对函数的返回感到困惑

Question

好的，我在第一个程序上遇到了麻烦。 如果您不看这两个函数，那么我想要的是如果“ cur”值不在最小值和最大值之间，因为它将“目标”变量更改为“ 0”，因此我可以在程序中使用它。 。但这似乎不可行...我使用“ return”错误吗？ 基本上，函数需要把target从1到0 ，这样ri_fit变为0，让我最后if thingys触发...

对不起，代码不好，我才刚刚开始！

#Clearance calculator
#clearances are in clearances.txt

targets = open("clearances.txt", "r")
lines = targets.readlines()   #get target clearances from file

in_min_target = float(lines[2])     #minimum intake clearance
in_max_target = float(lines[4])     #maximum intake clearance

ex_min_target = float(lines[8])    #miminum exhaust clearances
ex_max_target = float(lines[10])   #maximum exhaust clearances
targets.close

target_intake = (in_min_target + in_max_target) / 2     #find the ideal intake
target_exhaust = (ex_min_target + ex_max_target) / 2    #find the ideal exhaust

print "Intake Min: ", in_min_target
print "Intake Max: ", in_max_target
print "Exhaust Min: ", ex_min_target
print "Exhaust Max: ", ex_max_target
print """Target intake: %r
Target Exhaust: %r""" % (target_intake, target_exhaust)

print""
print "Enter current RIGHT side Intake clearance"
cur_r_in = float(raw_input(">"))
print ""
print "Enter current RIGHT side Exhaust clearance"
cur_r_ex = float(raw_input(">"))
print ""
print "Enter current LEFT side Intake clearance"
cur_l_in = float(raw_input(">"))
print ""
print "Enter current LEFT side Exhaust clearance"
cur_l_ex = float(raw_input(">"))


target = 1

def in_range(min, max, cur, valve, target):  #figures if intake valves are correct
    if min <= cur <= max:
        print "%r is in range." % valve
        target=1

    else:
        print "%r is OUT OF RANGE." %valve
        target=0

    return target

def ex_range(min, max, cur, valve, target):   #figures if exhaust valves are correct
    if min <= cur <= max:
        print "%r is in range." % valve
        target=1

    else:
        print "%r is OUT OF RANGE." %valve
        target=0
    return target


ri_fit = 1
re_fit = 1        #Assumes all valves are right, until further notice...
li_fit = 1
le_fit = 1

valve = "Right Intake"
print in_range(in_min_target, in_max_target, cur_r_in, valve, target)
print target
if target == 0:
    ri_fit = 0


print ""

valve = "Right Exhaust"
print ex_range(ex_min_target, ex_max_target, cur_r_ex, valve, target)

print ""

valve = "Left Intake"
print in_range(in_min_target, in_max_target, cur_l_in, valve, target)

print ""

valve = "Left Exhaust"
print ex_range(ex_min_target, ex_max_target, cur_l_ex, valve, target)

print ri_fit

if ri_fit==0:
    print "Right intake is out of range."
    print "Enter current right intake shim size."
    ri_cur_shim = int(raw_input(">"))

if re_fit==0:
    print "Right exhaust is out of range."
    print "Enter current right exhaust shim size."
    re_cur_shim = int(raw_input(">"))

if li_fit==0:
    print "Right intake is out of range."
    print "Enter current right intake shim size."
    li_cur_shim = int(raw_input(">"))

if le_fit==0:
    print "Right exhaust is out of range."
    print "Enter current right exhaust shim size."
    le_cur_shim = int(raw_input(">"))

Answer 1

在其中一个函数中分配给target时，您将在其中替换名为target的参数的值，而不接触全局函数。 因此，如果要将函数分配给全局目标，则无法在函数中定义新目标。

但是，如果您不想更改函数，则可以使用return返回一个值的事实（由于打印了它，您似乎已经明白了）。 如果不是将其打印出来，而是将其分配给目标，那么您将在所需的位置获得所需的值，并且无论最终将计算的值放在何处，函数都将起作用。

Answer 2

return从函数返回一个值。 您可以将结果存储在变量中。

在您的代码中，您尝试将target作为参数传递，然后设置其值，但是由于整数是不可变的，因此您将无法在函数内更改target 。

就您而言，我会将您的代码更改为以下形式：

def ex_range(min, max, cur, valve):  # You don't need `target` here
    if min <= cur <= max:
        print "%r is in range." % valve
        target=1

    else:
        print "%r is OUT OF RANGE." %valve
        target=0
    return target

target = ex_range(...)  # Store the results in a variable

更好的是，使用布尔值：

def ex_range(min, max, cur, valve):  # You don't need `target` here
    if min <= cur <= max:
        print "%r is in range." % valve
        return True

    else:
        print "%r is OUT OF RANGE." %valve
        return False

Answer 3

我认为问题在于您将target作为一种价值而不是作为参考传递。 在函数中，应删除target参数，而应编写：

def ex_range(min, max, cur, valve):
    global target
    # ...
    # rest of your code

这样，解释器会识别出您正在尝试设置全局target变量，而不是本地实例。

否则，您可以执行以下操作：

def in_range(min, max, cur, valve):  #figures if intake valves are correct
    if min <= cur <= max:
        print "%r is in range." % valve
        return 1

    else:
        print "%r is OUT OF RANGE." %valve
        return 0

target = in_range(...) # fill in args
print target

请参阅此相关问题。

Answer 4

首先，我估计close后需要() ：

target.close()

第二： in_range和ex_range完全相同！

第三，无论如何都可以，因此可以正常运行：

in_range(1,10,7,'Right Intake',1)
in_range(1,10,17,'Right Intake',1)

输出：

>>> 'Right Intake' is in range.
>>> 'Right Intake' is OUT OF RANGE.

因此，可能是您传递给函数的参数不正确 ！ 检查您是否从文件中正确读取了它们。

最后，避免将min ， max用作参数，因为它们是built-in Python函数。