PHP. Pass variable by reference vs string. How to works with these two different arguments?
问题 I'm writing my own debug functions and I need some help to fix the code below. I'm trying to print a variable and its name, the file where the variable and the function was declared and the line of the function call. The first part I did, the variable, the variable name, the file and the line is printed correctly. At the code, a($variable) works good. The problem is I'd like this function accepts a string too, out of a variable. But PHP returns with a fatal error ( PHP Fatal error: Only