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5f56a0388

Denys Vlasenko
Denys Vlasenko
2023-06-15
SUCCESS Duration: 154s

Subject

shell/math: fix parsing of ?: and explain why it's parsed that way

Binary

No binary (build failed)

Architecture & Compiler Flags

Architecture: x86_64-linux-gnu Compiler: gcc (Ubuntu 11.4.0-1ubuntu1~22.04.3) 11.4.0

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Build Log

(log yok)

Commit Diff GitHub →

shell/ash_test/ash-arith/arith-ternary_nested1.right +1 -0shell/ash_test/ash-arith/arith-ternary_nested1.tests +2 -0shell/ash_test/ash-arith/arith.right +1 -1shell/hush_test/hush-arith/arith-precedence1.right +4 -0shell/hush_test/hush-arith/arith-precedence1.tests +15 -0shell/hush_test/hush-arith/arith-ternary2.right +0 -3shell/hush_test/hush-arith/arith-ternary2.tests +0 -7shell/hush_test/hush-arith/arith-ternary_nested1.right +1 -0shell/hush_test/hush-arith/arith-ternary_nested1.tests +2 -0shell/hush_test/hush-arith/arith.right +1 -1shell/math.c +34 -14
shell/ash_test/ash-arith/arith-ternary_nested1.right+1 -0
@@ -0,0 +1 @@
+3:3
shell/ash_test/ash-arith/arith-ternary_nested1.tests+2 -0
@@ -0,0 +1,2 @@
+exec 2>&1
+echo 3:$((1?(2?(3):4):5))
shell/ash_test/ash-arith/arith.right+1 -1
@@ -92,7 +92,7 @@ ghi
./arith.tests: line 190: arithmetic syntax error
16 16
./arith.tests: line 195: arithmetic syntax error
-./arith.tests: line 196: malformed ?: operator
+./arith.tests: line 196: arithmetic syntax error
./arith.tests: line 197: arithmetic syntax error
9 9
./arith.tests: line 204: arithmetic syntax error
shell/hush_test/hush-arith/arith-precedence1.right+4 -0
@@ -0,0 +1,4 @@
+4:4
+4:4
+4:4
+4:4
shell/hush_test/hush-arith/arith-precedence1.tests+15 -0
@@ -0,0 +1,15 @@
+exec 2>&1
+# bash documentation says that precedence order is:
+# ...
+# expr ? expr1 : expr2
+# = *= /= %= += -= <<= >>= &= ^= |=
+# exprA , exprB
+# but in practice, the rules for expr1 and expr2 are different:
+# assignments and commas in expr1 have higher precedence than :?,
+# but in expr2 they haven't:
+# "v ? 1,2 : 3,4" is parsed as "(v ? (1,2) : 3),4"
+# "v ? a=2 : b=4" is parsed as "(v ? (a=1) : b)=4" (thus, this is a syntax error)
+echo 4:$((0 ? 1,2 : 3,4))
+echo 4:$((1 ? 1,2 : 3,4))
+echo 4:"$((0 ? 1,2 : 3,4))"
+echo 4:"$((1 ? 1,2 : 3,4))"
shell/hush_test/hush-arith/arith-ternary2.right+0 -3
@@ -1,3 +0,0 @@
-6:6
-a=b=+err+
-b=6
shell/hush_test/hush-arith/arith-ternary2.tests+0 -7
@@ -1,7 +0,0 @@
-exec 2>&1
-a='b=+err+'
-b=5
-# The not-taken branch should not parse variables
-echo 6:$((0 ? a : ++b))
-echo "a=$a"
-echo "b=$b"
shell/hush_test/hush-arith/arith-ternary_nested1.right+1 -0
@@ -0,0 +1 @@
+3:3
shell/hush_test/hush-arith/arith-ternary_nested1.tests+2 -0
@@ -0,0 +1,2 @@
+exec 2>&1
+echo 3:$((1?(2?(3):4):5))
shell/hush_test/hush-arith/arith.right+1 -1
@@ -94,7 +94,7 @@ ghi
hush: arithmetic syntax error
16 16
