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glsl: Copy function out to temp if we don't directly ref a variable
Otherwise we can end up with IR that looks like this:
(
(declare (temporary ) vec4 f@8)
(assign (xyzw) (var_ref f@8) (var_ref f) )
(call f16 ((swiz y (var_ref f@8) )))
(assign (xyzw) (var_ref f) (var_ref f@8) )
))
When we really need:
(declare (temporary ) float inout_tmp)
(assign (x) (var_ref inout_tmp) (swiz y (var_ref f) ))
(call f16 ((var_ref inout_tmp) ))
(assign (y) (var_ref f) (swiz y (swiz xxxx (var_ref inout_tmp) )))
(declare (temporary ) void void_var)
The GLSL IR function inlining code seemed to produce correct code
even without this but we need the correct IR for GLSL IR -> NIR to
be able to understand whats going on.
Reviewed-by: Kenneth Graunke <kenneth@whitecape.org>
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parent
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commit
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1 changed files with 3 additions and 2 deletions
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@ -402,7 +402,8 @@ fix_parameter(void *mem_ctx, ir_rvalue *actual, const glsl_type *formal_type,
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* nothing needs to be done to fix the parameter.
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*/
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if (formal_type == actual->type
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&& (expr == NULL || expr->operation != ir_binop_vector_extract))
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&& (expr == NULL || expr->operation != ir_binop_vector_extract)
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&& actual->as_dereference_variable())
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return;
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/* An array index could also be an out variable so we need to make a copy
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@ -456,7 +457,7 @@ fix_parameter(void *mem_ctx, ir_rvalue *actual, const glsl_type *formal_type,
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ir_dereference_variable *const deref_tmp_1 =
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new(mem_ctx) ir_dereference_variable(tmp);
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ir_assignment *const assignment =
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new(mem_ctx) ir_assignment(deref_tmp_1, actual);
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new(mem_ctx) ir_assignment(deref_tmp_1, actual->clone(mem_ctx, NULL));
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before_instructions->push_tail(assignment);
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}
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