I want to retrieve the index in the array where the value is stored. I know the value of the item at that point in the array. I'm thinking it's similar to the findIndex function in c#.
For example, array[2] = {4, 7, 8}. I know the value is 7, how do I get the value of the index, 1, if I know it is at array[1]?
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2Write a loop from 0 to the size of the array, compare current element to the given value, stop when found.The Paramagnetic Croissant– The Paramagnetic Croissant2014-07-28 20:43:34 +00:00Commented Jul 28, 2014 at 20:43
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1If you know the array is sorted, you can also use binary searchDrew McGowen– Drew McGowen2014-07-28 20:47:46 +00:00Commented Jul 28, 2014 at 20:47
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@user1798299 May I append my post to show how to use bsearch?:)Vlad from Moscow– Vlad from Moscow2014-07-28 21:11:00 +00:00Commented Jul 28, 2014 at 21:11
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@VladfromMoscow - just do it.ryyker– ryyker2014-07-28 21:18:50 +00:00Commented Jul 28, 2014 at 21:18
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@VladfromMoscow of course!user1798299– user17982992014-07-28 21:22:50 +00:00Commented Jul 28, 2014 at 21:22
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3 Answers
For example you can define the corresponding function the following way
size_t FindIndex( const int a[], size_t size, int value )
{
size_t index = 0;
while ( index < size && a[index] != value ) ++index;
return ( index == size ? -1 : index );
}
Also instead of type size_t you can use type int.
But the better way is to use standard algorithm std::find or std::find_if declared in header <algorithm> provided that you use C++
For example
#include <algorithm>
#include <iterator>
int main()
{
int a[] = { 4, 7, 8 };
auto it = std::find( std::begin( a ), std::end( a ), 7 );
if ( it != std::end( a ) )
{
std::cout << "The index of the element with value 7 is "
<< std::distance( std::begin( a ), it )
<< std::endl;
}
}
The output is
The index of the element with value 7 is 1
Otherwise you have to write the function yourself as I showed abve.:)
If the array is sorted you can use standard C function bsearch declared in header <stdlib.h>
For example
#include <stdio.h>
#include <stdlib.h>
int cmp( const void *lhs, const void *rhs )
{
if ( *( const int * )lhs < *( const int * )rhs ) return -1;
else if ( *( const int * )rhs < *( const int * )lhs ) return 1;
else return 0;
}
int main()
{
int a[] = { 4, 7, 8 };
int x = 7;
int *p = ( int * )bsearch( &x, a, 3, sizeof( int ), cmp );
if ( p != NULL ) printf( "%d\n", p - a );
return 0;
}
answered Jul 28, 2014 at 20:45
Vlad from Moscow
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13 Comments
ryyker
In your first example, you have return type of
size_t. This is typedef unsigned int size_t;, so returning a -1 may be problematic. Your second example looks like C++. (tagged C)ryyker
OP is tagged C, not C++ though. Other than that I like both approaches, except size_t cannot be negative.
Drew McGowen
@ryyker IIRC
-1 is implicitly converted to size_t, which is the same as SIZE_MAX (since overflow/underflow for unsigned integers is not UB)Drew McGowen
@VladfromMoscow given that the question is tagged as C, then while a C++ answer is demonstrative, it cannot be used as C code.
Drew McGowen
@VladfromMoscow true, but it's irrelevant to this particular question.
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First its important that the argument list contain size information for the array, i.e. passing a pointer to an array only does not provide enough information to know how many elements the array has. The argument decays into a pointer type with no size information to the function.
So given that, you could do something like this:
int findIndex(int *array, size_t size, int target)
{
int i=0;
while((i<size) && (array[i] != target)) i++;
return (i<size) ? (i) : (-1);
}
For small arrays this approach will be fine. For very large arrays, some sorting and a binary search would improve performance
Comments
Here's my version without a additional variable.
// Return index of element starting
// Return -1 if element is not present
int indexOf(const int elm, const int *ar, int ar_cnt)
{
// decreasing array count till it reaches negative
// arr_cnt - 1 to 0
while (ar_cnt--)
{
// Return array index if current element equals provided element
if (ar[ar_cnt] == elm)
return ar_cnt;
}
// Element not present
return -1; // Should never reaches this point
}
Hope the comments are self explanatory!
