I tried this following code but it clearly didn't work:

    void main3()
{
    int n;
    printf("Input the dimension of the array");
    scanf("%d", &n);
    int* testMatrix2 = malloc(sizeof(int) * n);

    testMatrix2 = {512, 51, 642}; //is there a way to do this? "expected an expression" error
}

As the comment shows, is there a way to initialize an array as you would normally do without using arrays? I know it's a risky move, but it would make my life easier during testing.

The use of for cycles would not be feasible for me as I need a sort of randomness in the array.

CodePudding user response：

As @EugeneSh. stated it is not possible for any size array. If the array has a fixed size you can wrap it into the structure:

typedef struct
{
    int x[10];
}wrappedArray;

void foo(void)
{
    wrappedArray *wa = malloc(sizeof(*wa));

    *wa = (wrappedArray){1,2,3,4,5,6,7,8,9,10};
}

CodePudding user response：

Arrays are "non-modifiable" L-values. Therefore they cannot be assigned with = operator.

However, you could memcpy from a compound literal.

memcpy(testMatrix2, (const int[]){512, 51, 642}, sizeof(int[3]));

Just ensure that n >= 3 before doing this to avoid Undefined Behavior.

CodePudding user response：

You can use compound literal:

#include <stdio.h>

int main (void) {
    int* testMatrix2 = (int []){512, 51, 642};

    for (int i = 0; i < 3;   i) {
        printf ("%d ", testMatrix2[i]);
    }
    printf ("\n");
    return 0;
}

Output:

# ./a.out 
512 51 642 

Compound literals Constructs an unnamed object of specified type in-place.
In this statement of above program

int* testMatrix2 = (int []){512, 51, 642};

(int []){512, 51, 642} is a compound literal. It will create an unnamed static array of type int [3], initialise the array to the value 512, 51 and 642 and the pointer to first element of this array object will be assigned to testMatrix2.