c10.c

unix v6 的c compiler 很老了

#
/*

	    	C compiler, part 2


*/

#include "c1h.c"

char	maprel[] {	EQUAL, NEQUAL, GREATEQ, GREAT, LESSEQ,
			LESS, GREATQP, GREATP, LESSEQP, LESSP
};

char	notrel[] {	NEQUAL, EQUAL, GREAT, GREATEQ, LESS,
			LESSEQ, GREATP, GREATQP, LESSP, LESSEQP
};

struct tconst czero { CON, INT, 0, 0};
struct tconst cone  { CON, INT, 0, 1};
struct tconst fczero { SFCON, DOUBLE, 0, 0 };

struct	table	*cregtab;

int	nreg	3;
int	isn	10000;
int	namsiz	8;

main(argc, argv)
char *argv[];
{
	extern fout;

	if (argc<4) {
		error("Arg count");
		exit(1);
	}
	if(fopen(argv[1], ascbuf)<0) {
		error("Missing temp file");
		exit(1);
	}
	if ((fout = creat(argv[3], 0666)) < 0) {
		error("Can't create %s", argv[3]);
		exit(1);
	}
	spacep = treespace;
	getree();
	/*
	 * If any floating-point instructions
	 * were used, generate a reference which
	 * pulls in the floating-point part of printf.
	 */
	if (nfloat)
		printf(".globl	fltused\n");
	/*
	 * tack on the string file.
	 */
	close(ascbuf[0]);
	if (fopen(argv[2], ascbuf)<0) {
		error("Missing temp file");
		exit(1);
	}
	printf(".globl\n.data\n");
	getree();
	flush();
	exit(nerror!=0);
}

/*
 * Given a tree, a code table, and a
 * count of available registers, find the code table
 * for the appropriate operator such that the operands
 * are of the right type and the number of registers
 * required is not too large.
 * Return a ptr to the table entry or 0 if none found.
 */
char *match(atree, table, nrleft)
struct tnode *atree;
struct table *table;
{
	int op, d1, d2, t1, t2, dope;
	struct tnode *p2;
	register struct tnode *p1, *tree;
	register struct optab *opt;

	if ((tree=atree)==0)
		return(0);
	if (table==lsptab)
		table = sptab;
	if ((op = tree->op)==0)
		return(0);
	dope = opdope[op];
	if ((dope&LEAF) == 0)
		p1 = tree->tr1;
	else
		p1 = tree;
	t1 = p1->type;
	d1 = dcalc(p1, nrleft);
	if ((dope&BINARY)!=0) {
		p2 = tree->tr2;
		/*
		 * If a subtree starts off with a conversion operator,
		 * try for a match with the conversion eliminated.
		 * E.g. int = double can be done without generating
		 * the converted int in a register by
		 * movf double,fr0; movfi fr0,int .
		 */
		if(opdope[p1->op]&CNVRT && (opdope[p2->op]&CNVRT)==0) {
			tree->tr1 = p1->tr1;
			if (opt = match(tree, table, nrleft))
				return(opt);
			tree->tr1 = p1;
		} else if (opdope[p2->op]&CNVRT && (opdope[p1->op]&CNVRT)==0) {
			tree->tr2 = p2->tr1;
			if (opt = match(tree, table, nrleft))
				return(opt);
			tree->tr2 = p2;
		}
		t2 = p2->type;
		d2 = dcalc(p2, nrleft);
	}
	for (; table->op!=op; table++)
		if (table->op==0)
			return(0);
	for (opt = table->tabp; opt->tabdeg1!=0; opt++) {
		if (d1 > (opt->tabdeg1&077)
		 || (opt->tabdeg1 >= 0100 && (p1->op != STAR)))
			continue;
		if (notcompat(p1, opt->tabtyp1, op)) {
			continue;
		}
		if ((opdope[op]&BINARY)!=0 && p2!=0) {
			if (d2 > (opt->tabdeg2&077)
			 || (opt->tabdeg2 >= 0100) && (p2->op != STAR) )
				continue;
			if (notcompat(p2,opt->tabtyp2, 0))
				continue;
		}
		return(opt);
	}
	return(0);
}

