This error happens if a unknown object is used.

Variables

Not compiling:

#include <iostream>

int main(int argc, char *argv[])
{
    {
        int i = 2;
    }

    std::cout << i << std::endl; // i is not in the scope of the main function

    return 0;
}

Fix:

#include <iostream>

int main(int argc, char *argv[])
{
    {
        int i = 2;
        std::cout << i << std::endl;
    }

    return 0;
}

Functions

Most of the time this error occurs if the needed header is not included (e.g. using std::cout without #include <iostream>)

Not compiling:

#include <iostream>

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

    return 0;
}

void doCompile()
{
    std::cout << "No!" << std::endl;
}

Fix:

#include <iostream>

void doCompile(); // forward declare the function

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

    return 0;
}

void doCompile()
{
    std::cout << "No!" << std::endl;
}

Or:

#include <iostream>

void doCompile() // define the function before using it
{
    std::cout << "No!" << std::endl;
}

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

    return 0;
}

Note: The compiler interprets the code from top to bottom (simplification). Everything must be at least declared (or defined) before usage.