Python Cython中有函数类型吗？

有没有办法告诉Cython编译器param是函数。差不多

  cpdef float calc_class_re(list data, func callback)

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应该不言自明……？：）

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如果所有其他方法都失败了，您可能需要使用C

typedef

。也许有一种更好的纯Cython方法。你是指python函数还是c函数？当函数签名已知时，“delnan”的注释将适用于c。对于

cdef

或

cpdef

函数，c样式的functype应该可以工作。比如

ctypedef（*my_func_type）（object、int、float、str）

。对于纯python函数，您需要使用

对象

类型。“对于cdef或cpdef函数，C风格的functype应该可以工作。”@NiklasR，您能给出详细的示例吗？

# Define a new type for a function-type that accepts an integer and
# a string, returning an integer.
ctypedef int (*f_type)(int, str)

# Extern a function of that type from foo.h
cdef extern from "foo.h":
    int do_this(int, str)

# Passing this function will not work.
cpdef int do_that(int a, str b):
    return 0

# However, this will work.
cdef int do_stuff(int a, str b):
    return 0

# This functio uses a function of that type. Note that it cannot be a
# cpdef function because the function-type is not available from Python.
cdef void foo(f_type f):
    print f(0, "bar")

# Works:
foo(do_this)   # the externed function
foo(do_stuff)  # the cdef function

# Error:
# Cannot assign type 'int (int, str, int __pyx_skip_dispatch)' to 'f_type'
foo(do_that)   # the cpdef function