invalid conversion from 'char' to 'cha

Question

Ok so here are the parts of my code that I'm having trouble with:

char * historyArray;
historyArray = new char [20];

//get input
cin.getline(readBuffer, 512);       
cout << readBuffer <<endl;

//save to history
for(int i = 20; i > 0; i--){
    strcpy(historyArray[i], historyArray[i-1]); //ERROR HERE//  
}

strcpy(historyArray[0], readBuffer); //and here but it's the same error//

The error that i'm receiving is:

"invalid conversion from 'char' to 'char*' 
           initializing argument 1 of 'char* strcpy(char*, const char*)'

The project is to create a psudo OS Shell that will catch and handle interrupts as well as run basic unix commands. The issue that I'm having is that I must store the past 20 commands into a character array that is dynamically allocated on the stack. (And also de-allocated)

When I just use a 2d character array the above code works fine:

char historyArray[20][];

but the problem is that it's not dynamic...

And yes I do know that strcpy is supposed to be used to copy strings.

Any help would be greatly appreciated!

Answer 1

historyArray[i] is a char. It is a single character. You want to use a sting. Your fundemental problem is that historyArray is a char* which means that it points to a memory range containing characters. You want it to be a char** which is a pointer to a pointer to a string. Your initialization code would be

char** historyArray;
historyArray = new char* [20];
for (int i = 0; i < 20; i++)
{
    historyArray[i] = new char [512];  //Big enough to have a 512 char buffer copied in
}

Answer 2

historyArray points to (the first element of) an array of 20 chars. You can only store one string in that array.

In C, you could create a char** object and have it point to the first element of an array of char* objects, where each element points to a string. This is what the argv argument to main() does.

But since you're using C++, it makes a lot more sense to use a vector of strings and let the library do the memory management for you.

Answer 3

char * historyArray;
historyArray = new char [20];

//get input
cin.getline(readBuffer, 512);       
cout << readBuffer <<endl;

//save to history
for(int i = 20; i > 0; i--){
   strcpy(&(historyArray[i]), &(historyArray[i-1])); //ERROR HERE//  
}

strcpy(historyArray, readBuffer); //and here but it's the same error//

But that will only fix the compiler errors, not the logical errors in the code. Your using C++ so the string solution:

vector<string> history;

cin.getline(readBuffer,512);

history.push_back(readBuffer);

Alternatively if you want one long string containing everything from readBuffer:

string history;

cin.getline(readBuffer,512);
history = history += string(readBuffer);

For example...

Answer 4

Error 1: You're indexing past your array bounds with i being set to 20.

Error 2: historyArray[i] is a char, not a char *. You need &historyArray[i].

Answer 5

Two solutions. The first is if you for some reason really want arrays, the other is more recommended and more "C++"ish using std::strings.

char * historyArray[20]; // Create an array of char pointers

// ...

historyArray[i] = new char[SIZE]; // Do this for each element in historyArray

Then you can use strcpy on the elements in historyArray.

Second solution which I repeat is recommended (I've fixed a few other things):

string historyArray[20];

getline(cin, readBuffer); // Make readbuffer an std::string as well
cout << readBuffer << endl;

for(int i = 19; i > 0; i--){ // I think you meant 19 instead of 20
    historyArray[i] = historyArray[i-1];
}

historyArray[0] = readBuffer;

Answer 6

strcpy(&historyArray[i], &historyArray[i-1]);

Array notation gives references while strcopy wants pointers. Convert references to pointers with address-of (&) operator.