Added explanation of code

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Clyne 4 years ago committed by GitHub
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@ -25,3 +25,45 @@ Try it [on Compiler Explorer](https://godbolt.org/z/T-MFoh).
**Known issues:** **Known issues:**
* With C++17 GCC, `to_string` must be used to initialize variables; otherwise, the integer-string conversion is done at run-time. * 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;
}())];
```

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