我是个学生,用cpp开发了一个个人电脑客户端。我不知道如何处理哪个rapidjson与编码Unicode。我总是有个乱七八糟的密码。我对cpp很在行,怎么才能得到正确的结果呢?我会非常感激的!
举一个例子:
class Test {
// I have got the string of json
// eg: { "name" : "小明" }
public : void test(const std::string& data) {
rapidjson::Document json;
json.
我已经为.substr函数创建了一个包装器:
wstring MidEx(wstring u, long uStartBased1, long uLenBased1)
{
//Extracts a substring. It is fail-safe. In case we read beyond the string, it will just read as much as it has
// For example when we read from the word HELLO , and we read from position 4, len 5000, it wi
编译器告诉我,我的代码中有3个错误: Linking specification incompatible with "SpeakInternal" (declare in line 13 of voice.cpp) (previously)
wstring: not declared identifier
Syntax error: Missing ")" before identifier "uText" 代码是: __declspec(dllexport) void __cdecl SpeakInternal(wstring uTex
我想使用以下代码: wstring s = L"GetLockCount " + g_SpeakerAlias.c_str();
MessageBoxW(NULL, unsignedinttowstring(iLockCount).c_str(), s, MB_OK); 编译器告诉我: The expression must have an integer enumeration type or an enumeration type without range restriction 我尝试连接两个wstring的方式似乎是错误的: L"GetLockCoun