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("\tbefore-->++numArray=%d\n",*numArray);
*numArray+=4;
printf("\t after-->++numArray=%d\n",*numArray);
return(1);
}
int updintArray(int *numArray[5])
{
int i=0;
printf("\nupdintArray\n");
for ( i=0; i<ASIZE; i++ )
{
printf("numArray[%d]=%d\n",i,numArray);
*numArray+=1;
printf("numArray[%d]=%d\n\n",i,numArray);
}
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("numArray[%d]=%d\n",i,numArray);
}

printf("before->calcfldpos:\tnumArray[2] = %d\n",numArray[2]);
calcfldpos(&row,&column,&numArray[2]);
printf(" after->calcfldpos:\tnumArray[2] = %d\n",numArray[2]);
for ( i=0; i<ASIZE; i++ )
{
printf("numArray[%d]=%d\n",i,numArray);
}
printf("Now updintArray\n");
updintArray(&numArray);
return(0);
}

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

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

 

[Barry Schwarz]

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

C passes by value exclusively. The technique used for arrays does
allow the called function to updated the arrays directly.

Here is my code:

snip

int updintArray(int *numArray[5])
{
int i=0;
printf("\nupdintArray\n");
for ( i=0; i<ASIZE; i++ )
{
printf("numArray[%d]=%d\n",i,numArray);
*numArray+=1;
printf("numArray[%d]=%d\n\n",i,numArray);
}
return(0);
}
int main(int argc,char** argv)
{
int row=10;
int column=0;
int i=0;
int numArray[ASIZE];

snip

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

Please learn to indent consistently. It will save you (and us) a lot
of aggravation.

updintArray(&numArray);
return(0);
}

I'm getting a core dumb on this statement:
*numArray+=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.

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

<<Remove the del for email>>

 

[Suman]

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

C passes by value exclusively. The technique used for arrays does
allow the called function to updated the arrays directly.

Here is my code:

snip

int updintArray(int *numArray[5])
{
int i=0;
printf("\nupdintArray\n");
for ( i=0; i<ASIZE; i++ )
{
printf("numArray[%d]=%d\n",i,numArray);
*numArray+=1;
printf("numArray[%d]=%d\n\n",i,numArray);
}
return(0);
}
int main(int argc,char** argv)
{
int row=10;
int column=0;
int i=0;
int numArray[ASIZE];

snip

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

Please learn to indent consistently. It will save you (and us) a lot
of aggravation.

updintArray(&numArray);
return(0);
}

I'm getting a core dumb on this statement:
*numArray+=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.

He didn't.
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("\tbefore-->++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("\nupdintArray\n");
for (i = 0; i < ASIZE; i++) {
/* was: format '%d' expects type 'int', but argument 3
has type 'int *' */
printf("numArray[%d]=%d\n", i, numArray);
/* *numArray += 1; */
numArray += 1;
/* was: format '%d' expects type 'int', but argument 3
has type 'int *' */
printf("numArray[%d]=%d\n\n", i, numArray);
}
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("numArray[%d]=%d\n", i, numArray);
}

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

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

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.

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("\tbefore-->++numArray=%d\n",*numArray);
*numArray+=4;
printf("\t after-->++numArray=%d\n",*numArray);
return(1);
}
int updintArray(int *numArray[5])

^^^^^^^^^^^
change to *numArray // Why specifing index ?? when you
// are passing the base address of
the
// array in the main function.

{

int *numArrayPtr = numArray; // Declare the Pointer
// Which holds the
address
// of the passed
argument.

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

^^^^^^^^^^^^^^
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

printf("numArray[%d]=%d\n\n",i,numArray);
}
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("numArray[%d]=%d\n",i,numArray);
}

printf("before->calcfldpos:\tnumArray[2] = %d\n",numArray[2]);
calcfldpos(&row,&column,&numArray[2]);
printf(" after->calcfldpos:\tnumArray[2] = %d\n",numArray[2]);
for ( i=0; i<ASIZE; i++ )
{
printf("numArray[%d]=%d\n",i,numArray);
}
printf("Now updintArray\n");
updintArray(&numArray);
return(0);
}

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

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

 

[Emmanuel Delahaye]

Jeff K wrote on 14/07/05 :

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

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]

Jeff K wrote on 14/07/05 :

<snipped some buggy code>

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);
numArray += 1;
printf ("numArray[%d]=%d\n\n", i, numArray);
}
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);
}

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

int updintArray(int *numArray[5])
{ snip
}
int main(int argc,char** argv)
{
int row=10;
int column=0;
int i=0;
int numArray[ASIZE]; snip
updintArray(&numArray);
return(0);
}

I'm getting a core dumb on this statement:
*numArray+=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.

He didn't.

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]

int updintArray(int *numArray[5])
{ snip
}
int main(int argc,char** argv)
{
int row=10;
int column=0;
int i=0;
int numArray[ASIZE]; snip
updintArray(&numArray);
return(0);
}

I'm getting a core dumb on this statement:
*numArray+=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.

He didn't.

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.

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.