How to use varargs for small logger? - c

Question

I am looking to make a small logger to be used like this:
logger log;
const char* text = "World";
log.write("Hello %s", text);
log.write("Step %d", 1);
This is my code, which doesn't work correctly:
class logger
{
public:
void write(const char* msg, ...)
{
FILE* file = fopen("c:/test.txt", "a");
if(file != NULL)
{
va_list args;
va_start(args, msg);
fprintf(file, "%s\n", msg, args);
va_end(args);
fclose(file);
}
}
};
This is what I get:
Hello %s
Step %d
I never used varargs before so I am not sure if I am using it correctly.

Answer 1

You're thinking that C behaves in a manner similar to Java or Python or other languages that can "splat" an array argument to a function that accepts varargs, but C isn't that sophisticated. When you pass args to fprintf, C literally pushes the value of args (a variable of type va_list) onto the stack. What you need instead is to push the contents of args onto the stack.
Or, instead, you could use a function that accepts a va_list as a parameter. That function is vprintf (and friends vsprintf, vfprintf, etc.).
You have another issue, which is that you're fprintf-ing the caller's "msg" parameter using "%s" but then apparently expecting fprintf to recursively use the result as a format string and fprintf the args too. That's not going to work. Instead just use the caller's msg as the format string.
va_list args;
va_start(args, msg);
vfprintf(file, msg, args);
fputc('\n', file);
va_end(args);