Added explanation of code

README.md

@@ -25,3 +25,45 @@ Try it [on Compiler Explorer](https://godbolt.org/z/T-MFoh).
 **Known issues:**
 
 * With C++17 GCC, `to_string` must be used to initialize variables; otherwise, the integer-string conversion is done at run-time.
+
+# How it works
+
+The basic structure of `to_string` is shown below:
+
+```cpp
+template<auto N, unsigned int base, /* N type-check and base bounds-check */>
+struct to_string_t {
+    char buf[];                          // Size selection explained later.
+    constexpr to_string_t() {}           // Converts the integer to a string stored in buf.
+    constexpr operator char *() {}       // These allow for the object to be implicitly converted
+    constexpr operator const char *() {} // to a character pointer.
+};
+
+template<auto N, unsigned int base = 10>
+to_string_t<N, base> to_string;          // Simplifies usage: to_string_t<N, base>() becomes to_string<N, base>.
+```
+
+Since the number and base are template parameters, each differing `to_string` use will get its own character buffer.
+
+The integer/string conversion is done using a simple method I learned over the years, where the string is built in reverse using `n % base` to calculate the value of the lowest digit:
+
+```cpp
+constexpr to_string_t() {
+    auto ptr = buf + sizeof(buf) / sizeof(buf[0]);
+    *--ptr = '\0';
+    for (auto n = N < 0 ? -N : N; n; n /= base)
+        *--ptr = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n % base];
+    if (N < 0)
+        *--ptr = '-';
+}
+```
+
+As you may have noticed, `buf` needs to be given a size for all this to work; in fact, the above code relies on the buffer having a size equal to the generated string (or else `buf[0]` would still be uninitialized). This is actually the case: a lambda is used within `buf`'s declaration to count how many characters long the string will ultimately be. This counting is done in a manner similar to conversion loop shown above:
+
+```cpp
+char buf[([] {
+              unsigned int len = N >= 0 ? 1 : 2; // Need one byte for '\0', two if there'll be a minus
+              for (auto n = N < 0 ? -N : N; n; len++, n /= base);
+              return len;
+          }())];
+```