Foreach function in C - c

Question

I want to create a function that takes a array and a function and calls that function on each element in the array
I've looked around for a solution but they all seem to use macros whereas I would prefer a function if at all possible.
I'm looking for something that would work something like the following
void print_string()
{
printf("%d\n", num);
}
int array[] = { 1, 2, 3, 4, NULL }; // So the foreach function knows when it has reached the end
for_each_function(array, print_number);
Outputting:
1
2
3
4
Edit: It needs to be generic so it will probably require a macro

Answer 1

You should probably use function pointers. One complete implementation using function pointers is like this.
#include <stdio.h>
void for_each(int * arr, int size, void (*fun)(int))
{
for (int i = 0; i < size; i++)
{
(*fun)(arr[i]);
}
}
void print_num(int num)
{
printf("%d\n", num);
}
int main()
{
int array [] = {1, 2, 3, 4};
for_each(array, 4, print_num);
return 0;
}

Answer 2

Use a function pointer, like this:
#include <stdio.h>
void for_each_function(int* arr, size_t size, void(*fun)(int)) {
for(size_t i = 0; i < size; ++i)
(*fun)(arr[i]);
}
void print_number(int);
int main(void) {
int array[] = { 1, 2, 3, 4 };
for_each_function(array, 4, &print_number);
return 0;
}
void print_number(int num)
{
printf("%d\n", num);
}
Output:
1
2
3
4
Remarks:
The third parameter of for_each_function, namely void(*fun)(int),
is a function pointer named fun, with a return type void, and a
parameter of type int.
Inside the for each body, you call the function pointer like this
(*fun)(arr[i]);, which dereferences the function pointer fun, and
would pass arr[i] as its parameter. You could also use fun(arr[i]).
In the call of the for each function in main function, you need to
use the address of the function you want to point to, like this
for_each_function(array, 4, &print_number);. You could also use for_each_function(array, 4, print_number);.
PS: In your example array, I don't think you wanted to use NULL as the last element (you would probably get a warning like warning: initialization of 'int' from 'void *' makes integer from pointer without a cast [-Wint-conversion], unless NULL was defined as 0, as commented by #EricPostpischil), so I got rid of it in my example.

Answer 3

A modification of the answer of davidlowryduda / gsamaras to be compatible with any type of array, not just int, however that impacts the called function which gets a pointer rather than a value :
#include <stdio.h>
void for_each(void * arr, size_t nElts, size_t sizeElt, void (*fun)(void *))
{
char * p = arr;
for (size_t i = 0; i < nElts; i++)
{
(*fun)(p);
p += sizeElt;
}
}
/* for_each can be called directly or via the following macro,
the array can be an expression without any problem : is it given
in argument and the other uses are only for sizeof */
#define FOR_EACH(a, f) for_each(a, sizeof(a)/sizeof(*a), sizeof(*a), f)
void print_int(void * p)
{
printf("%d\n", *((int *) p));
}
void print_short(void * p)
{
printf("%d\n", *((short *) p));
}
int main()
{
int iarray [] = {1, 2, 3, 4};
FOR_EACH(iarray, print_int); /* for_each(iarray, sizeof(iarray)/sizeof(*iarray), sizeof(*iarray), print_int); */
short sarray [] = {5, 6, 7, 8};
FOR_EACH(sarray, print_short); /* for_each(sarray, sizeof(sarray)/sizeof(*sarray), sizeof(*sarray), print_short); */
return 0;
}
Compilation and execution:
pi#raspberrypi:~ $ gcc -pedantic -Wall -Wextra i.c
pi#raspberrypi:~ $ ./a.out
1
2
3
4
5
6
7
8

Answer 4

You can define a macro for this.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define FOR_EACH(type, target, size, body) \
\
for (size_t i = 0; i < size; i++) \
{ \
type value = target[i]; \
body; \
}
int main(int argc, char const *argv[])
{
int arr[4] = {0, 1, 2, 3};
char *name = "My Name";
FOR_EACH(int, arr, 4, {
printf("%d\n", value);
})
size_t len = strlen(name);
FOR_EACH(char, name, len, {
printf("%c-", value);
})
return 0;
}
In this case you don't even need a function and it can be used to iterate over any array of any type as long as you know its length.

Answer 5

You could do something like that using function pointers. But using macros would be very more powerful, it will allow your foreach to be generic.
#include <stdio.h>
typedef struct int_array{
int *data;
int len;
}Int_array;
void print_number(int num)
{
printf("%d\n", num);
}
void for_each_function(Int_array a, void (*f)(int)){
for(int i =0; i < a.len;i++){
(*f)(a.data[i]);
}
}
int main(){
int data[4] = { 1, 2, 3, 4 };
Int_array x;
x.len = 4;
x.data = data;
for_each_function(x, print_number);
}

Answer 6

I took a solution's friend on forum and I increased.
I got the idea that foreach doesn't have to have a value to stop the loop.
#include <iostream>
#include <stdio.h>
void for_each(int * arr, void (*fun)(int)){
for (int i = 0; i < arr[i]; i++){
(*fun)(arr[i]);
}
}
void print_num(int num){
printf("%d\n", num);
}
int main(){
int array [] = {1, 2, 3, 4, 5, 6};
for_each(array, print_num);
}