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[qfcc] Move alias expr inside call block expression 

This fixes the trampled return value when the first expression aliases
the return result.
This commit is contained in:
Bill Currie 2020-03-26 20:16:52 +09:00
parent 98eac2afbc
commit 4de2c6b30e
1 changed files with 16 additions and 0 deletions
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16
tools/qfcc/source/expr_binary.c
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@ -952,6 +952,11 @@ check_precedence (int op, expr_t *e1, expr_t *e2)
return 0;
}
static int is_call (expr_t *e)
{
return e->type == ex_block && e->e.block.is_call;
}
expr_t *
binary_expr (int op, expr_t *e1, expr_t *e2)
{
@ -964,8 +969,19 @@ binary_expr (int op, expr_t *e1, expr_t *e2)
e1 = convert_vector (e1);
// FIXME this is target-specific info and should not be in the
// expression tree
if ((e1->type == ex_expr || e1->type == ex_uexpr) && e1->e.expr.op == 'A'
&& is_call (e1->e.expr.e1)) {
// move the alias expression inside the block so the following check
// can detect the call and move the temp assignment into the block
expr_t *block = e1->e.expr.e1;
e1->e.expr.e1 = block->e.block.result;
block->e.block.result = e1;
e1 = block;
}
if (e1->type == ex_block && e1->e.block.is_call
&& has_function_call (e2) && e1->e.block.result) {
// the temp assignment needs to be insided the block so assignment
// code generation doesn't see it when applying right-associativity
expr_t *tmp = new_temp_def_expr (get_type (e1->e.block.result));
e = assign_expr (tmp, e1->e.block.result);
append_expr (e1, e);