# direct methods .method public static constructor <clinit>()v .registers 16 .line 1 new-instance v0, lcom/alipay/sdk/m/j/c; const-string v1, "succeeded" const/4 v2, 0x0 const/16 v3, 0x2328 const-string v4, "处理成功" invoke-direct {v0, v1, v2, v3, v4}, lcom/alipay/sdk/m/j/c;-><init>(ljava/lang/string;iiljava/lang/string;)v sput-object v0, lcom/alipay/sdk/m/j/c;->c:lcom/alipay/sdk/m/j/c; .line 3 new-instance v1, lcom/alipay/sdk/m/j/c; const-st
这段代码是Android平台上的Smali代码,用于初始化一个名为`c`的静态字段。这个字段的类型是`com.alipay.sdk.m.j.c`。在这段代码中,首先创建了一个新的`com.alipay.sdk.m.j.c`实例,然后将其赋值给静态字段`c`。 [2024-08-09 20:30:24 | AI写代码神器 | 76点数解答]
相关提问
- [讯飞星火] # direct methods .method public static constructor <clinit>()v .registers 16 .line 1 new-instance v0, lcom/alipay/sdk/m/j/c; const-string v1, "succeeded" const/4 v2, 0x0 const/16 v3, 0x2328 const-string v4, "处理成功" invoke-direct {v0, v1, v2, v3, v4}, lcom/alipay/sdk/m/j/c;-><init>(ljava/lang/string;iiljava/lang/string;)v sput-object v0, lcom/alipay/sdk/m/j/c;->c:lcom/alipay/sdk/m/j/c; .line 3 new-instance v1, lcom/alipay/sdk/m/j/c; const-st(76点数解答 | 2024-08-09 20:30:24)294
- [GPT] .method public 按钮1$被弹起(ii)v .locals 4 .param p1, "\u6a2a\u5750\u6807" # i .param p2, "\u7eb5\u5750\u6807" # i .prologue .line 20 new-instance v0, ljava/lang/stringbuilder; invoke-direct {v0}, ljava/lang/stringbuilder;-><init>()v iget-object v1, p0, lcom/chuchenqi/窗口29;->手机1:lcom/e4a/runtime/components/impl/android/n9/手机; invoke-interface {v1}, lcom/e4a/runtime/components/impl/android/n9/手机;->取imei码()ljava/lang/string; move-result-object v1(208点数解答 | 2024-09-12 16:36:31)329
- [字节豆包] 给以下代码进行注释: fn czzb a b c=(--已知三点a、b、c,求c点在ab直线上的垂足坐标 ab=b-a; ac=c-a; lab=length ab; lac=length ac; cosct=((dot ab ac)/(lab*lac));--求出cosct abxl=if cosct>=0 then ab/lab;else -1*(ab/lab)--ab的单位向量 lty=abs (lac*cosct);--求出投影长度 xl=lty*abxl;--求出偏移向量 return (a+xl);--返回垂足坐标 ) -- fn qiumianji v1 v2 v3=( -- local chang=length (v3-v1); -- local v0=czzb v1 v3 v2; -- local gao=length (v0-v2) -- local mj=0.5*chang*gao; -- return mj; -- ) fn qiumianji v1 v2 v3=( local a=length (v1-v2); local(324点数解答 | 2024-10-16 18:51:32)284
- [字节豆包] .method public setupplayer(lcn/subat/music/model/spplayconfig;ljava/lang/string;i)v .registers 6 .line 299 iget-object v0, p0, lcn/subat/music/fragment/player/spmovieplayerfragment;->covermask:lcn/subat/music/base/spconstraintlayout; const/16 v1, 0x8 invoke-virtual {v0, v1}, lcn/subat/music/base/spconstraintlayout;->setvisibility(i)v .line 300 iget-object v0, p0, lcn/subat/music/fragment/player/spmovieplayerfragment;->sharetip:lcn/subat/music/base/sptextview;(29点数解答 | 2024-11-05 01:36:35)277
- [字节豆包] 给以下代码进行注释: fn qiumianji v1 v2 v3=( local a=length (v1-v2); local b=length (v2-v3); local c=length (v1-v3); local p=0.5*(a+b+c); local s=sqrt (p*(p-a)*(p-b)*(p-c)); return s; )(224点数解答 | 2024-10-16 18:56:35)227
- [字节豆包] public partial class form1 : form { public string vintext = ""; public int vin_number = 0; public struct vin_info { public int row; public string vin_number; public string vin_type; public string vin_car; public string vin_enter; public string vin_out; public string vin_go; public string vin_arrive; } public list<vin_info> vin_infos=new list<vin_info>(); public form1() { initializecomponent()(39点数解答 | 2024-11-05 15:09:08)285