/*
 * Given a tree, a code table, and a register,
 * produce code to evaluate the tree with the appropriate table.
 * Registers reg and upcan be used.
 * If there is a value, it is desired that it appear in reg.
 * The routine returns the register in which the value actually appears.
 * This routine must work or there is an error.
 * If the table called for is cctab, sptab, or efftab,
 * and tree can't be done using the called-for table,
 * another try is made.
 * If the tree can't be compiled using cctab, regtab is
 * used and a "tst" instruction is produced.
 * If the tree can't be compiled using sptab,
 * regtab is used and the register is pushed on the stack.
 * If the tree can't be compiled using efftab,
 * just use regtab.
 * Regtab must succeed or an "op not found" error results.
 *
 * A number of special cases are recognized, and
 * there is an interaction with the optimizer routines.
 */
rcexpr(atree, atable, reg)
struct tnode *atree;
struct table *atable;
{
	register r;
	int modf, nargs, recurf;
	register struct tnode *tree;
	register struct table *table;

	table = atable;
	recurf = 0;
	if (reg<0) {
		recurf++;
		reg = ~reg;
		if (reg>=020) {
			reg =- 020;
			recurf++;
		}
	}
	if((tree=atree)==0)
		return(0);
	switch (tree->op)  {

	/*
	 * A conditional branch
	 */
	case CBRANCH:
		cbranch(optim(tree->btree), tree->lbl, tree->cond, 0);
		return(0);

	/*
	 * An initializing expression
	 */
	case INIT:
		if (tree->tr1->op == AMPER)
			tree->tr1 = tree->tr1->tr1;
		if (tree->tr1->op==NAME)
			pname(tree->tr1);
		else if (tree->tr1==CON)
			psoct(tree->tr1->value);
		else
			error("Illegal initialization");
		putchar('\n');
		return(0);

	/*
	 * Put the value of an expression in r0,
	 * for a switch or a return
	 */
	case RFORCE:
		if((r=rcexpr(tree->tr1, regtab, reg)) != 0)
			printf("mov%c	r%d,r0\n", isfloat(tree->tr1), r);
		return(0);

	/*
	 * sequential execution
	 */
	case COMMA:
		rcexpr(tree->tr1, efftab, reg);
		atree = tree = tree->tr2;
		break;

	/*
	 * In the generated &~ operator,
	 * fiddle things so a PDP-11 "bit"
	 * instruction will be produced when cctab is used.
	 */
	case NAND:
		if (table==cctab) {
			tree->op = TAND;
			tree->tr2 = optim(block(1, COMPL, INT, 0, tree->tr2));
		}
		break;


	/*
	 * Handle a subroutine call. It has to be done
	 * here because if cexpr got called twice, the
	 * arguments might be compiled twice.
	 * There is also some fiddling so the
	 * first argument, in favorable circumstances,
	 * goes to (sp) instead of -(sp), reducing
	 * the amount of stack-popping.
	 */
	case CALL:
		r = 0;
		nargs = 0;
		modf = 0;
		if (tree->tr1->op!=NAME) {	/* get nargs right */
			nargs++;
			nstack++;
		}
		tree = tree->tr2;
		if(tree->op) {
			while (tree->op==COMMA) {
				r =+ comarg(tree->tr2, &modf);
				tree = tree->tr1;
				nargs++;
			}
			r =+ comarg(tree, &modf);
			nargs++;
		}
		tree = atree;
		tree->op = CALL2;
		if (modf && tree->tr1->op==NAME && tree->tr1->class==EXTERN)
			tree->op = CALL1;
		cexpr(tree, regtab, reg);
		popstk(r);
		nstack =- nargs;
		if (table==efftab || table==regtab)
			return(0);
		r = 0;
		goto fixup;

	/*
	 * Longs need special treatment.
	 */
	case ASLSH:
	case LSHIFT:
		if (tree->type==LONG) {
			if (tree->tr2->op==ITOL)
				tree->tr2 = tree->tr2->tr1;
			if (tree->op==ASLSH)
				tree->op = ASLSHL;
			else
				tree->op = LLSHIFT;
		}
		break;

