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-rw-r--r--sysdeps/sparc/sparc32/divrem.m4234
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diff --git a/sysdeps/sparc/sparc32/divrem.m4 b/sysdeps/sparc/sparc32/divrem.m4
deleted file mode 100644
index c08c530020..0000000000
--- a/sysdeps/sparc/sparc32/divrem.m4
+++ /dev/null
@@ -1,234 +0,0 @@
-/*
- * Division and remainder, from Appendix E of the Sparc Version 8
- * Architecture Manual, with fixes from Gordon Irlam.
- */
-
-/*
- * Input: dividend and divisor in %o0 and %o1 respectively.
- *
- * m4 parameters:
- *  NAME	name of function to generate
- *  OP		OP=div => %o0 / %o1; OP=rem => %o0 % %o1
- *  S		S=true => signed; S=false => unsigned
- *
- * Algorithm parameters:
- *  N		how many bits per iteration we try to get (4)
- *  WORDSIZE	total number of bits (32)
- *
- * Derived constants:
- *  TOPBITS	number of bits in the top `decade' of a number
- *
- * Important variables:
- *  Q		the partial quotient under development (initially 0)
- *  R		the remainder so far, initially the dividend
- *  ITER	number of main division loop iterations required;
- *		equal to ceil(log2(quotient) / N).  Note that this
- *		is the log base (2^N) of the quotient.
- *  V		the current comparand, initially divisor*2^(ITER*N-1)
- *
- * Cost:
- *  Current estimate for non-large dividend is
- *	ceil(log2(quotient) / N) * (10 + 7N/2) + C
- *  A large dividend is one greater than 2^(31-TOPBITS) and takes a
- *  different path, as the upper bits of the quotient must be developed
- *  one bit at a time.
- */
-
-define(N, `4')dnl
-define(WORDSIZE, `32')dnl
-define(TOPBITS, eval(WORDSIZE - N*((WORDSIZE-1)/N)))dnl
-dnl
-define(dividend, `%o0')dnl
-define(divisor, `%o1')dnl
-define(Q, `%o2')dnl
-define(R, `%o3')dnl
-define(ITER, `%o4')dnl
-define(V, `%o5')dnl
-dnl
-dnl m4 reminder: ifelse(a,b,c,d) => if a is b, then c, else d
-define(T, `%g1')dnl
-define(SC, `%g2')dnl
-ifelse(S, `true', `define(SIGN, `%g3')')dnl
-
-dnl
-dnl This is the recursive definition for developing quotient digits.
-dnl
-dnl Parameters:
-dnl  $1	the current depth, 1 <= $1 <= N
-dnl  $2	the current accumulation of quotient bits
-dnl  N	max depth
-dnl
-dnl We add a new bit to $2 and either recurse or insert the bits in
-dnl the quotient.  R, Q, and V are inputs and outputs as defined above;
-dnl the condition codes are expected to reflect the input R, and are
-dnl modified to reflect the output R.
-dnl
-define(DEVELOP_QUOTIENT_BITS,
-`	! depth $1, accumulated bits $2
-	bl	LOC($1.eval(2**N+$2))
-	srl	V,1,V
-	! remainder is positive
-	subcc	R,V,R
-	ifelse($1, N,
-	`	b	9f
-		add	Q, ($2*2+1), Q
-', `	DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2+1)')')
-LOC($1.eval(2**N+$2)):
-	! remainder is negative
-	addcc	R,V,R
-	ifelse($1, N,
-	`	b	9f
-		add	Q, ($2*2-1), Q
-', `	DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2-1)')')
-ifelse($1, 1, `9:')')dnl
-
-#include <sysdep.h>
-#include <sys/trap.h>
-
-ENTRY(NAME)
-ifelse(S, `true',
-`	! compute sign of result; if neither is negative, no problem
-	orcc	divisor, dividend, %g0	! either negative?
-	bge	2f			! no, go do the divide
-ifelse(OP, `div',
-`	xor	divisor, dividend, SIGN	! compute sign in any case',
-`	mov	dividend, SIGN		! sign of remainder matches dividend')
-	tst	divisor
-	bge	1f
-	tst	dividend
-	! divisor is definitely negative; dividend might also be negative
-	bge	2f			! if dividend not negative...
-	sub	%g0, divisor, divisor	! in any case, make divisor nonneg
-1:	! dividend is negative, divisor is nonnegative
-	sub	%g0, dividend, dividend	! make dividend nonnegative
-2:
-')
-	! Ready to divide.  Compute size of quotient; scale comparand.
