CallOp not using operands

MLIR
1

Hi,

I want to create a func::CallOp via C++, but even though I call

mlir::func::CallOp::create(*builder_, *location_, callee, *operands_)

with one operand, I get

error: 'func.call' op incorrect number of operands for callee

Here’s the output MLIR

"builtin.module"() <{sym_name = "root"}> ({
  "func.func"() <{function_type = (tensor<f64>) -> tensor<f64>, sym_name = "f1"}> ({
  ^bb0(%arg0: tensor<f64>):
    %2 = "stablehlo.add"(%arg0, %arg0) : (tensor<f64>, tensor<f64>) -> tensor<f64>
    "func.return"(%2) : (tensor<f64>) -> ()
  }) : () -> ()
  "func.func"() <{function_type = () -> tensor<f64>, sym_name = "main"}> ({
    %0 = "stablehlo.constant"() <{value = dense<1.000000e+00> : tensor<f64>}> : () -> tensor<f64>
    %1 = "func.call"() <{callee = @f1}> : () -> tensor<f64>
    "func.return"(%1) : (tensor<f64>) -> ()
  }) : () -> ()
}) : () -> ()

which makes no mention of %0 (or indeed any value) in func.call. Debugging,

printf(“%zu\n”, operands_->getTypes().size());
operands_->getTypes()[0].dump();

produces 1 and tensor<f64>, as expected.

2

I improved debugging, and operands_->size() produces 1, while (*operands_)[0].dump() confirms I’m passing the correct Value.

%0 = "stablehlo.constant"() <{value = dense<1.000000e+00> : tensor<f64>}> : () -> tensor<f64>
3

I resolved it by using a different overload, where I specified the function result types TypeRange. Not sure why that helped, but I suspect it was using the default, empty set for operands.