When I try to compile, I get the error:
error: expected expression before 'char'
Here is the part it talks about:
Why is this not working?Code:int get_names(int students, char names[students][20]) { int i; for (i = 0; i < students; ++i) { printf("Enter name for Student %d: ", i); scanf("%s", names[i]); } return char names[students]; }
> return char names[students];
You've got a type here, and there shouldn't be.
Secondly, names[students] is both a char pointer, and a buffer overflow.
Yet the function says it is returning an int.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
