/* * Copyright (C) 2018 Intel Corporation. All rights reserved. * * SPDX-License-Identifier: BSD-3-Clause */ #include /* * Convert a string to a long integer - decimal support only. */ int64_t strtol_deci(const char *nptr) { const char *s = nptr; char c; uint64_t acc, cutoff, cutlim; int32_t neg = 0, any; uint64_t base = 10UL; /* * Skip white space and pick up leading +/- sign if any. */ do { c = *s; s++; } while (is_space(c)); if (c == '-') { neg = 1; c = *s; s++; } else if (c == '+') { c = *s; s++; } else { /* No sign character. */ } /* * Compute the cutoff value between legal numbers and illegal * numbers. That is the largest legal value, divided by the * base. An input number that is greater than this value, if * followed by a legal input character, is too big. One that * is equal to this value may be valid or not; the limit * between valid and invalid numbers is then based on the last * digit. For instance, if the range for longs is * [-2147483648..2147483647] and the input base is 10, * cutoff will be set to 214748364 and cutlim to either * 7 (neg==0) or 8 (neg==1), meaning that if we have accumulated * a value > 214748364, or equal but the next digit is > 7 (or 8), * the number is too big, and we will return a range error. * * Set any if any `digits' consumed; make it negative to indicate * overflow. */ cutoff = (neg != 0) ? LONG_MIN : LONG_MAX; cutlim = cutoff % base; cutoff /= base; acc = 0UL; any = 0; while ((c >= '0') && (c <= '9')) { c -= '0'; if ((acc > cutoff) || ((acc == cutoff) && ((uint64_t)c > cutlim))) { any = -1; break; } else { acc *= base; acc += (uint64_t)c; } c = *s; s++; } if (any < 0) { acc = (neg != 0) ? LONG_MIN : LONG_MAX; } else if (neg != 0) { acc = ~acc + 1UL; } else { /* There is no overflow and no leading '-' exists. In such case * acc already holds the right number. No action required. */ } return (long)acc; } int32_t atoi(const char *str) { return (int32_t)strtol_deci(str); }