and us) a lot
> of aggravation.
>[color=green]
> > updintArray(&nu mArray);
> > return(0);
> >}
> >
> >I'm getting a core dumb on this statement:
> > *numArray[i]+=1;[/color]
>
> How did you get past the compilation stage. updint is expecting an
> array of 5 pointer to int (int *[5]). You pass it a pointer to an
> array of 5 int (int (*)[5]). These types are not compatible an should
> result in some diagnostic.[/color]

He didn't.[color=blue][color=green]
> >
> >I do not know how the pointer are messed up?
> >Any comment would be appreciated.[/color][/color]

Here are my efforts at fixing the code:
The warnings I found with gcc 4.0.0 are embedded in the code as well.

#include <stdio.h>

#define COLUMNSIZE 30
#define ASIZE 5
int
calcfldpos(int *row, int *column, int *numArray)
{
/* was: unused variable
int i = 0;
*/
printf("\tbefor e-->++numArray= %d\n", *numArray);
*numArray += 4;
printf("\t after-->++numArray=%d\ n", *numArray);
return (1);

}

int
/* updintArray(int *numArray[5]) */
updintArray(int numArray[5])
{
int i = 0;
printf("\nupdin tArray\n");
for (i = 0; i < ASIZE; i++) {
/* was: format '%d' expects type 'int', but argument 3
has type 'int *' */
printf("numArra y[%d]=%d\n", i, numArray[i]);
/* *numArray[i] += 1; */
numArray[i] += 1;
/* was: format '%d' expects type 'int', but argument 3
has type 'int *' */
printf("numArra y[%d]=%d\n\n", i, numArray[i]);
}
return (0);
}

int
/* main(int argc, char **argv) */
main(void)
{
int row = 10;
int column = 0;
int i = 0;
int numArray[ASIZE];
numArray[0] = 0;
numArray[1] = 1;
numArray[2] = 2;
numArray[3] = 3;
numArray[4] = 4;

for (i = 0; i < ASIZE; i++) {
printf("numArra y[%d]=%d\n", i, numArray[i]);
}

printf("before->calcfldpos:\ tn umArray[2] = %d\n",
numArray[2]);
calcfldpos(&row , &column, &numArray[2]);
printf(" after->calcfldpos:\tn umArray[2 ] = %d\n",
numArray[2]);
for (i = 0; i < ASIZE; i++) {
printf("numArra y[%d]=%d\n", i, numArray[i]);
}
printf("Now updintArray\n") ;
/* passing argument 1 of 'updintArray' from incompatible
pointer type
updintArray(&nu mArray);
*/
updintArray(num Array);
return (0);
}

This was compiled with:
gcc -std=c99 -Wall -pedantic -ansi test.c

Whether this is what the OP wanted to try out, I am not too sure.

[ranjeet.gupta@gmail.com]

Re: Passing INT Array by reference



Jeff K wrote:[color=blue]
> Can you pass an int array by reference to a function and modify
> selective elements?[/color]

Call by refrence is In C++ not in C, In C their is only call by value
and below also in your code you have called by value :-)

you have passed the copy of address, so passing the address (call
by value) we modify the content in the passed address
(call by value),

I hope now you are clear that there is nothing such as call by
refrence in C, Yes In C++ we have the Call by Refrence, Dont get
confuse....any way i did some modification in the code and
placed my comment also.
[color=blue]
>
> Here is my code:
>
> #include <stdio.h>
>
> #define COLUMNSIZE 30
> #define ASIZE 5
> int calcfldpos(int *row, int *column, int *numArray)
> {
> int i=0;
> printf("\tbefor e-->++numArray=%d\ n",*numArray );
> *numArray+=4;
> printf("\t after-->++numArray=%d\ n",*numArray );
> return(1);
> }
> int updintArray(int *numArray[5])[/color]
^^^^^^^^^^^
change to *numArray // Why specifing index ?? when you
// are passing the base address of
the
// array in the main function.[color=blue]
> {[/color]

int *numArrayPtr = numArray; // Declare the Pointer
// Which holds the
address
// of the passed
argument.[color=blue]
> int i=0;
> printf("\nupdin tArray\n");
> for ( i=0; i<ASIZE; i++ )
> {
> printf("numArra y[%d]=%d\n",i,numArr ay[i]);
> *numArray[i]+=1;[/color]
^^^^^^^^^^^^^^
This will compile But Give the
memory allocation problem......

