Question

我使用了一个函数来计算我在地图中初始化的某些指令的信息，比如

void get_objectcode(char*&token1,const int &y)
{
map<string,int> operations;
    operations["ADD"] = 18;
    operations["AND"] = 40;
    operations["COMP"] = 28;
    operations["DIV"] = 24;
    operations["J"] = 0X3c;
    operations["JEQ"] =30;
    operations["JGT"] =34;
    operations["JLT"] =38;
    operations["JSUB"] =48;
    operations["LDA"] =00;
    operations["LDCH"] =50;
    operations["LDL"] =55;
    operations["LDX"] =04;
    operations["MUL"] =20;
    operations["OR"] =44;
    operations["RD"] =0xd8;
    operations["RSUB"] =0x4c;
    operations["STA"] =0x0c;
    operations["STCH"] =54;
    operations["STL"] =14;
    operations["STSW"] =0xe8;
    operations["STX"] =10;
    operations["SUB"] =0x1c;
    operations["TD"] =0xe0;
    operations["TIX"] =0x2c;
    operations["WD"] =0xdc;

         if  ((operations.find("ADD")->first==token1)||(operations.find("AND")->first==token1)||(operations.find("COMP")->first==token1)
            ||(operations.find("DIV")->first==token1)||(operations.find("J")->first==token1)||(operations.find("JEQ")->first==token1)
            ||(operations.find("JGT")->first==token1)||(operations.find("JLT")->first==token1)||(operations.find("JSUB")->first==token1)
            ||(operations.find("LDA")->first==token1)||(operations.find("LDCH")->first==token1)||(operations.find("LDL")->first==token1)
            ||(operations.find("LDX")->first==token1)||(operations.find("MUL")->first==token1)||(operations.find("OR")->first==token1)
            ||(operations.find("RD")->first==token1)||(operations.find("RSUB")->first==token1)||(operations.find("STA")->first==token1)||(operations.find("STCH")->first==token1)||(operations.find("STCH")->first==token1)||(operations.find("STL")->first==token1)
            ||(operations.find("STSW")->first==token1)||(operations.find("STX")->first==token1)||(operations.find("SUB")->first==token1)
            ||(operations.find("TD")->first==token1)||(operations.find("TIX")->first==token1)||(operations.find("WD")->first==token1))

            {
                int y=operations.find(token1)->second;
                //cout<<hex<<y<<endl;
            }

        return ;
}

如果我在函数中输入y给我一个答案就好了，这就是我需要的但是有一个问题是从函数中返回值以便我可以在函数外部使用它，它给出了一个完全不同的答案，问题是什么

Answer 1

函数中的第二个参数是常量引用。尝试更换 -

void get_objectcode(char*&token1,const int &y)

与

void get_objectcode(char*&token1,int &y)

并在if条件中，删除y的新声明并将其替换为 -

y=operations.find(token1)->second;

希望这有帮助！

Answer 2

这可能更接近你想要的东西：

void get_objectcode(const char *token, int &y)
{
    typedef std::map<std::string,int> OpMap;
    OpMap operations;
    operations["ADD" ] = 18;
    operations["AND" ] = 40;
    operations["COMP"] = 28;
    operations["DIV" ] = 24;
    // etc.
    operations["SUB" ] = 0x1c;
    operations["TD"  ] = 0xe0;
    operations["TIX" ] = 0x2c;
    operations["WD"  ] = 0xdc;

    OpMap::iterator result = operations.find(token);

    // note: assigns 0 to y if token is not found
    y = (result == operations.end()) ? 0 : result->second;

    //std::cout << std::hex << y << std::endl;
}

Answer 3

看一下boost assign map_list_of here。它可用于分配地图。然后使用地图中的find方法（引用为here）

Answer 4

您的地图仅存在于该功能中。因此，元素仅存在于该函数中。如果在您调用函数的位置，您正在使用y初始化引用，那么这将是对将不复存在的元素的引用。

每次调用函数时都不应该真正创建映射，至少最好只从函数中返回值：

std::map<std::string,int> operations;
operations["ADD"]  = 18;
operations["AND"]  = 40;
operations["COMP"] = 28;
operations["DIV"]  = 24;
operations["J"]    = 0X3c;
operations["JEQ"]  = 30;
operations["JGT"]  = 34;
operations["JLT"]  = 38;
operations["JSUB"] = 48;
operations["LDA"]  = 00;
operations["LDCH"] = 50;
operations["LDL"]  = 55;
operations["LDX"]  = 04;
operations["MUL"]  = 20;
operations["OR"]   = 44;
operations["RD"]   = 0xd8;
operations["RSUB"] = 0x4c;
operations["STA"]  = 0x0c;
operations["STCH"] = 54;
operations["STL"]  = 14;
operations["STSW"] = 0xe8;
operations["STX"]  = 10;
operations["SUB"]  = 0x1c;
operations["TD"]   = 0xe0;
operations["TIX"]  = 0x2c;
operations["WD"]   = 0xdc;

int get_objectcode(const std::string& key)
{
    std::map<std::string, int>::iterator it = operations.find(key);
    if (it == operations.end())
        return -1;
    else
        return it->second;
}

如何从函数返回值

4 个答案: