User input array size C

Question

Write a program that asks the users to enter a value for array size “n” and fill up the array with n integers. Then reverse the array and print it on the screen. I'm using Visual Studio and have this so far: I am having a problem with "size" in "int arr1[size]". It says it must be a constant value.

#include <stdio.h>
int main(void)
{
int size, i;
printf("Enter the size of the arrays:\n");
scanf("%d", &size);

int arr1[size];
printf("Enter the elements of the array:\n");
for (i = 0; i < size; i++) {
    scanf_s("%d", arr1[size]);
}
printf("The current array is:\n %d", arr1[i]);
}

Answer 1

Some dialects of C have variable-length arrays (VLA), notably C99 (and GCC accepts them....)

But your teacher probably wants you to understand C dynamic memory allocation, and you should also test when scanf fails, so you might code:

int main(void) {
    int size, i;
    printf("Enter the size of the arrays:\n");
    if (scanf("%d", &size)<1) {
      perror("scan size");
      exit(EXIT_FAILURE);
    };

^{Actually, you'll better test against the case of size being negative! I leave that up to you.}

    int *arr1 = calloc(size, sizeof(int));
    if (arr1==NULL) {
       perror("calloc arr1");
       exit(EXIT_FAILURE);
    }

^{You always need to handle failure of calloc or malloc}

 printf("Enter the elements of the array:\n");
 for (i = 0; i < size; i++) {
    if(scanf("%d", &arr1[size])<1) {
      perror("scanf arr1[size]");
      exit(EXIT_FAILURE);
    }
 }

^{scanf_s is only existing in C11, and on a C11 compliant implementation you could use VLAs}

I leave the rest up to you. You want to code a for loop to print every element of the array.

Of course you should call free appropriately (and I am leaving that to you). Otherwise, you have a memory leak. Some systems have valgrind to help hunting them. Don't forget to compile with all warnings & debug info (with GCC use gcc -Wall -g) then use the debugger (e.g. gdb).

Don't forget that a programming language is a specification (written in some document), not a software. You may want to glance into n1570. And you need to understand what undefined behavior means.

Answer 2

You can't use static memory allocation to allocate an array with a dynamic size. You'll need to dynamically request memory to be allocated to your program, by the operating system, of whatever dynamic size the use asks for.

Here's an example to get started:

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

void printArray(int *array, int size) {
    // Sample array: [1, 2, 3, 4, 5]
    printf("["); // [
    for (int i = 0; i < size - 1; i++) { // [1, 2, 3, 4, 
        printf("%i, ", array[i]);
    }
    if (size >= 1) printf("%i", array[size-1]); // [1, 2, 3, 4, 5
    printf("]\n"); // [1, 2, 3, 4, 5]
}

int main(void) {
    int count;
    printf("Enter the size of the array:\n");
    scanf("%d", &count);

    // ask for enough memory to fit `count` elements,
    // each having the size of an `int`
    int *array = malloc(count * sizeof(*array));
    if (!array) {
        printf("There was a problem with malloc.");
        exit(EXIT_FAILURE);
    }

    printf("Enter the elements of the array:\n");
    for (int i = 0; i < count; i++) scanf("%d", &array[i]);

    printArray(array, count);

    //do stuff

    free(array);
}

Answer 3

In C , arrays can not be defined with user defined size in this fashion .

Incorrect way :

int size, i;
printf("Enter the size of the arrays:\n");
scanf("%d", &size);   
int arr1[size];

Correct way :

int size ;
int *arr1 ;
printf("Enter the size of the arrays:\n");
scanf("%d", &size);  
 arr1 = (int *) malloc(sizeof(int) * size);