	/*
	 * Try to change * and / to shifts.
	 */
	case TIMES:
	case DIVIDE:
	case ASTIMES:
	case ASDIV:
		tree = pow2(tree);
	}
	/*
	 * Try to find postfix ++ and -- operators that can be
	 * pulled out and done after the rest of the expression
	 */
	if (table!=cctab && table!=cregtab && recurf<2
	 && (opdope[tree->op]&LEAF)==0) {
		if (r=delay(&atree, table, reg)) {
			tree = atree;
			table = efftab;
			reg = r-1;
		}
	}
	/*
	 * Basically, try to reorder the computation
	 * so  reg = x+y  is done as  reg = x; reg =+ y
	 */
	if (recurf==0 && reorder(&atree, table, reg)) {
		if (table==cctab && atree->op==NAME)
			return(reg);
	}
	tree = atree;
	if (table==efftab && tree->op==NAME)
		return(reg);
	if ((r=cexpr(tree, table, reg))>=0)
		return(r);
	if (table!=regtab)  {
		if((r=cexpr(tree, regtab, reg))>=0) {
	fixup:
			modf = isfloat(tree);
			if (table==sptab || table==lsptab) {
				if (tree->type==LONG) {
					printf("mov\tr%d,-(sp)\n",r+1);
					nstack++;
				}
				printf("mov%c	r%d,%c(sp)\n", modf, r,
					table==sptab? '-':0);
				nstack++;
			}
			if (table==cctab)
				printf("tst%c	r%d\n", modf, r);
			return(r);
		}
	}
	if (tree->op>0 && tree->op<RFORCE && opntab[tree->op])
		error("No code table for op: %s", opntab[tree->op]);
	else
		error("No code table for op %d", tree->op);
	return(reg);
}

/*
 * Try to compile the tree with the code table using
 * registers areg and up.  If successful,
 * return the register where the value actually ended up.
 * If unsuccessful, return -1.
 *
 * Most of the work is the macro-expansion of the
 * code table.
 */
cexpr(atree, table, areg)
struct tnode *atree;
struct table *table;
{
	int c, r;
	register struct tnode *p, *p1, *tree;
	struct table *ctable;
	struct tnode *p2;
	char *string;
	int reg, reg1, rreg, flag, opd;
	char *opt;

	tree = atree;
	reg = areg;
	p1 = tree->tr2;
	c = tree->op;
	opd = opdope[c];
	/*
	 * When the value of a relational or a logical expression is
	 * desired, more work must be done.
	 */
	if ((opd&RELAT||c==LOGAND||c==LOGOR||c==EXCLA) && table!=cctab) {
		cbranch(tree, c=isn++, 1, reg);
		rcexpr(&czero, table, reg);
		branch(isn, 0);
		label(c);
		rcexpr(&cone, table, reg);
		label(isn++);
		return(reg);
	}
	if(c==QUEST) {
		if (table==cctab)
			return(-1);
		cbranch(tree->tr1, c=isn++, 0, reg);
		flag = nstack;
		rreg = rcexpr(p1->tr1, table, reg);
		nstack = flag;
		branch(r=isn++, 0);
		label(c);
		reg = rcexpr(p1->tr2, table, rreg);
		if (rreg!=reg)
			printf("mov%c	r%d,r%d\n",
			    isfloat(tree),reg,rreg);
		label(r);
		return(rreg);
	}
	reg = oddreg(tree, reg);
	reg1 = reg+1;
	/*
	 * long values take 2 registers.
	 */
	if (tree->type==LONG && tree->op!=ITOL)
		reg1++;
	/*
	 * Leaves of the expression tree
	 */
	if ((r = chkleaf(tree, table, reg)) >= 0)
		return(r);
	/*
	 * x + (-1) is better done as x-1.
	 */

	if ((tree->op==PLUS||tree->op==ASPLUS) &&
	    (p1=tree->tr2)->op == CON && p1->value == -1) {
		p1->value = 1;
		tree->op =+ (MINUS-PLUS);
	}
	if (table==cregtab)
		table = regtab;
	/*
	 * The following peculiar code depends on the fact that
	 * if you just want the codition codes set, efftab
	 * will generate the right code unless the operator is
	 * postfix ++ or --. Unravelled, if the table is
	 * cctab and the operator is not special, try first
	 * for efftab;  if the table isn't, if the operator is,
	 * or the first match fails, try to match
	 * with the table actually asked for.
	 */
	if (table!=cctab || c==INCAFT || c==DECAFT
	 || (opt = match(tree, efftab, nreg-reg)) == 0)
		if ((opt=match(tree, table, nreg-reg))==0)
			return(-1);
	string = opt->tabstring;
	p1 = tree->tr1;
	p2 = 0;
	if (opdope[tree->op]&BINARY)
		p2 = tree->tr2;
loop:
	/*
	 * The 0200 bit asks for a tab.
	 */
	if ((c = *string++) & 0200) {
		c =& 0177;
		putchar('\t');
	}
	switch (c) {

	case '\0':
		if (!isfloat(tree))
			if (tree->op==DIVIDE || tree->op==ASDIV)