[Solved] ‘No suitable conversion function’ with cout [closed]

Question

The problem is in the definition of your main function. You need to add a parameter list. Either of these are valid:

int main() { .... }
int main(int argc, char** argv) { ... }

Accepted Answer

The problem is in the definition of your main function. You need to add a parameter list. Either of these are valid:

int main() { .... }
int main(int argc, char** argv) { ... }