Studying note of GCC-3.4.6 source (10 cont1)

1.6.3. Create node for array type

When invoking build_array_type, the second parameter index_type should be node of index created in previous sector.

 

3790 tree

3791 build_array_type (tree elt_type, tree index_type)                                                 in tree.c

3792 {

3793   tree t;

3794   unsigned int hashcode;

3795

3796   if (TREE_CODE (elt_type) == FUNCTION_TYPE)

3797   {

3798     error ("arrays of functions are not meaningful");

3799     elt_type = integer_type_node;

3800   }

3801

3802   /* Make sure TYPE_POINTER_TO (elt_type) is filled in.  */

3803   build_pointer_type (elt_type);

3804

3805   /* Allocate the array after the pointer type,

3806     in case we free it in type_hash_canon.  */

3807   t = make_node (ARRAY_TYPE);

3808   TREE_TYPE (t) = elt_type;

3809   TYPE_DOMAIN (t) = index_type;

3810

3811   if (index_type == 0)

3812   {

3813     return t;

3814   }

3815

3816   hashcode = TYPE_HASH (elt_type) + TYPE_HASH (index_type);

3817   t = type_hash_canon (hashcode, t);

3818

3819   if (!COMPLETE_TYPE_P (t))

3820     layout_type (t);

3821   return t;

3822 }

 

In C++, array name can be used as the starting address, and array name plus index is the pointer the element. For example:

int a[5];

int *p = a + 3;         // but can’t be int &p = a +3;

So at line 3803, it creates pointer type for the element (at the time of creating array type, the type of element should be ready, but may not for its pointer type).

1.6.3.1.    Layout the type

For array type, the extra information need be recorded includes size of array in bits, size of array in bytes and alignment, tegother with that information of element.

 

1528 void

1529 layout_type (tree type)                                                                              in stor-layout.c

1530 {

1531   if (type == 0)

1532     abort ();

1533

1534   /* Do nothing if type has been laid out before.  */

1535   if (TYPE_SIZE (type))

1536     return;

1537

1538   switch (TREE_CODE (type))

1539   {

        

1619     case ARRAY_TYPE:

1620     {

1621       tree index = TYPE_DOMAIN (type);

1622       tree element = TREE_TYPE (type);

1623

1624       build_pointer_type (element);

1625

1626       /* We need to know both bounds in order to compute the size.  */

1627       if (index && TYPE_MAX_VALUE (index) && TYPE_MIN_VALUE (index)

1628          && TYPE_SIZE (element))

1629       {

1630         tree ub = TYPE_MAX_VALUE (index);

1631         tree lb = TYPE_MIN_VALUE (index);

1632         tree length;

1633         tree element_size;

1634

1635         /* The initial subtraction should happen in the original type so

1636           that (possible) negative values are handled appropriately.  */

1637         length = size_binop (PLUS_EXPR, size_one_node,

1638                         convert (sizetype,

1639                                fold (build (MINUS_EXPR,

1640                                         TREE_TYPE (lb),

1641                                    ub, lb))));

 

C++ allows array definition like “int a[] = {…};”. For such definition, laying out the array type should be deferred to the time of finishing parsing the part of “{…}”. At the same time, as the advent of template, the element can be a template. The size of template usually is unfixed and depends on template argments at instantiation. So compiler will define this size as 0, under which situation, the layout should also be deferred to the time of instantiation.

1.6.3.1.1.            Get array’s boundary

However, even we defer the time of layout; we will still come here eventually. Otherwise, the compiler will give out an error message sooner or later and stop. Since without layout information, it doesn’t know how to create instance of the type in memory.

At line 1637 above, size_binop combines operands arg0 and arg1 with arithmetic operation code. Here the expression to be evaluated is (high boundary – low boundary)+1, it is the number of elements in the array. Notice that size_one_node at line 1637 is the tree node of numeric value 1 of integer mode. It is used as constant 1.

