Problems with a simple linear search

[nembo kid]

I have an issue with a simple function that has to make a linear search
for a key into an array.

If the key is found in the array, the function it has to return 1 to the
caller and pass array index through a out parameter.

The issue is that the out parameter is not being updated.

If I return the position to the caller (instead to use a out parameter)
all is ok.

But I would to use a boolean (1 or 0) value as return value to the
caller: 1 stays for found, 0 stays for not found.

Thanks in advance and apologies for my bad english.

Here it is my piece of code.

/*** Code starts here ***/

/* Search for number 'x' in array v[] */
/* If 'x' found pass array position through 'pos' parameter */
/* and returns 1 to caller; else returns 0 */

int linsearch (int v[], int nmax, int x, int pos)
{

int i;

for (i=0; i<nmax; i++)
{
if(v[i]==x)
{
pos=i; /* pass the index where found */
return 1; /* returns 1 (found) */
}
}

return 0; /* Exit from for...so returns 0 */
}

int main (void)
{
int myvet[] = {5,4,3,2,1};

int key = 2; /* Number to find */

int index=0; /* Actual parameter that holds index of element */

if(linsearch(my vet, 5, 2, index))
printf ("\nNumber %d found at position %d ", key, index);
else
printf ("\nNumber %d not found", key);

return 0;
}

/*** Code ends here ***/

[Jim Langston]

Re: Problems with a simple linear search

nembo kid wrote:

I have an issue with a simple function that has to make a linear
search for a key into an array.
>
If the key is found in the array, the function it has to return 1 to
the caller and pass array index through a out parameter.
>
The issue is that the out parameter is not being updated.
>
If I return the position to the caller (instead to use a out
parameter) all is ok.
>
But I would to use a boolean (1 or 0) value as return value to the
caller: 1 stays for found, 0 stays for not found.
>
Thanks in advance and apologies for my bad english.
>
Here it is my piece of code.
>
/*** Code starts here ***/
>
/* Search for number 'x' in array v[] */
/* If 'x' found pass array position through 'pos' parameter */
/* and returns 1 to caller; else returns 0 */
>
int linsearch (int v[], int nmax, int x, int pos)

pos is being passed by value. A copy of it is made and passed into the
function. pos becomes a local variable here, only visible during the
lifetime of the function. Any changes to this local pos will never be seen
by the caller.

{
>
int i;
>
for (i=0; i<nmax; i++)
{
if(v[i]==x)
{
pos=i; /* pass the index where found */

This only changes the local copy of pos, the local variable. The local
variable pos disappears when this function is returned (the next line).

return 1; /* returns 1 (found) */
}
}
>
return 0; /* Exit from for...so returns 0 */
}
>
int main (void)
{
int myvet[] = {5,4,3,2,1};
>
int key = 2; /* Number to find */
>
int index=0; /* Actual parameter that holds index of element */
>
if(linsearch(my vet, 5, 2, index))

index is passed by value. The contents of index (0) is copied and passed to
the function.

printf ("\nNumber %d found at position %d ", key, index);
else
printf ("\nNumber %d not found", key);
>
return 0;
}
>
/*** Code ends here ***/

Okay, so the question is, how to fix this? You need to pass a pointer to
the function you wish to change. Change the signature of linsearch to:
int linsearch (int v[], int nmax, int x, int* pos)
now pos is a pointer to an integer. Since you want to change what the
pointer points to, not the pointer itself, you need a level of indirection.
*pos=i; /* pass the index where found */

* used in this method is to dereference the pointer. Basically saying,
"what pos points to".

One other change is needed when you call linsearch. Instead of passing the
variable, you need to pass a pointer to the variable:
if(linsearch(my vet, 5, 2, &index))

& used this way is "adress off" basically saying, "pass the address of
index".

--
Jim Langston
tazmaster@rocke tmail.com

[Eric Sosman]

Re: Problems with a simple linear search

nembo kid wrote:

I have an issue with a simple function that has to make a linear search
for a key into an array.
>
If the key is found in the array, the function it has to return 1 to the
caller and pass array index through a out parameter.
>
The issue is that the out parameter is not being updated.
>
If I return the position to the caller (instead to use a out parameter)
all is ok.
[...]

This is Question 4.8 in the comp.lang.c Frequently
Asked Questions (FAQ) list, <http://www.c-faq.com/>. The
question's title in the FAQ doesn't seem at first blush to
have much to do with your problem, but trust me and read on.

--
Eric Sosman
esosman@ieee-dot-org.invalid

[nembo kid]

Re: Problems with a simple linear search

Jim Langston ha scritto:

& used this way is "adress off" basically saying, "pass the address of
index".

Very clear. I fix it by using a pointer.
Thanks again to others too.