Why does array with variable-sized work but not dynamic array?

Question

First of all, I am new to programming and don't know what "variable-sized array" and "dynamic array" mean. I am just using those words to describe my situation. Feel free to edit with correct nomenclature. Now to the question, I have following pieces of code:-

...
int n;
printf("Enter number of rows/columns = ");
scanf(" %d",n);
int a[n][n];
...

and

...
int n;
n=6;
int a[n][n];
...
n=7;
...
n=4;
...

Both of them are compiled successfully.

• In first case, upon execution program crashes after scanf() function.
• Second case works perfectly.

Why?

Answer 1

Use of & in scanf()

When you scan for user input using scanf from <stdio.h>, you need to pass a pointer to it. You can read more about why that's the case here: Use of & in scanf() but not in printf()

int n;
printf("Integer: ");
scanf("%d", &n);
int a[n][n];

Variable-length array

A variable-length array is just an array whose length is determined at run time instead of at compile time. In your case the length is determined by the variable n which makes it a variable-length array. And as Eugene Sh. mentioned in the comments, changing the variable n doesn't modify the array's length after defining it.

Answer 2

In the both code snippets there are used variable length arrays because in their declarations there are not used integer constant expressions that specify the number of elements in the arrays.

There is no dynamically allocated array.

The first code snippet has a bug in the statement with the call of scanf.

scanf(" %d",n);

The argument shall be a pointer to the variable n.

scanf(" %d", &n);

That is to change the variable n by means of the function scanf you have to pass it to the function by reference.

In C passing by reference means passing an object indirectly through a pointer to it. So dereferencing the pointer the function will have a direct access to the object that must be modified.

Pay attention to that in the second code snippet changing the value of the variable n after the array declaration does not change the number of elements in the array a.

int n;
n=6;
int a[n][n];
...
n=7;
...
n=4;

The array a will have exactly 6 elements according to the value stored in the variable n before the array declaration.

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    size_t n;
    
    do
    {
        printf( "Enter number of rows/columns (greater than 0 ) = " );
        
    } while ( !( scanf( "%zu", &n ) == 1 && n != 0 ) );
    
    int a[n][n];
    
    printf( "The size of the array declared like int a[%zu][%zu] is %zu.\n", 
            n, n, sizeof( a ) );
    
    return 0;
}

The program output might look like

Enter number of rows/columns (greater than 0 ) = 10
The size of the array declared like int a[10][10] is 400.

A variable length array shall have automatic storage duration (that is it may not be declared for example in a file scope or have the storage specifier static; it may be declared in a block scope).

From the C Standard (6.7.6.2 Array declarators)

4 If the size is not present, the array type is an incomplete type. If the size is * instead of being an expression, the array type is a variable length array type of unspecified size, which can only be used in declarations or type names with function prototype scope; such arrays are nonetheless complete types. If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type; otherwise, the array type is a variable length array type. (Variable length arrays are a conditional feature that implementations need not support; see 6.10.8.3.)