Saving some variables in an array using for loop

[cppfighter]

I want to save some variables in an an array(a) using for loop. here those variables are even numbers.

I used below code using malloc function in stdlib.h

#include<stdio.h>
#include<stdlib.h>

int main(){
    int *a; //single dimension array
    int i;
    for(i=0;i<=20;i++){
        if(i%2==0){
        a=malloc(i * sizeof(int));
        printf("%d\n",i);
        }
    }
    return 0;
    
}

But its returning this error

error: invalid conversion from 'void*' to 'int*' [-fpermissive]

[sheel]

malloc function is of type void* which returns no data. you need to cast it to int* first

a = (int*) malloc(i*sizeof(int));

Refer http://www.cplusplus.com/reference/cstdlib/malloc/

[cppfighter]

It dosent seems to work

#include<stdio.h>
#include<stdlib.h>

int main(){
    int *a; //single dimension array
    int i;
    for(i=0;i<=20;i++){
        if(i%2==0){
        a = (int*) malloc(i*sizeof(int));
        printf("%d\n",i);
        }
    }
    
     printf("%d\n",a[2]);
    
    return 0;
    
}

output I expected is 4 in the last but it give zero

0
2
4
6
8
10
12
14
16
18
20
0

[LA48b7Xj]

Define an array instead of a pointer, and don't use malloc. I modified your program a little. You were creating a new array in dynamic memory every other loop iteration and not deleting them.

#include<cstdio>

using namespace std;

int main(){
    int a[11];
    int top = 0;

    for(int i=0;i<=20;i+=2)
        a[top++] = i;

    for(auto e : a)
        printf("%d\n",e);
}

[kamal choudhary]

Yes as Krako said do not use malloc function. You can easily store even integers in an array using for loop. Krako has shown the code.

[cppfighter]

@krako @kamal this works when array size is known but what if the size of the array is unknown. what should then we do ?