python - 返回字典 get 方法中的递归

Question

使用 MWE。函数 foo 工作正常，bar 则不行，并引发“RuntimeError:超出最大递归深度”异常。有什么提示吗？

def foo(i1, i2):
    inverse_map = {"<": ">"}
    if i1[1] < i2[0]: return "<"
    return inverse_map[foo(i2, i1)]

def bar(i1, i2):
    inverse_map = {"<": ">"}
    return {
        i1[1] < i2[0]: "<"
    }.get(True, inverse_map[bar(i2, i1)])

i1, i2 = (0, 2), (4, 6)
print foo(i2, i1), ">"
print bar(i2, i1), ">"

Answer 1

Python字典get方法首先检查默认值，因此，在bar函数中，它会在之前递归调用bar >返回值。您的代码可以转换为以下格式，这会导致死递归:

def foo(i1, i2):
    inverse_map = {"<": ">"}
    if i1[1] < i2[0]: return "<"
    return inverse_map[foo(i2, i1)]

def bar(i1, i2):
    inverse_map = {"<": ">"}
    d = {i1[1] < i2[0] : "<"}
    inverse = inverse_map[bar(i2,i1)]
    return d.get(True,inverse)

i1, i2 = (0, 2), (4, 6)
print foo(i1, i2), ">"
print bar(i2, i1), ">"

字典get实现的快照如下所示:

static PyObject *
dict_get(register PyDictObject *mp, PyObject *args)
{
    PyObject *key;
    PyObject *failobj = Py_None;
    PyObject *val = NULL;
    long hash;
    PyDictEntry *ep;
    #commented by myself, get function first check default value
    if (!PyArg_UnpackTuple(args, "get", 1, 2, &key, &failobj))
        return NULL;

    if (!PyString_CheckExact(key) ||
        (hash = ((PyStringObject *) key)->ob_shash) == -1) {
        hash = PyObject_Hash(key);
        if (hash == -1)
            return NULL;
    }
    ep = (mp->ma_lookup)(mp, key, hash);
    if (ep == NULL)
        return NULL;
    val = ep->me_value;
    if (val == NULL)
        val = failobj;
    Py_INCREF(val);
    return val;
}

python dictionary implementation