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f0ddadcf2e
To make it clearer that Py_BUILD_ASSERT_EXPR(cond) cannot be used as assert(cond).
56 lines
1.7 KiB
C
56 lines
1.7 KiB
C
#ifndef Py_PYMACRO_H
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#define Py_PYMACRO_H
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#define Py_MIN(x, y) (((x) > (y)) ? (y) : (x))
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#define Py_MAX(x, y) (((x) > (y)) ? (x) : (y))
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/* Argument must be a char or an int in [-128, 127] or [0, 255]. */
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#define Py_CHARMASK(c) ((unsigned char)((c) & 0xff))
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/* Assert a build-time dependency, as an expression.
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Your compile will fail if the condition isn't true, or can't be evaluated
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by the compiler. This can be used in an expression: its value is 0.
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Example:
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#define foo_to_char(foo) \
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((char *)(foo) \
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+ Py_BUILD_ASSERT_EXPR(offsetof(struct foo, string) == 0))
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Written by Rusty Russell, public domain, http://ccodearchive.net/ */
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#define Py_BUILD_ASSERT_EXPR(cond) \
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(sizeof(char [1 - 2*!(cond)]) - 1)
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/* Get the number of elements in a visible array
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This does not work on pointers, or arrays declared as [], or function
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parameters. With correct compiler support, such usage will cause a build
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error (see Py_BUILD_ASSERT_EXPR).
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Written by Rusty Russell, public domain, http://ccodearchive.net/ */
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#if defined(__GNUC__)
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/* Two gcc extensions.
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&a[0] degrades to a pointer: a different type from an array */
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#define Py_ARRAY_LENGTH(array) \
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(sizeof(array) / sizeof((array)[0]) \
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+ Py_BUILD_ASSERT_EXPR(!__builtin_types_compatible_p(typeof(array), \
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typeof(&(array)[0]))))
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#else
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#define Py_ARRAY_LENGTH(array) \
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(sizeof(array) / sizeof((array)[0]))
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#endif
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/* Define macros for inline documentation. */
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#define PyDoc_VAR(name) static char name[]
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#define PyDoc_STRVAR(name,str) PyDoc_VAR(name) = PyDoc_STR(str)
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#ifdef WITH_DOC_STRINGS
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#define PyDoc_STR(str) str
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#else
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#define PyDoc_STR(str) ""
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#endif
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#endif /* Py_PYMACRO_H */
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