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Searched refs:rd2 (Results 1 – 7 of 7) sorted by relevance

/linux/arch/arm64/kernel/
H A Dcompat_alignment.c71 unsigned int rd2; in do_alignment_ldrdstrd() local
76 rd2 = (instr >> 8) & 0xf; in do_alignment_ldrdstrd()
82 rd2 = rd + 1; in do_alignment_ldrdstrd()
92 regs->regs[rd2] = val2; in do_alignment_ldrdstrd()
95 put_user(regs->regs[rd2], (u32 __user *)(addr + 4))) in do_alignment_ldrdstrd()
/linux/Documentation/translations/zh_CN/scheduler/
H A Dsched-energy.rst92 RDs: |----rd1----|-----rd2-----|
95 每个根域包含6个CPU。这两个根域在上图中被表示为rd1和rd2。由于pd4与rd1和rd2
99 * rd2->pd: pd4 -> pd8
/linux/arch/arm/mm/
H A Dalignment.c392 unsigned int rd2; in do_alignment_ldrdstrd() local
397 rd2 = (instr >> 8) & 0xf; in do_alignment_ldrdstrd()
403 rd2 = rd + 1; in do_alignment_ldrdstrd()
416 regs->uregs[rd2] = val; in do_alignment_ldrdstrd()
419 put32_unaligned_check(regs->uregs[rd2], addr + 4); in do_alignment_ldrdstrd()
435 regs->uregs[rd2] = val2; in do_alignment_ldrdstrd()
439 put32t_unaligned_check(regs->uregs[rd2], addr + 4); in do_alignment_ldrdstrd()
/linux/drivers/mtd/nand/raw/
H A Dnand_bbt.c955 struct nand_bbt_descr *rd, *rd2; in check_create() local
967 rd2 = NULL; in check_create()
985 rd2 = md; in check_create()
1028 if (rd2) { in check_create()
1029 res2 = read_abs_bbt(this, buf, rd2, chipsel); in check_create()
1032 rd2->pages[i] = -1; in check_create()
1033 rd2->version[i] = 0; in check_create()
/linux/net/wireless/
H A Dreg.c1337 const struct ieee80211_regdomain *rd2, in reg_rules_intersect() argument
1373 max_bandwidth2 = reg_get_max_bandwidth(rd2, rule2); in reg_rules_intersect()
1502 const struct ieee80211_regdomain *rd2) in regdom_intersect() argument
1511 if (!rd1 || !rd2) in regdom_intersect()
1524 for (y = 0; y < rd2->n_reg_rules; y++) { in regdom_intersect()
1525 rule2 = &rd2->reg_rules[y]; in regdom_intersect()
1526 if (!reg_rules_intersect(rd1, rd2, rule1, rule2, in regdom_intersect()
1541 for (y = 0; y < rd2->n_reg_rules; y++) { in regdom_intersect()
1542 rule2 = &rd2->reg_rules[y]; in regdom_intersect()
1543 r = reg_rules_intersect(rd1, rd2, rule1, rule2, in regdom_intersect()
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/linux/Documentation/scheduler/
H A Dsched-energy.rst104 RDs: |----rd1----|-----rd2-----|
108 containing 6 CPUs. The two root domains are denoted rd1 and rd2 in the
109 above figure. Since pd4 intersects with both rd1 and rd2, it will be
113 * rd2->pd: pd4 -> pd8
/linux/Documentation/admin-guide/
H A Dmd.rst604 So for a 3 drive array there will be rd0, rd1, rd2.