So solution is that you have to take the pointer varibale and replace
the above with this
*numArrayPtr +=1;

Regards
Ranjeet
[color=blue]
> printf("numArra y[%d]=%d\n\n",i,numA rray[i]);
> }
> return(0);
> }
> int main(int argc,char** argv)
> {
> int row=10;
> int column=0;
> int i=0;
> int numArray[ASIZE];
>
> numArray[0]=0; numArray[1]=1; numArray[2]=2; numArray[3]=3;
> numArray[4]=4;
>
> for ( i=0; i<ASIZE; i++ )
> {
> printf("numArra y[%d]=%d\n",i,numArr ay[i]);
> }
>
> printf("before->calcfldpos:\tn umArray[2] = %d\n",numArray[2]);
> calcfldpos(&row ,&column,&numAr ray[2]);
> printf(" after->calcfldpos:\tn umArray[2] = %d\n",numArray[2]);
> for ( i=0; i<ASIZE; i++ )
> {
> printf("numArra y[%d]=%d\n",i,numArr ay[i]);
> }
> printf("Now updintArray\n") ;
> updintArray(&nu mArray);
> return(0);
> }
>
> I'm getting a core dumb on this statement:
> *numArray[i]+=1;
>
> I do not know how the pointer are messed up?
> Any comment would be appreciated.[/color]

[Emmanuel Delahaye]

Re: Passing INT Array by reference

Jeff K wrote on 14/07/05 :[color=blue]
> Can you pass an int array by reference to a function and modify
> selective elements?[/color]

Well, all parameters are passed by value in C. Arrays are kinda
special. WHat is passed is a copy of its address or of the address of
it's first element via some pointer to array or to element
rescpectively.

#1

f ((T *)arr[10])
{
}

#2
f (T arr[10])
{
}
or
f (T arr[])
{
}
or
f (T *arr)
{
}


The first way (pointer to array) keeps the size info, and the second
one (pointer to element) has no size info. An extra size parameter
should help...


#include <stdio.h>

typedef int T;

int f (T (*arr)[10])
{
printf ("sizeof arr[0] = %u\n", (unsigned) sizeof *arr[0]);
printf ("sizeof arr = %u\n", (unsigned) sizeof *arr);
return 0;
}

int g (T *arr)
{
printf ("sizeof arr[0] = %u\n", (unsigned) sizeof arr[0]);
printf ("sizeof arr = %u\n", (unsigned) sizeof arr);
return 0;
}


int main (void)
{

T a[10];

f (&a);
g (a);
return 0;
}

--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html

"Clearly your code does not meet the original spec."
"You are sentenced to 30 lashes with a wet noodle."
-- Jerry Coffin in a.l.c.c++

[Emmanuel Delahaye]

Re: Passing INT Array by reference

Jeff K wrote on 14/07/05 :[color=blue]
> <snipped some buggy code>[/color]

Code fixed and enhanced. Ask for details if you don't understand...

#include <stdio.h>

/* useful and reusable trick to get the number of elements of an array
* (I insist : *array*) ...
*/
#define NELEM(a) (sizeof(a)/sizeof*(a))

int calcfldpos (int *numArray)
{
printf ("\tbefore-->++numArray=%d\ n", *numArray);
*numArray += 4;
printf ("\t after-->++numArray=%d\ n", *numArray);
return (1);
}

int updintArray (int *numArray, size_t size)
{
size_t i;
printf ("\nupdintArray \n");
for (i = 0; i < size; i++)
{
printf ("numArray[%d]=%d\n", i, numArray[i]);
numArray[i] += 1;
printf ("numArray[%d]=%d\n\n", i, numArray[i]);
}
return (0);
}

