c05564c4d8
Android 13
88 lines
2 KiB
ArmAsm
Executable file
88 lines
2 KiB
ArmAsm
Executable file
/* SPDX-License-Identifier: GPL-2.0 */
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#include <linux/linkage.h>
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/*
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* Unsigned modulo operation for 32 bit integers.
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* Input : op1 in Reg r5
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* op2 in Reg r6
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* Output: op1 mod op2 in Reg r3
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*/
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.text
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.globl __umodsi3
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.type __umodsi3, @function
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.ent __umodsi3
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__umodsi3:
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.frame r1, 0, r15
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addik r1, r1, -12
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swi r29, r1, 0
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swi r30, r1, 4
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swi r31, r1, 8
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beqi r6, div_by_zero /* div_by_zero - division error */
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beqid r5, result_is_zero /* result is zero */
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addik r3, r0, 0 /* clear div */
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addik r30, r0, 0 /* clear mod */
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addik r29, r0, 32 /* initialize the loop count */
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/* check if r6 and r5 are equal /* if yes, return 0 */
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rsub r18, r5, r6
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beqi r18, return_here
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/* check if (uns)r6 is greater than (uns)r5. in that case, just return r5 */
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xor r18, r5, r6
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bgeid r18, 16
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addik r3, r5, 0
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blti r6, return_here
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bri $lcheckr6
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rsub r18, r5, r6 /* microblazecmp */
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bgti r18, return_here
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/* if r6 [bit 31] is set, then return result as r5-r6 */
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$lcheckr6:
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bgtid r6, div0
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addik r3, r0, 0
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addik r18, r0, 0x7fffffff
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and r5, r5, r18
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and r6, r6, r18
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brid return_here
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rsub r3, r6, r5
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/* first part: try to find the first '1' in the r5 */
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div0:
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blti r5, div2
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div1:
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add r5, r5, r5 /* left shift logical r5 */
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bgeid r5, div1
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addik r29, r29, -1
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div2:
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/* left shift logical r5 get the '1' into the carry */
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add r5, r5, r5
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addc r3, r3, r3 /* move that bit into the mod register */
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rsub r31, r6, r3 /* try to subtract (r3 a r6) */
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blti r31, mod_too_small
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/* move the r31 to mod since the result was positive */
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or r3, r0, r31
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addik r30, r30, 1
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mod_too_small:
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addik r29, r29, -1
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beqi r29, loop_end
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add r30, r30, r30 /* shift in the '1' into div */
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bri div2 /* div2 */
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loop_end:
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bri return_here
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div_by_zero:
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result_is_zero:
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or r3, r0, r0 /* set result to 0 */
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return_here:
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/* restore values of csrs and that of r3 and the divisor and the dividend */
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lwi r29, r1, 0
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lwi r30, r1, 4
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lwi r31, r1, 8
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rtsd r15, 8
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addik r1, r1, 12
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.size __umodsi3, . - __umodsi3
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.end __umodsi3
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