Pass string array to method

Question

I am a beginner to C++, I have used C before but never used C++. This is one of my first programs, and it's supposed to do something really simple, however I'm unable to even pass strings between methods... When I call the method setMode with a string array, the method instance recieves an empty array, and not the one I've sent. Why is this happening?

#include <iostream>
#include <fstream>
#include <string>
#include <sstream>

using namespace std;

#define LED_PATH "sys/class/leds/beaglebone:green:usr"

class LED{

    private:
        string path;
        int number;

    public:
        LED(int number);
        virtual void setMode(string mode[]);
        virtual ~LED();
};


LED::LED(int number){
    ostringstream fs;
    this->number = number;
    fs << LED_PATH << number;
    this->path = string(fs.str());

    cout << this->path << endl;
}

LED::~LED(){

}

void LED::setMode(string mode[]){
    //Will use all fields of mode[] in the future
    cout << "setMode: mode: " << mode[0].c_str() << endl;

}


int main(int argc, char* argv[]){

    LED LEDs[4] = {LED(0), LED(1), LED(2), LED(3)};
    string mode[argc-1];

    //TODO Perform argc value safety check
    for(int i=1; i<argc; i++){
        mode[i] = string(argv[i]);
        cout << mode[i].c_str()<<endl;
    }

    for(int i=0; i<4; i++){
            LEDs[i].setMode(mode);
    }

    return 0;

}

Output:

debian@beaglebone:~/Desktop/LED_Cpp$ ./led on 1
sys/class/leds/beaglebone:green:usr0
sys/class/leds/beaglebone:green:usr1
sys/class/leds/beaglebone:green:usr2
sys/class/leds/beaglebone:green:usr3
on
1
setMode: mode:
setMode: mode:
setMode: mode:
setMode: mode:

Also, why are you passing the whole array, when setMode() only needs a single string parameter? — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Oct 31, 2015 at 12:29
The method is not yet implemented, it will used the whole array in the future. — AmiguelS
– AmiguelS, Commented Oct 31, 2015 at 12:30

Christian Hackl · Accepted Answer · 2015-10-31 15:31:08Z

string mode[argc-1];

This uses a proprietary GCC extension. In standard C++, a raw array's size must be known at compile time.

You need something like this instead:

if (argc < 4) {
    std::cerr << "error\n";
    return EXIT_FAILURE;
}
string mode[4];

Or, what would be very much preferable, use std::vector:

std::vector<std::string> mode(argc-1);

for(int i=1; i<argc; i++){
    mode[i] = string(argv[i]);
    cout << mode[i].c_str()<<endl;
}

This will leave mode[0] empty.

for(int i=0; i<4; i++){
        LEDs[i].setMode(mode);
}

The array (or std::vector) passed to setMode will thus always be such that the first element is an empty string.

void LED::setMode(string mode[]){

This is an attempt to pass an array to a function. What happens really is that a pointer to the first element is passed and the size information is lost.

The correct way of passing a raw array including its size information would be to use a reference:

 void LED::setMode(string (&mode)[4])

But as I mentioned previously, just use a std::vector and you'll be fine. When you need to modify the vector, pass it via &, else via const&:

 void LED::setMode(std::vector<std::string> const& mode)

In either case, inside of the method, you currenly just access the first element:

cout << "setMode: mode: " << mode[0].c_str() << endl;

As we've established before, mode[0] is always empty. That's why nothing is printed.

Captain Giraffe · Accepted Answer · 2015-10-31 12:37:24Z

2

You are off by one. You are writing to

mode[1]

in

for(int i=1; i<argc; i++){
    mode[i] = string(argv[i]);

and you use

mode[0]

for output.

answered Oct 31, 2015 at 12:37

Captain Giraffe

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