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 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 ``` ``````/* math.h - interface to shell math "library" -- this allows shells to share * the implementation of arithmetic \$((...)) expansions. * * This aims to be a POSIX shell math library as documented here: * http://www.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_04 * * See math.c for internal documentation. */ /* The math library has just one function: * * arith_t arith(const char *expr, int *perrcode, arith_eval_hooks_t *hooks); * * The first argument is the math string to parse. All normal expansions must * be done already. i.e. no dollar symbols should be present. * * The second argument is a semi-detailed error description in case something * goes wrong in the parsing steps. Currently, those values are (for * compatibility, you should assume all negative values are errors): * 0 - no errors (yay!) * -1 - unspecified problem * -2 - divide by zero * -3 - exponent less than 0 * -5 - expression recursion loop detected * * The third argument is a struct pointer of hooks for your shell (see below). * * The function returns the answer to the expression. So if you called it * with the expression: * "1 + 2 + 3" * You would obviously get back 6. */ /* To add support to a shell, you need to implement three functions: * * lookupvar() - look up and return the value of a variable * * If the shell does: * foo=123 * Then the code: * const char *val = lookupvar("foo"); * Will result in val pointing to "123" * * setvar() - set a variable to some value * * If the arithmetic expansion does something like: * \$(( i = 1)) * Then the math code will make a call like so: * setvar("i", "1", 0); * The storage for the first two parameters are not allocated, so your * shell implementation will most likely need to strdup() them to save. * * endofname() - return the end of a variable name from input * * The arithmetic code does not know about variable naming conventions. * So when it is given an experession, it knows something is not numeric, * but it is up to the shell to dictate what is a valid identifiers. * So when it encounters something like: * \$(( some_var + 123 )) * It will make a call like so: * end = endofname("some_var + 123"); * So the shell needs to scan the input string and return a pointer to the * first non-identifier string. In this case, it should return the input * pointer with an offset pointing to the first space. The typical * implementation will return the offset of first char that does not match * the regex (in C locale): ^[a-zA-Z_][a-zA-Z_0-9]* */ /* To make your life easier when dealing with optional 64bit math support, * rather than assume that the type is "signed long" and you can always * use "%ld" to scan/print the value, use the arith_t helper defines. See * below for the exact things that are available. */ #ifndef SHELL_MATH_H #define SHELL_MATH_H 1 PUSH_AND_SET_FUNCTION_VISIBILITY_TO_HIDDEN #if ENABLE_SH_MATH_SUPPORT_64 typedef long long arith_t; #define arith_t_fmt "%lld" #define strto_arith_t strtoull #else typedef long arith_t; #define arith_t_fmt "%ld" #define strto_arith_t strtoul #endif typedef const char* FAST_FUNC (*arith_var_lookup_t)(const char *name); typedef void FAST_FUNC (*arith_var_set_t)(const char *name, const char *val); typedef char* FAST_FUNC (*arith_var_endofname_t)(const char *name); typedef struct arith_eval_hooks { arith_var_lookup_t lookupvar; arith_var_set_t setvar; arith_var_endofname_t endofname; } arith_eval_hooks_t; arith_t arith(const char *expr, int *perrcode, arith_eval_hooks_t*); POP_SAVED_FUNCTION_VISIBILITY #endif ``````