 

1594 tree

1595 size_binop (enum tree_code code, tree arg0, tree arg1)                          in fold-const.c

1596 {

1597   tree type = TREE_TYPE (arg0);

1598

1599   if (TREE_CODE (type) != INTEGER_TYPE || ! TYPE_IS_SIZETYPE (type)

1600       || type != TREE_TYPE (arg1))

1601     abort ();

1602

1603   /* Handle the special case of two integer constants faster.  */

1604   if (TREE_CODE (arg0) == INTEGER_CST && TREE_CODE (arg1) == INTEGER_CST)

1605   {

1606     /* And some specific cases even faster than that.  */

1607     if (code == PLUS_EXPR && integer_zerop (arg0))

1608       return arg1;

1609     else if ((code == MINUS_EXPR || code == PLUS_EXPR)

1610            && integer_zerop (arg1))

1611       return arg0;

1612     else if (code == MULT_EXPR && integer_onep (arg0))

1613       return arg1;

1614

1615     /* Handle general case of two integer constants.  */

1616     return int_const_binop (code, arg0, arg1, 0);

1617   }

1618

1619   if (arg0 == error_mark_node || arg1 == error_mark_node)

1620     return error_mark_node;

1621

1622   return fold (build (code, type, arg0, arg1));

1623 }

 

If arg0 and arg1 are not both integer constant, the expression requires more complex processing, and it is handed to fold at line 1622.

1.6.3.1.1.1.      Operate integer constants

First it checks if there is any operand can be cancelled. Routine integer_zerop returns 1 if argument expr is the integer constant zero or a complex constant of zero.

 

579  int

580  integer_zerop (tree expr)                                                                                    in tree.c

581  {

582    STRIP_NOPS (expr);

583 

584    return ((TREE_CODE (expr) == INTEGER_CST

585            && ! TREE_CONSTANT_OVERFLOW (expr)

586            && TREE_INT_CST_LOW (expr) == 0

587            && TREE_INT_CST_HIGH (expr) == 0)

588           || (TREE_CODE (expr) == COMPLEX_CST

589               && integer_zerop (TREE_REALPART (expr))

590               && integer_zerop (TREE_IMAGPART (expr))));

591  }

 

While STRIP_NOPS, given an expression as a tree, strip any outmost NOP_EXPRs and NON_LVALUE_EXPRs that don't change the machine mode.

 

405  #define STRIP_NOPS(EXP)                                    /                                         in tree.h

406    while ((TREE_CODE (EXP) == NOP_EXPR                      /

407           || TREE_CODE (EXP) == CONVERT_EXPR               /

408           || TREE_CODE (EXP) == NON_LVALUE_EXPR)        /

409         && TREE_OPERAND (EXP, 0) != error_mark_node             /

410         && (TYPE_MODE (TREE_TYPE (EXP))                     /

411              == TYPE_MODE (TREE_TYPE (TREE_OPERAND (EXP, 0))))) /

412      (EXP) = TREE_OPERAND (EXP, 0)

 

NOP_EXPR represents a conversion expected to require no code to be generated. NON_LVALUE_EXPR returns value is same as argument EXP, but guarantees it is not a l-value. See that expressions of above types will not affect the value of their single argument.

Then at line 1612, integer_onep returns 1 if expr is the integer constant 1 or the corresponding complex constant.

 

596  int

597  integer_onep (tree expr)                                                                                    in tree.c

598  {

599    STRIP_NOPS (expr);

600 

601    return ((TREE_CODE (expr) == INTEGER_CST

602            && ! TREE_CONSTANT_OVERFLOW (expr)

603            && TREE_INT_CST_LOW (expr) == 1

604            && TREE_INT_CST_HIGH (expr) == 0)

605           || (TREE_CODE (expr) == COMPLEX_CST

606               && integer_onep (TREE_REALPART (expr))

607               && integer_zerop (TREE_IMAGPART (expr))));

608  }

 

If no operants can be cancelled, it needs int_const_binop at line 1616 does the calcultaion.

 

1185 static tree

1186 int_const_binop (enum tree_code code, tree arg1, tree arg2, int notrunc)        in fold-const.c

1187 {

1188   unsigned HOST_WIDE_INT int1l, int2l;

1189   HOST_WIDE_INT int1h, int2h;

1190   unsigned HOST_WIDE_INT low;

1191   HOST_WIDE_INT hi;

1192   unsigned HOST_WIDE_INT garbagel;

1193   HOST_WIDE_INT garbageh;

1194   tree t;

1195   tree type = TREE_TYPE (arg1);

1196   int uns = TREE_UNSIGNED (type);

1197   int is_sizetype

1198     = (TREE_CODE (type) == INTEGER_TYPE && TYPE_IS_SIZETYPE (type));

1199   int overflow = 0;

1200   int no_overflow = 0;

1201

1202   int1l = TREE_INT_CST_LOW (arg1);

1203   int1h = TREE_INT_CST_HIGH (arg1);

1204   int2l = TREE_INT_CST_LOW (arg2);

