我不知道这是否可行,但我会试着解释一下:
我有以下变量:
const char * VAR1 = "var1";
const char * VAR2 = "var2";
const char * VAR1_A = "etc1";
const char * VAR2_A = "etc2";
然后我宣布一个std:vector
并将变量放入:
vector <const char *> v1;
v1.push_back(VAR1);
v1.push_back(VAR2);
然后我迭代向量以这种方式找到匹配:
const char * param = "var1"; //Example parameter
for(int x=0; x<v1.size(); x++){
//Found a match in the vector
if(param == v1[x]){
//HERE is the point. I need to get the value of variable "[v1[x](NAME)]_A"
//Something like:
const char * varname = getVarName(v1[x]) + "_A"; //This would contain VAR1_A
const char * varvalue = getVarValueByVarName(varname); //This would contain "etc1";
break;
}
}
这对我的问题是一个很好的推理还是存在这种方法?
答案 0 :(得分:5)
只需使用std::map<std::string, std::string>
。
这样你可以这样做:
var geoCoder: CLGeocoder = CLGeocoder()
如果密钥(在本例中为字符串)不存在,则运算符map<string, string> myMap;
myMap["var1"] = "var1_value";
myMap["var2"] = "var2_value";
//say now you get input into a variable named 'str' for what variable the user wants to print
cout << myMap[str] << endl;
将添加到地图中,如果密钥存在,则会更改该值。