Passing structs to functions

Question

I am having trouble understanding how to pass in a struct (by reference) to a function so that the struct's member functions can be populated. So far I have written:

bool data(struct *sampleData)
{

}

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

      struct sampleData {

        int N;
        int M;
        string sample_name;
        string speaker;
     };
         data(sampleData);

}

The error I get is:

C++ requires a type specifier for all declarations bool data(const &testStruct)

I have tried some examples explained here: Simple way to pass temporary struct by value in C++?

Hope someone can Help me.

Answer 1

Passing structs to functions by reference: simply :)

#define maxn 1000

struct solotion
{
    int sol[maxn];
    int arry_h[maxn];
    int cat[maxn];
    int scor[maxn];

};

void inser(solotion &come){
    come.sol[0]=2;
}

void initial(solotion &come){
    for(int i=0;i<maxn;i++)
        come.sol[i]=0;
}

int main()
{
    solotion sol1;
    inser(sol1);
    solotion sol2;
    initial(sol2);
}

Answer 2

First, the signature of your data() function:

bool data(struct *sampleData)

cannot possibly work, because the argument lacks a name. When you declare a function argument that you intend to actually access, it needs a name. So change it to something like:

bool data(struct sampleData *samples)

But in C++, you don't need to use struct at all actually. So this can simply become:

bool data(sampleData *samples)

---

Second, the sampleData struct is not known to data() at that point. So you should declare it before that:

struct sampleData {
    int N;
    int M;
    string sample_name;
    string speaker;
};

bool data(sampleData *samples)
{
    samples->N = 10;
    samples->M = 20;
    // etc.
}

And finally, you need to create a variable of type sampleData. For example, in your main() function:

int main(int argc, char *argv[]) {
    sampleData samples;
    data(&samples);
}

Note that you need to pass the address of the variable to the data() function, since it accepts a pointer.

However, note that in C++ you can directly pass arguments by reference and don't need to "emulate" it with pointers. You can do this instead:

// Note that the argument is taken by reference (the "&" in front
// of the argument name.)
bool data(sampleData &samples)
{
    samples.N = 10;
    samples.M = 20;
    // etc.
}

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

    // No need to pass a pointer here, since data() takes the
    // passed argument by reference.
    data(samples);
}

Answer 3

bool data(sampleData *data)
{
}

You need to tell the method which type of struct you are using. In this case, sampleData.

Note: In this case, you will need to define the struct prior to the method for it to be recognized.

Example:

struct sampleData
{
   int N;
   int M;
   // ...
};

bool data(struct *sampleData)
{

}

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

      sampleData sd;
      data(&sd);

}

Note 2: I'm a C guy. There may be a more c++ish way to do this.