non_constant_relational_pattern_expression

The relational pattern expression must be a constant.

The analyzer produces this diagnostic when the value in a relational pattern expression isn't a constant expression.

The following code produces this diagnostic because the operand of the > operator, a, isn't a constant:

final a = 0;

void f(int x) {
  if (x case > a) {}
}

Replace the value with a constant expression:

const a = 0;

void f(int x) {
  if (x case > a) {}
}