hush: arithmetic syntax error
-hush: malformed ?: operator
+hush: arithmetic syntax error
hush: arithmetic syntax error
9 9
hush: arithmetic syntax error
shell/math.c+34 -14
@@ -157,17 +157,17 @@ typedef unsigned char operator;
#define fix_assignment_prec(prec) do { if (prec == 3) prec = 2; } while (0)
/* Ternary conditional operator is right associative too */
-// FIXME:
-// bash documentation says that precedence order is:
-// ...
-// expr ? expr1 : expr2
-// = *= /= %= += -= <<= >>= &= ^= |=
-// exprA , exprB
-// but in practice, the rules for expr1 and expr2 are different:
-// assignments and commas in expr1 have higher precedence than ?:,
-// but in expr2 they haven't:
-// "v ? 1,2 : 3,4" is parsed as "(v ? (1,2) : 3),4"
-// "v ? a=2 : b=4" is parsed as "(v ? (a=1) : b)=4" (thus, this is a syntax error)
+/*
+ * bash documentation says that precedence order is:
+ * ...
+ * expr ? expr1 : expr2
+ * = *= /= %= += -= <<= >>= &= ^= |=
+ * exprA , exprB
+ * What it omits is that expr1 is parsed as if parenthesized
+ * (this matches the rules of ?: in C language):
+ * "v ? 1,2 : 3,4" is parsed as "(v ? (1,2) : 3),4"
+ * "v ? a=2 : b=4" is parsed as "(v ? (a=1) : b)=4" (thus, this is a syntax error)
+ */
#define TOK_CONDITIONAL tok_decl(4,0)
#define TOK_CONDITIONAL_SEP tok_decl(4,1)
@@ -629,6 +629,7 @@ evaluate_string(arith_state_t *math_state, const char *expr)
/* Stack of operator tokens */
operator *const opstack = alloca(expr_len * sizeof(opstack[0]));
operator *opstackptr = opstack;
+ operator insert_op = 0xff;
/* Start with a left paren */
dbg("(%d) op:TOK_LPAREN", (int)(opstackptr - opstack));
@@ -751,11 +752,24 @@ evaluate_string(arith_state_t *math_state, const char *expr)
goto err;
}
}
+ /* NB: expr now points past the operator */
tok_found:
op = p[1]; /* fetch TOK_foo value */
- tok_found1:
- /* NB: expr now points past the operator */
+ /* Special rule for "? EXPR :"
+ * "EXPR in the middle of ? : is parsed as if parenthesized"
+ * (this quirk originates in C grammar, I think).
+ */
+ if (op == TOK_CONDITIONAL) {
+ insert_op = TOK_LPAREN;
+ dbg("insert_op=%02x", insert_op);
+ }
+ if (op == TOK_CONDITIONAL_SEP) {
+ insert_op = op;
+ op = TOK_RPAREN;
+ dbg("insert_op=%02x op=%02x", insert_op, op);
+ }
+ tok_found1:
/* post grammar: a++ reduce to num */
if (lasttok == TOK_POST_INC || lasttok == TOK_POST_DEC)
lasttok = TOK_NUM;
@@ -865,9 +879,15 @@ dbg(" numstack:%d val:%lld '%s'", (int)(numstackptr - numstack), numstackptr[
/* else: LPAREN or UNARY: push it on opstack */
push_op:
/* Push this operator to opstack */
- dbg("(%d) op:%02x", (int)(opstackptr - opstack), op);
+ dbg("(%d) op:%02x insert_op:%02x", (int)(opstackptr - opstack), op, insert_op);
*opstackptr++ = lasttok = op;
next: ;
+ if (insert_op != 0xff) {
+ op = insert_op;
+ insert_op = 0xff;
+ dbg("inserting %02x", op);
+ goto tok_found1;
+ }
} /* while (1) */
err: