利用空闲写了一个模拟GDB输出call stack的程序。
#include
by picktracy - C/C++ - 2008-09-09 13:54:36 阅读(4345) 回复(7)
1, Listen outgoing call Register a broadcast receiver with action android.intent.action.NEW_OUTGOING_call, but please request to use permission android.permission.PROCESS_OUTGOING_callS. we can get outgoing phone number by calling String strPhoneNumber = intent.getStringExtra(Intent.EXTRA_PHONE_NUMBER); in BroadcastReceiver.onReceive(xxx); 2, Listen incoming call Use TelephonyManager and P...
遇到一个棘手的问题,不知道大家遇到过没有 当一个程序a call 另一个程序b时(动态),当b返回以后此时a没有结束,如果在这个时候修改b的源码 然后重新编译b,可是a再次调用的时候还是调用的原来的b,这种情况有法子避免吗? 还是必须重新调用a才行? :em14: :em14:
#!/usr/bin/env python # add stack datastructure # __author__ = "lynn lin" class UnderFlowError(Exception): pass class stack(object): def __init__(self): self.sta = [] def __len__(self): return len(self.sta) def stackEmpty(self): return self.__len__() def push(self,value): self.sta.append(value) def pop(self): try: ...
PGM (&INPUT &REMAINDER) /* SEARCH */ DCL &INPUT *CHAR LEN(50) DCL &REMAINDER *CHAR LEN(50) DCL &X *DEC LEN(2 0) VALUE(03) DCL &L *DEC LEN(2 0) /* REMAINING LENGTH */ SCAN: IF (%SST(&INPUT &X 1) *EQ '.') THEN(DO) CHGVAR VAR(&L) VALUE(50-&X) CHGVAR VAR(&X) VALUE(&X+1) CHGVAR VAR(&REM...
PGM (&INPUT &REMAINDER) /* SEARCH */ DCL &INPUT *CHAR LEN(50) DCL &REMAINDER *CHAR LEN(50) DCL &X *DEC LEN(2 0) VALUE(03) DCL &L *DEC LEN(2 0) /* REMAINING LENGTH */ SCAN: IF (%SST(&INPUT &X 1) *EQ '.') THEN(DO) CHGVAR VAR(&L) VALUE(50-&X) CHGVAR VAR(&X) VALUE(&X+1) CHGVAR VAR(&REM...