-	orcc	divisor, %g0, V
-	bne	1f
-	mov	dividend, R
-
-		! Divide by zero trap.  If it returns, return 0 (about as
-		! wrong as possible, but that is what SunOS does...).
-		ta	ST_DIV0
-		retl
-		clr	%o0
-
-1:
-	cmp	R, V			! if divisor exceeds dividend, done
-	blu	LOC(got_result)		! (and algorithm fails otherwise)
-	clr	Q
-	sethi	%hi(1 << (WORDSIZE - TOPBITS - 1)), T
-	cmp	R, T
-	blu	LOC(not_really_big)
-	clr	ITER
-
-	! `Here the dividend is >= 2**(31-N) or so.  We must be careful here,
-	! as our usual N-at-a-shot divide step will cause overflow and havoc.
-	! The number of bits in the result here is N*ITER+SC, where SC <= N.
-	! Compute ITER in an unorthodox manner: know we need to shift V into
-	! the top decade: so do not even bother to compare to R.'
-	1:
-		cmp	V, T
-		bgeu	3f
-		mov	1, SC
-		sll	V, N, V
-		b	1b
-		add	ITER, 1, ITER
-
-	! Now compute SC.
-	2:	addcc	V, V, V
-		bcc	LOC(not_too_big)
-		add	SC, 1, SC
-
-		! We get here if the divisor overflowed while shifting.
-		! This means that R has the high-order bit set.
-		! Restore V and subtract from R.
-		sll	T, TOPBITS, T	! high order bit
-		srl	V, 1, V		! rest of V
-		add	V, T, V
-		b	LOC(do_single_div)
-		sub	SC, 1, SC
-
-	LOC(not_too_big):
-	3:	cmp	V, R
-		blu	2b
-		nop
-		be	LOC(do_single_div)
-		nop
-	/* NB: these are commented out in the V8-Sparc manual as well */
-	/* (I do not understand this) */
-	! V > R: went too far: back up 1 step
-	!	srl	V, 1, V
-	!	dec	SC
-	! do single-bit divide steps
-	!
-	! We have to be careful here.  We know that R >= V, so we can do the
-	! first divide step without thinking.  BUT, the others are conditional,
-	! and are only done if R >= 0.  Because both R and V may have the high-
-	! order bit set in the first step, just falling into the regular
-	! division loop will mess up the first time around.
-	! So we unroll slightly...
-	LOC(do_single_div):
-		subcc	SC, 1, SC
-		bl	LOC(end_regular_divide)
-		nop
-		sub	R, V, R
-		mov	1, Q
-		b	LOC(end_single_divloop)
-		nop
-	LOC(single_divloop):
-		sll	Q, 1, Q
-		bl	1f
-		srl	V, 1, V
-		! R >= 0
-		sub	R, V, R
-		b	2f
-		add	Q, 1, Q
-	1:	! R < 0
-		add	R, V, R
-		sub	Q, 1, Q
-	2:
-	LOC(end_single_divloop):
-		subcc	SC, 1, SC
-		bge	LOC(single_divloop)
-		tst	R
-		b,a	LOC(end_regular_divide)
-
-LOC(not_really_big):
-1:
-	sll	V, N, V
-	cmp	V, R
-	bleu	1b
-	addcc	ITER, 1, ITER
-	be	LOC(got_result)
-	sub	ITER, 1, ITER
-
-	tst	R	! set up for initial iteration
-LOC(divloop):
-	sll	Q, N, Q
-	DEVELOP_QUOTIENT_BITS(1, 0)
-LOC(end_regular_divide):
-	subcc	ITER, 1, ITER
-	bge	LOC(divloop)
-	tst	R
-	bl,a	LOC(got_result)
-	! non-restoring fixup here (one instruction only!)
-ifelse(OP, `div',
-`	sub	Q, 1, Q
-', `	add	R, divisor, R
-')
-
-LOC(got_result):
-ifelse(S, `true',
-`	! check to see if answer should be < 0
-	tst	SIGN
-	bl,a	1f
-	ifelse(OP, `div', `sub %g0, Q, Q', `sub %g0, R, R')
-1:')
-	retl
-	ifelse(OP, `div', `mov Q, %o0', `mov R, %o0')
-
-END(NAME)
-ifelse(OP, `div', ifelse(S, `false', `strong_alias (.udiv, __wrap_.udiv)
-'))dnl