int main ()
{

int numArray[5] = {0,1,2,3,4};
size_t i;

for (i = 0; i < NELEM(numArray) ; i++)
{
printf ("numArray[%d]=%d\n", i, numArray[i]);
}

printf ("before->calcfldpos:\tn umArray[2] = %d\n", numArray[2]);
calcfldpos (numArray);
printf (" after->calcfldpos:\tn umArray[2] = %d\n", numArray[2]);
for (i = 0; i < NELEM(numArray) ; i++)
{
printf ("numArray[%d]=%d\n", i, numArray[i]);
}
printf ("Now updintArray\n") ;
updintArray (numArray, NELEM(numArray) );
return (0);
}

--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html

"Mal nommer les choses c'est ajouter du malheur au
monde." -- Albert Camus.

[Barry Schwarz]

Re: Passing INT Array by reference

On 14 Jul 2005 21:59:13 -0700, "Suman" <skarpio@gmail. com> wrote:
[color=blue]
>
>
>Barry Schwarz wrote:[color=green]
>> On 14 Jul 2005 13:59:51 -0700, "Jeff K" <jfkearnssr@gma il.com> wrote:[/color][/color]
snip[color=blue][color=green][color=darkred]
>> >int updintArray(int *numArray[5])
>> >{[/color][/color][/color]
snip[color=blue][color=green][color=darkred]
>> >}
>> >int main(int argc,char** argv)
>> >{
>> > int row=10;
>> > int column=0;
>> > int i=0;
>> > int numArray[ASIZE];[/color][/color][/color]
snip[color=blue][color=green][color=darkred]
>> > updintArray(&nu mArray);
>> > return(0);
>> >}
>> >
>> >I'm getting a core dumb on this statement:
>> > *numArray[i]+=1;[/color]
>>
>> How did you get past the compilation stage. updint is expecting an
>> array of 5 pointer to int (int *[5]). You pass it a pointer to an
>> array of 5 int (int (*)[5]). These types are not compatible an should
>> result in some diagnostic.[/color]
>
>He didn't.[/color]

Look again. numArray is an array of int. &numArray has type pointer
to array of int. The call to updintArray has an argument with an
incompatible type which requires a diagnostic.


<<Remove the del for email>>

[Suman]

Re: Passing INT Array by reference



Barry Schwarz wrote:[color=blue]
> On 14 Jul 2005 21:59:13 -0700, "Suman" <skarpio@gmail. com> wrote:
>[color=green]
> >
> >
> >Barry Schwarz wrote:[color=darkred]
> >> On 14 Jul 2005 13:59:51 -0700, "Jeff K" <jfkearnssr@gma il.com> wrote:[/color][/color]
> snip[color=green][color=darkred]
> >> >int updintArray(int *numArray[5])
> >> >{[/color][/color]
> snip[color=green][color=darkred]
> >> >}
> >> >int main(int argc,char** argv)
> >> >{
> >> > int row=10;
> >> > int column=0;
> >> > int i=0;
> >> > int numArray[ASIZE];[/color][/color]
> snip[color=green][color=darkred]
> >> > updintArray(&nu mArray);
> >> > return(0);
> >> >}
> >> >
> >> >I'm getting a core dumb on this statement:
> >> > *numArray[i]+=1;
> >>
> >> How did you get past the compilation stage. updint is expecting an
> >> array of 5 pointer to int (int *[5]). You pass it a pointer to an
> >> array of 5 int (int (*)[5]). These types are not compatible an should
> >> result in some diagnostic.[/color]
> >
> >He didn't.[/color]
>
> Look again. numArray is an array of int. &numArray has type pointer
> to array of int. The call to updintArray has an argument with an
> incompatible type which requires a diagnostic.[/color]

What I meant was, he didn't get through the compilation stage.
I quoted more than necessary, which caused the confusion. I compiled
his code as is, and found some errors/warnings, which I *tried* to
fix (since I am not too sure of his intent), and posted.

Any confusion caused is regretted.

Suman.