实例讲解Python中global语句下全局变量的值的修改
Python的全局变量:int string, list, dic(map) 如果存在global就能够修改它的值。而不管这个global是否是存在于if中,也不管这个if是否能够执行到。 if bGlobal: global g_strVal; int string 将会报错。而list dic(map)是ok的。 #!/usr/bin/dev python import sys import os g_nVal = 0; g_strVal = "aaaa"; g_map = { "aaa" : "111","bbb" : "222","ccc" : "333","ddd" : "444" } g_ls = ['a','b','c'] def FixInt(bGlobal = False): if bGlobal: global g_nVal; g_nVal = g_nVal + 1; def FixString(bGlobal = False): if bGlobal: global g_strVal; #fix string value g_strVal = g_strVal + 'b'; def FixMap(bGlobal = False): if bGlobal: global g_map; #fix map value g_map['aaa'] = 'aaa__' + g_strVal; g_map['bbb'] = 'bbb__' + g_strVal; g_map['ccc'] = 'ccc__' + g_strVal; g_map['ddd'] = 'ddd__' + g_strVal; def FixList(bGlobal = False): if bGlobal: global g_ls; g_ls.append('1'); def PrintVal(strInfo): if strInfo: print("==== %s =====" %strInfo); print("int value:%d" %g_nVal); print("string value:%s" %g_strVal); print("map value:%s" %g_map); print("list value:%s" %g_ls); print("nn"); if "__main__" == __name__: PrintVal("The orgin vlaue"); FixInt(); FixString(); FixMap(); FixList(); PrintVal("print all bGlobal = False vlaue"); FixInt(True); FixString(True); FixMap(True); FixList(True); PrintVal("print all bGlobal = True vlaue"); 结果: ==== The orgin vlaue ===== int value:0 string value:aaaa map value:{'aaa': '111','bbb': '222','ccc': '333','ddd': '444'} list value:['a','c'] g_nVal src:0 g_nVal dst:1 ==== print all bGlobal = False value ===== int value:1 string value:aaaab map value:{'aaa': 'aaa__aaaab','bbb': 'bbb__aaaab','ccc': 'ccc__aaaab','ddd': 'ddd__aaaab'} list value:['a','c','1'] g_nVal src:1 g_nVal dst:2 ==== print all bGlobal = True value ===== int value:2 string value:aaaabb map value:{'aaa': 'aaa__aaaabb','bbb': 'bbb__aaaabb','ccc': 'ccc__aaaabb','ddd': 'ddd__aaaabb'} list value:['a','1','1'] 为什么修改全局的dict变量不用global关键字 s = 'foo' d = {'a':1} def f(): s = 'bar' d['b'] = 2 f() print s print d 为什么修改字典d的值不用global关键字先声明呢? 但是如果是下面这样 d = {'a':1} def f(): d = {} d['b'] = 2 f() print d 在d = {}这句,它是”有歧义的“了,所以它是创建了局部变量d,而不是引用全局变量d,所以d['b']=2也是操作的局部变量。 推而远之,这一切现象的本质就是”它是否是明确的“。 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |