Understanding implicit conversions for printf

Question

The C99 Standard differentiate between implicit and explicit type conversions (6.3 Conversions). I guess, but could not found, that implicit casts are performed, when the target

Answer 1

printf("INT_MIN as double (or float?): %e\n", a);

Above line has problem You can not use %e to print ints. The behavior is undefined.

You should use

printf("INT_MIN as double (or float?): %e\n", (double)a);

or

double t = a;
printf("INT_MIN as double (or float?): %e\n", t);

Related post: This post explains how using incorrect print specifiers in printf can lead to UB.