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How to pass an array by value in C ?

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In C, array name represents address and when we pass an array, we actually pass address and the parameter receiving function always accepts them as pointers (even if we use [], refer this for details).

How to pass array by value, i.e., how to make sure that we have a new copy of array when we pass it to function?
This can be done by wrapping the array in a structure and creating a variable of type of that structure and assigning values to that array. After that, passing the variable to some other function and modifying it as per requirements. Note that array members are copied when passed as parameter, but dynamic arrays are not. So this solution works only for non-dynamic arrays (created without new or malloc).

Let’s see an example to demonstrate the above fact using a C program:

// C program to demonstrate passing an array
// by value using structures.
# define SIZE 5
// A wrapper for array to make sure that array
// is passed by value.
struct ArrayWrapper
    int arr[SIZE];
// An array is passed by value wrapped in temp
void modify(struct ArrayWrapper temp)
    int *ptr = temp.arr;
    int i;
    // Display array contents
    printf("In 'modify()', before modification\n");
    for (i = 0; i < SIZE; ++i)
        printf("%d ", ptr[i]);
    // Modify the array
    for (i = 0; i < SIZE; ++i)
        ptr[i] = 100; // OR *(ptr + i)
    printf("\nIn 'modify()', after modification\n");
    for (i = 0; i < SIZE; ++i)
        printf("%d ", ptr[i]); // OR *(ptr + i)
// Driver code
int main()
    int i;
    struct ArrayWrapper obj;
    for (i=0; i<SIZE; i++)
        obj.arr[i] = 10;
    // Display array contents
    printf("\n\nIn 'Main', after calling modify() \n");
    for (i = 0; i < SIZE; ++i)
        printf("%d ", obj.arr[i]); // Not changed
    return 0;


In 'modify()', before modification
10 10 10 10 10 

In 'modify()', after modification
100 100 100 100 100 

In 'Main', after calling modify() 
10 10 10 10 10 


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Last Updated : 21 Dec, 2018
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