Multiple assignment

[simendsjo]

On 26.02.2011 01:56, bearophile wrote:
> Is this program showing a bug in multiple assignments (DMD 2.052)?
>
>
> void main() {
>      int i;
>      int[2] x;
>      i, x[i] = 1;
>      assert(x == [1, 0]); // OK
>
>      int j;
>      int[2] y;
>      y[j], j = 1;
>      assert(y == [0, 0]); // Not OK
> }
>
>
> At the end of the program I expect y to be [1,0] instead of [0,0].
>
> Yet this C program with GCC:
>
> #include "stdio.h"
> int main() {
>      int i = 0;
>      int x[2] = {0, 0};
>      i, x[i] = 1;
>      printf("%d %d\n", x[0], x[1]);
>
>      int j = 0;
>      int y[2] = {0, 0};
>      y[j], j = 1;
>      printf("%d %d\n", y[0], y[1]);
>
>      return 0;
> }
>
>
> has the same output as DMD:
> 1 0
> 0 0
>
> Bye,
> bearophile

I couldn't find any info on the comma expression in the language 
reference, but this was my first google hit:
"""
A comma expression contains two operands of any type separated by a 
comma and has *left-to-right* associativity. The left operand is fully 
evaluated, possibly producing side effects, and its value, if there is 
one, is *discarded*. The right operand is then evaluated. The type and 
value of the result of a comma expression are those of its right 
operand, after the usual unary conversions
"""