1205   int2h = TREE_INT_CST_HIGH (arg2);

1206

1207   switch (code)

1208   {

1209     case BIT_IOR_EXPR:

1210       low = int1l | int2l, hi = int1h | int2h;

1211       break;

1212

1213     case BIT_XOR_EXPR:

1214       low = int1l ^ int2l, hi = int1h ^ int2h;

1215       break;

1216

1217     case BIT_AND_EXPR:

1218       low = int1l & int2l, hi = int1h & int2h;

1219       break;

1220

1221     case RSHIFT_EXPR:

1222       int2l = -int2l;

1223     case LSHIFT_EXPR:

1224       /* It's unclear from the C standard whether shifts can overflow.

1225         The following code ignores overflow; perhaps a C standard

1226         interpretation ruling is needed.  */

1227       lshift_double (int1l, int1h, int2l, TYPE_PRECISION (type),

1228                   &low, &hi, !uns);

1229       no_overflow = 1;

1230       break;

1231

1232     case RROTATE_EXPR:

1233       int2l = - int2l;

1234     case LROTATE_EXPR:

1235       lrotate_double (int1l, int1h, int2l, TYPE_PRECISION (type),

1236                    &low, &hi);

1237       break;

 

For bit operation, it is straight forward. Just handles the low and high part respectively (remember the constant is 128 bits in front-end).

1.6.3.1.1.1.1.              Left shift

For shift operation, sign extension is required for signed number. In lshift_double, paramter ll is the lower part of the number being shifted, and hl is the high part, while count is the bits to shift. Then lv and hv hold the result. And arith if nonzero indicates it is arithmetic shifting; otherwise is logical shift.

 

364  void

365  lshift_double (unsigned HOST_WIDE_INT l1, HOST_WIDE_INT h1,    in fold-const.c

366                HOST_WIDE_INT count, unsigned int prec,

367                unsigned HOST_WIDE_INT *lv, HOST_WIDE_INT *hv, int arith)

368  {

369    unsigned HOST_WIDE_INT signmask;

370 

371    if (count < 0)

372    {

373      rshift_double (l1, h1, -count, prec, lv, hv, arith);

374      return;

375    }

376 

377  #ifdef SHIFT_COUNT_TRUNCATED

378    if (SHIFT_COUNT_TRUNCATED)

379      count %= prec;

380  #endif

381 

382    if (count >= 2 * HOST_BITS_PER_WIDE_INT)

383    {

384      /* Shifting by the host word size is undefined according to the

385        ANSI standard, so we must handle this as a special case.  */

386      *hv = 0;

387      *lv = 0;

388    }

389    else if (count >= HOST_BITS_PER_WIDE_INT)

390    {

391      *hv = l1 << (count - HOST_BITS_PER_WIDE_INT);

392      *lv = 0;

393    }

394    else

395    {

396      *hv = (((unsigned HOST_WIDE_INT) h1 << count)

397              | (l1 >> (HOST_BITS_PER_WIDE_INT - count - 1) >> 1));

398      *lv = l1 << count;

399    }

400 

401    /* Sign extend all bits that are beyond the precision.  */

402 

403    signmask = -((prec > HOST_BITS_PER_WIDE_INT

404                ? ((unsigned HOST_WIDE_INT) *hv

405                   >> (prec - HOST_BITS_PER_WIDE_INT - 1))

406                : (*lv >> (prec - 1))) & 1);

407 

408    if (prec >= 2 * HOST_BITS_PER_WIDE_INT)

409      ;

410    else if (prec >= HOST_BITS_PER_WIDE_INT)

411    {

412      *hv &= ~((HOST_WIDE_INT) (-1) << (prec - HOST_BITS_PER_WIDE_INT));

413      *hv |= signmask << (prec - HOST_BITS_PER_WIDE_INT);

414    }

415    else

416    {

417      *hv = signmask;

418      *lv &= ~((unsigned HOST_WIDE_INT) (-1) << prec);

419      *lv |= signmask << prec;

420    }

421  }

 

Though left shift hasn’t signed extension, as below figure demonstrates, when the precision is less than 2*HOST_WIDE_INT, bits beyond the precision needs be signed extended. If the result of shift overflows, the bits moved out of the precision should be signed extended too.

figure 2 Bits beyond precision is signed extended

 

 

Studying note of GCC-3.4.6 source (10 cont1)_第1张图片

figure 3 Bits moved out of precision is signed extended.

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