最后一个应该是是s[i]==‘;’,你漏掉了下标
1、std::string编译器是不认识的,只认识int,float,int*等类型,string在编译器里的类型是std::basic_string<char,std::char_traits<char>,std::allocator<char> > ; 2、pa(搜狗中文输入状态下输入“pa”,按下Enter是选中英文字符,按下空格键是选中“怕”,按下shift是选中英文字符,且切换到英文状态 ),如下...
include <iostream> include <string.h> int main(){ char a[100];int zimu=0,shuzi=0,kongge=0,qita=0,b,i=0;gets(a);b=strlen(a);while (i<=b){ if (('a'<=a[i-1]&&'z'>=a[i-1])||('A'<=a[i-1]&&'Z'>=a[i-1])){ zimu++;} else if ('0'<=a[i-1...
个参数是const char* data, 类型是const char*; 函数调用时 , 在const char* data参数位置 , 传入了unsigned char*类型的数据 ; 代码语言:javascript 复制 std::stringSearchCode(unsigned char*data,unsigned size){std::string strOut;strOut+=search_string(pModuleName,ver[i].address(),ver[i].realSize(...
没有可能从char*转化为int类型 意思是在你使用“==”做判断的时候，一边是char*类型，一边是int类型，这两个类型没有可比性，因为他们之间不能相互转化，所以出错了
1> strOut += search_string(pModuleName, ver[i].address(), ver[i].realSize() + ver[i].address(), 1> ^~~~ 1>./native/native.cpp:40:13: note: candidate function not viable: no known conversion from 'unsigned char *' to 'const char *' for 4th argument 1>std::string...
简介:【错误记录】Android NDK 编译报错 ( no known conversion from ‘unsigned char *‘ to ‘const char *‘ ) 文章目录 一、报错信息 二、解决方案 一、报错信息 在Visual Studio 2019 中编译 Android NDK , 构建方式参考 【Android 逆向】Android 进程注入工具开发 ( Visual Studio 开发 Android NDK 应用...
問題描述 字符轉換沒有字符串輸出 (No string output for char conversion) 我有一個 Char(1) 可空列(區),我想在 NULL 時輸出“未分配”; 我已經嘗試過 SELECT CASE District WHEN NULL THEN 'Not Assigned' ELSE District
include <stdio.h># include <windows.h># include <string.h># include <conio.h># define PASSWORD "123456"int main(){ char p[20],i=0; system("cls"); printf("请输入密码 = >：\n"); while(p[i]=getch()) { if(p[i]==13) break; if(p[i]!='\b') { ...
[7]" to "std::basic_string<char, std::char_traits<char>, std::allocator<char>>" 41 22 Fall2014monkeyBusiness 6 IntelliSense: no suitable constructor exists to convert from "std::string [7]" to "std::basic_string<char, std::char_traits<char>, std::allocator<char>>" 48 19 Fall...