在C ++中返回结构的向量时，没有用于调用[class]的匹配函数

Question

I get an error at the line indicated below: I get this error when i try to return a vector of structures. I used a template to be able to return these vectors. Im not where my error is. If i do not use the template, set the return type to vector.

Header File:

typedef unsigned long ulong_t;
class Tool {
public:
    Tool();
void toolInterface();
void run();
private:

ifstream allDevicesFile;
struct devStats {
    //structure variables,
};
template<class devStats>
vector<devStats> readDev();
vector<devStats> stats;
string fileNameAll;
Report report;
Commandline cmd;
Configuration conf;
Devices dev;
};

Tool.cpp:

Tool::Tool() {
fileNameAll = "/proc/diskstats";
    allDevicesFile.open((char*)fileNameAll.c_str());
    if (allDevicesFile.fail()) {
        cout << "Could not open /proc/diskstats\n";
    }// TODO Auto-generated constructor stub

}
template<class devStats>
vector<devStats> Tool::readDev() {
devStats dev;
while (!allDevicesFile.eof()) {
    allDevicesFile >> dev.decoy1;
    allDevicesFile >> dev.decoy2;
    allDevicesFile >> dev.devName;
    allDevicesFile >> dev.reads;
    allDevicesFile >> dev.readMerge;
    allDevicesFile >> dev.writes;
    allDevicesFile >> dev.secReading;
    allDevicesFile >> dev.mSecondsRead;
    allDevicesFile >> dev.writeCompleted;
    allDevicesFile >> dev.secWritting;
    allDevicesFile >> dev.mSecondWrite;
    allDevicesFile >> dev.currentI_O;
    allDevicesFile >> dev.mSecondsI_O;
    allDevicesFile >> dev.weightedI_O;
    dev.mSecondsRead = dev.mSecondsRead / 1000;
    dev.mSecondWrite = dev.mSecondWrite / 1000;
    dev.mSecondsI_O = dev.mSecondsI_O / 1000;
    stats.push_back(dev);
}
cout << stats[0].devName;
return stats;
}

void Tool::run() {
stats = readDev(); //error occurs here.
}

Answer 1

Ordinary overload resolution does not consider return type. The return type is not matched against anything. So there is no way that the template parameter can be automatically resolved.

As a practical solution, just remove the templatization.

---

^{There is a special case where a template parameter that is only used as return type, can be resolved by the calling context, and that's for a conversion operator.}