Is there a way to shorten this while condition?
Sure, you could just use a string of valid operators and search it.
#include <cstring>
// : :
const char* ops = "+-*/";
while(strchr(ops, temp->left->oper) || strchr(ops, temp->right->oper))
{
// do something
}
If you are concerned about performance, then maybe table lookups:
#include <climits>
// : :
// Start with a table initialized to all zeroes.
char is_op[1 << CHAR_BIT] = {0};
// Build the table any way you please. This way using a string is handy.
const char* ops = "+-*/";
for (const char* op = ops; *op; op++) is_op[*op] = 1;
// Then tests require no searching
while(is_op[temp->left->oper] || is_op[temp->right->oper])
{
// do something
}
Yes, indeed you can!
Store the valid-characters to a std::array
or even a plain array and apply the standard algorithm std::any_of
to it for checking the condition.
#include <array> // std::array
#include <algorithm> // std::any_of
static constexpr std::array<char, 4> options{ '+', '-', '*', '/' };
const auto tester = [temp](const char c) { return temp->left->oper == c || temp->right->oper == c; };
const bool isValid = std::any_of(options.cbegin(), options.cend(), tester);
while(isValid) // now the while-loop is simplified to
{
// do something
}
This can be more cleaned by packing into a function, which accepts the node
object to be checked.
#include <array> // std::array
#include <algorithm> // std::any_of
bool isValid(const node *const temp) /* noexcept */
{
static constexpr std::array<char, 4> options{ '+', '-', '*', '/' };
const auto tester = [temp](const char c) { return temp->left->oper == c || temp->right->oper == c; };
return std::any_of(options.cbegin(), options.cend(), tester);
}
which can be called in the while-loop
while (isValid(temp)) // pass the `node*` to be checked
{
// do something
}