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8b16be9185
Many of the license and Intel copyright headers include the "All rights reserved" string. It is not relevant in the context of the BSD-3-Clause license that the code is released under. This patch removes those strings throughout the code (hypervisor, devicemodel and misc). Tracked-On: #7254 Signed-off-by: Geoffroy Van Cutsem <geoffroy.vancutsem@intel.com>
88 lines
2.0 KiB
C
88 lines
2.0 KiB
C
/*
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* Copyright (C) 2018 Intel Corporation.
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*
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* SPDX-License-Identifier: BSD-3-Clause
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*/
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#include <types.h>
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#include <rtl.h>
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/*
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* Convert a string to a long integer - decimal support only.
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*/
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int64_t strtol_deci(const char *nptr)
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{
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const char *s = nptr;
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char c;
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uint64_t acc, cutoff, cutlim;
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int32_t neg = 0, any;
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uint64_t base = 10UL;
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/*
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* Skip white space and pick up leading +/- sign if any.
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*/
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do {
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c = *s;
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s++;
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} while (is_space(c));
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if (c == '-') {
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neg = 1;
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c = *s;
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s++;
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} else if (c == '+') {
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c = *s;
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s++;
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} else {
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/* No sign character. */
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}
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/*
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* Compute the cutoff value between legal numbers and illegal
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* numbers. That is the largest legal value, divided by the
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* base. An input number that is greater than this value, if
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* followed by a legal input character, is too big. One that
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* is equal to this value may be valid or not; the limit
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* between valid and invalid numbers is then based on the last
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* digit. For instance, if the range for longs is
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* [-2147483648..2147483647] and the input base is 10,
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* cutoff will be set to 214748364 and cutlim to either
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* 7 (neg==0) or 8 (neg==1), meaning that if we have accumulated
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* a value > 214748364, or equal but the next digit is > 7 (or 8),
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* the number is too big, and we will return a range error.
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*
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* Set any if any `digits' consumed; make it negative to indicate
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* overflow.
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*/
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cutoff = (neg != 0) ? LONG_MIN : LONG_MAX;
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cutlim = cutoff % base;
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cutoff /= base;
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acc = 0UL;
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any = 0;
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while ((c >= '0') && (c <= '9')) {
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c -= '0';
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if ((acc > cutoff) ||
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((acc == cutoff) && ((uint64_t)c > cutlim))) {
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any = -1;
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break;
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} else {
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acc *= base;
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acc += (uint64_t)c;
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}
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c = *s;
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s++;
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}
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if (any < 0) {
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acc = (neg != 0) ? LONG_MIN : LONG_MAX;
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} else if (neg != 0) {
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acc = ~acc + 1UL;
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} else {
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/* There is no overflow and no leading '-' exists. In such case
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* acc already holds the right number. No action required. */
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}
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return (long)acc;
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}
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