- [字节豆包] #include <stdio.h> #include <stdlib.h> #include <stdbool.h> #define maxsize 100 typedef char elemtype; typedef struct node { elemtype data; struct node* lchild; struct node* rchild; } btnode; typedef struct { btnode* data[maxsize]; int top; } stacktype; void initstack(stacktype* st) { st->top = -1; } bool stackempty(stacktype* st) { return st->top == -1; } bool push(stacktype* st, btnode* e) { if (st->top < maxsize - 1) { st->data[++st->top] = e;(95点数解答 | 2024-12-10 13:17:25)214
- [阿里通义] + make tini-static Scanning dependencies of target tini-static [100%] Building C object CMakeFiles/tini-static.dir/src/tini.c.o Linking C executable tini-static /usr/bin/ld: cannot find -lc collect2: error: ld returned 1 exit status make[3]: *** [tini-static] Error 1 make[2]: *** [CMakeFiles/tini-static.dir/all] Error 2 make[1]: *** [CMakeFiles/tini-static.dir/rule] Error 2 make: *** [tini-static] Error 2 error: Bad exit status from /var/tmp/rpm-tmp.0XoVnu (%build)(264点数解答 | 2025-05-24 02:02:15)209
- [讯飞星火] + make tini-static Scanning dependencies of target tini-static [100%] Building C object CMakeFiles/tini-static.dir/src/tini.c.o Linking C executable tini-static /usr/bin/ld: cannot find -lc collect2: error: ld returned 1 exit status make[3]: *** [tini-static] Error 1 make[2]: *** [CMakeFiles/tini-static.dir/all] Error 2 make[1]: *** [CMakeFiles/tini-static.dir/rule] Error 2 make: *** [tini-static] Error 2 error: Bad exit status from /var/tmp/rpm-tmp.0XoVnu (%build)(571点数解答 | 2025-05-24 02:03:08)203
- [字节豆包] #include <iostream> #include <stack> #include <string> using namespace std; // 判断字符串是否为合法括号序列 bool isValid(const string& s) { stack<char> st; for (char c : s) { //遍历字符串s中每个字符赋值给c if (c == '(' || c == '[' || c == '{') { st.push(c); //放到栈中 ,位于栈底 } else { if (st.empty()) return false; //空 char top = st.top(); //查询栈顶 st.pop(); (685点数解答 | 2025-12-04 18:51:20)27
- [字节豆包] 学生类型ST的定义如下: typedef struct student{ char name[10],id[10]; int gender; int age; double scored; } ST; 编写函数,返回指定学生数组中的男生或女生的最高分的地址(约定:整数0和1分别代表男和女)。 函数接口定义: ST* MaxST(ST d[],int n,int k);//k=0|1 其中 d 是学生数组的初地址, n是数组的长度, k 是查找的性别描述(值确保是0或1),函数须返回指定类型学生中的最高分者的地址,如果不存在,返回空地址。 裁判测试程序样例: 在这里给出函数被调用进行测试的例子。例如: #include <stdio.h> #include <stdlib.h> typedef struct student{ char name[10],id[10]; int gender; int age; double scored; } ST; void output(ST *d){//输出一个记录 i(101点数解答 | 2025-12-12 13:22:54)33
- 编写程序模拟实现如下教学管理场景中的应用: (1) 设计学生、老师和课程类见下图, (2) 老师能够开课,学生能选课 (3)在主方法中测试见图2: |老师创建一门课程,并添加58位学生来选择该课程。 1统计出58位学生中男女数量及比重。 public class testcourse ( static course cour;//静态变量,学生的课程 public static void testcreatecourse(int num)( //num学生数 //创建老师 //老师tea 开设了课程 //设置课程的老师tea 的信息; //输出课程信息 //课程添加学生 public static void teststatics()(//统计男女学生个数,比例 public static void main(string[] args)(testcreatecourse(58); teststatics();} 程序模拟选课,运行效果如下: 林老师开课了. 课程名称:《java 程序设计》 开课学期:2020第二学期 开课老师编号:课程1001:名字: 林老师 学生:李 选择了(2653点数解答 | 2024-04-02 11:00:16)444