why this code can run ? double a[3]; a[1,1]=1;

Question

int main()
{
    double a[3];  
    a[1,1]=1;
}

It passes the vs2013 compiler, and it is not 2D array.

Answer 1

In the expression inside square brackets there is so-called comma operator.

a[1,1]=1;

Its value is the value of the last subexpression.

So this statement is equivalent to

a[1]=1;

This syntax as

a[1,1]=1;

is also valid in C# but it sets an element of a two-dimensional array.

In C/C++ each index of a multidimensional array shall be enclosed in separate square brackets.

Here is a more interesting example with the comma operator

int main()
{
    double a[3];  
    size_t i = 0;  
    a[i++, i++]=1;
}

It is also equivalent to

    a[1]=1;

Answer 2

a[1,1]=1;

is equivalent to:

a[1]=1;

The expression 1,1 evaluates to 1, because the first 1 is discarded and only the second 1 is evaluated. Read up on the comma operator for more info.

Answer 3

You are invoking the comma operator. This evaluates its first operand, discards it, and returns the second one. So your code is the equivalent of

a[1] = 1;

The syntax for accessing an element of a 2D array would be

b[1][2] = 42;