I have an array int arr[5] = {10, 2, 3, 5, 1}, and I want to pass in the last 4 elements (basically from index 1 to index4) into an argument as an array (so: [2, 3, 5, 1]). Is there a way to do this very simply (like how in Ruby you would be able to do arr[1..4]), or would I have to use a for loop?
4 Answers
You can manually increment the pointer by 1:
your_function(arr + 1)
Pointer arithmetic in C implicitly accounts for the size of the elements, so adding 1 will actually add 1 * sizeof(int)
For a closer analogue to array slicing from other languages, try this function:
int *copy_array_slice(int *array, int start, int end) {
int numElements = (end - start + 1);
int numBytes = sizeof(int) * numElements;
int *slice = malloc(numBytes);
memcpy(slice, array + start, numBytes);
return slice;
}
It makes a slice of the array between the given start and end indices. Remember to free() the slice once you're done with it!
Comments
Answer
Given you current code:
int arr[5] = {10, 2, 3, 5, 1};
You can duplicate the range 1..4 by:
int arr_dup[4];
memcpy(arr_dup,arr+1,sizeof(int)*4);
Remember that your function definition should be a pointer, example:
void a_function(int *arr_arg); //Call via a_function(arr_dup);
Explanation
Arrays in c implemented as pointers (aka variables that hold memory addresses).
If you do arithmetic on the pointer, it will advance to the respective element. Example:
ptr + 1 //Next Element
ptr - 1 //Previous Element
Comments
#include <stdio.h>
#include <stdlib.h>
void my_function(int arr_size, int *arr)
{
int i;
for(i=0; i < arr_size; i++)
{
printf("[%d]:%d\n", i, arr[i]);
}
}
int main(int argc, char **argv)
{
int arr[] = { 10, 2, 3, 5, 1 };
(void)my_function(4, &arr[1]); /* TODO > use more flexible indexing */
return EXIT_SUCCESS;
}
4 Comments
Keith Thompson
Why are you using old-style pre-ANSI function definitions? Use prototypes;
void my_function(int arr_size, int *arr), int main(int argc, char **argv)
bluesawdust
Explain why there is an appreciable difference. No compiler nor standard was mentioned, so I picked a historical style. I work in older code bases and have come to like this style of function definition due to the ease of documenting parameters. Additionally, why would I use prototypes for something so simple?
Keith Thompson
Old-style function declarations and definitions have been obsolescent since 1989. Prototypes were added to the language because old-style declarations do not let the compiler check the correctness of calls. You could write
my_function("foo", 12.3, 42, NULL) and the compiler wouldn't complain (but it would probably blow up at run time).bluesawdust
I hadn't tried calling my functions incorrectly, heh. Updated the answer appropriately.
I think you can use memcpy,memcpy can copy data byte to byte.In memmory,our data is binary,so even int is 4 bytes,we can copy it byte to byte.
int dst[4];
memcpy(dst,&arr[1],size(int) * 4);
&arr[1]does the job — as long as you don't want to pass a copy (the function isn't going to modify the array, or it doesn't matter if it does). If you want to copy, you have to do the copying manually.