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fix `a = b = b + c;' (probably d + c, too). Caused by forgetting to emit

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an assignment for `a ='.
This commit is contained in:
Bill Currie 2001-07-12 18:56:43 +00:00
parent 7a06a2254a
commit 8bbef35197
1 changed files with 7 additions and 3 deletions
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10
tools/qfcc/source/expr.c
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@ -822,7 +822,7 @@ emit_function_call (expr_t *e, def_t *dest)
} }
def_t * def_t *
emit_assign_expr (expr_t *e) emit_assign_expr (expr_t *e, def_t *dest)
{ {
def_t *def_a, *def_b; def_t *def_a, *def_b;
opcode_t *op; opcode_t *op;
@ -854,6 +854,10 @@ emit_assign_expr (expr_t *e)
op = PR_Opcode_Find ("=", 5, def_a, def_b, def_b); op = PR_Opcode_Find ("=", 5, def_a, def_b, def_b);
emit_statement (op, def_b, def_a, 0); emit_statement (op, def_b, def_a, 0);
} }
if (dest) {
op = PR_Opcode_Find ("=", 5, dest, def_b, def_b);
emit_statement (op, def_b, dest, 0);
}
} }
return def_b; return def_b;
} }
@ -875,7 +879,7 @@ emit_sub_expr (expr_t *e, def_t *dest)
if (e->e.expr.op == 'c') if (e->e.expr.op == 'c')
return emit_function_call (e, dest); return emit_function_call (e, dest);
if (e->e.expr.op == '=') if (e->e.expr.op == '=')
return emit_assign_expr (e); return emit_assign_expr (e, dest);
def_a = emit_sub_expr (e->e.expr.e1, 0); def_a = emit_sub_expr (e->e.expr.e1, 0);
def_b = emit_sub_expr (e->e.expr.e2, 0); def_b = emit_sub_expr (e->e.expr.e2, 0);
switch (e->e.expr.op) { switch (e->e.expr.op) {
@ -1014,7 +1018,7 @@ emit_expr (expr_t *e)
case ex_expr: case ex_expr:
switch (e->e.expr.op) { switch (e->e.expr.op) {
case '=': case '=':
emit_assign_expr (e); emit_assign_expr (e, 0);
break; break;
case 'n': case 'n':
emit_branch (op_ifnot, e->e.expr.e1, e->e.expr.e2); emit_branch (op_ifnot, e->e.expr.e1, e->e.expr.e2);