Printing out table column names - c++

Question

I currently have this function to print out every rows of my tables
static int callback(void *NotUsed, int argc, char **argv, char **szColName)
{
for(int i = 0; i < argc; i++)
{
cout.width(17); cout << left << argv[i];
}
std::cout << "\n";
return 0;
}
How do I print out szColName, such that it only appear once on top, and not multiple occurences of it?
Tried this:
static int callback(void *NotUsed, int argc, char **argv, char **szColName)
{
int n = sizeof(szColName) / sizeof(szColName[0]);
for (int i = 0; i < n; i++)
{
cout.width(17); cout << left << szColName[i];
}
printf("\n");
for(int i = 0; i < argc; i++)
{
cout.width(17); cout << left << argv[i];
}
std::cout << "\n";
return 0;
}
But it outputs everytime after a outputting the row values

Answer 1

You might want to declare a static bool inside callback to record whether it has already printed out the column names. Or, if you want to be able to reset it...
such as:
static bool firstline = true;
static int callback(void *NotUsed, int argc, char **argv, char **szColName)
{
if (firstline){
int n = sizeof(szColName) / sizeof(szColName[0]);//this is incorrect but fixing
// it requires changing the prototype.
//See the comments below
for (int i = 0; i < n; i++)
{
cout.width(17); cout << szColName[i] << left;
}
printf("\n");
firstline=false;
}
for(int i = 0; i < argc; i++)
{
cout.width(17); cout << argv[i] << left;
}
std::cout << "\n";
return 0;
}
int main(){
for(int x=0;x<10;++x)callback( ... , ... , ... ); // give whatever argument you need to give
firstline = true; //reset the variable so that next time you call it, the col names will appear
for(int x=0;x<10;++x)callback(...,...,...);// now the col names will appear again.
}
I assume that what you provide would print out the rows and the column names correctly. I only added a variable to check if printing column names are needed.