Replacing a string with 2 other strings separatly [closed] - c++

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I was looking for replacing a character in a string with a character in another string comparing to another string.
Well, hard to describe this, a code would be better :
std::string alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
std::string keyboard = "AZERTYUIOPQSDFGHJKLMWXCVBNazertyuiopqsdfghjklmwxcvbn";
std::string change(std::string str)
{
for (short i = 0; i < alphabet.length(); i++)
{
std::cout << str[i] << std::endl;
}
std::cout << str << std::endl;
system("PAUSE");
return str;
}
Well, I don't think it is understandable, so what I try to do is :
Find (maybe Find() ?) the letter in str and get the i-variable in alphabet.
With the i-variable want to get the pos and then get the letter in the keyboard string.
Hope you understand what I mean, and thank you.
I though about replace(), find() and swap(), but didn't work, I maybe did something wrong.
Thank you in advance !
EDIT : Well, I was sure you won't understand anything, imagine str = "Hello", I want to replace H per Y (alphabet = H -> keyboard = I)? (alphabet get the position of the H in the alphabet string, we memorise this position in i, and we keyboard[i]), this is what I want to do.

Answer 1

I propose
template <typename InputIterator,
typename BidirIterator,
typename ForwardIterator>
InputIterator substitute( InputIterator first, InputIterator last,
BidirIterator key_first, BidirIterator key_last,
ForwardIterator val_first )
{
for (; first != last; ++first)
{
auto&& ref = *first;
auto i = std::lower_bound( key_first, key_last, ref );
if (i == key_last || *i != ref)
return first;
ref = *std::next(val_first, std::distance(key_first, i));
}
return last;
}
This assumes that [key_first, key_last) is sorted and that val_first refers to a range that is at least as long as [key_first, key_last).
Demo.
The idea is that we iterate through all elements of a range using iterators whose type is to be determined - i.e. we can use both char const[]s and std::vector<char>s. We pick the current element and use a binary search to find that element within the key set. Then we use the offset to access the substitution counterpart in the value set, and assign that to our original element.
We use lower_bound to do the binary search. It doesn't itself constitute one, so we have to apply some checks to the return value to ensure the iterator refers to the correct value. If we couldn't find the key then the iterator to the first non-working element is returned. (That can be the past-the-end iterator.)
The avove is a general method that works for arbitrary key/value sets: For alphabetic shuffling there are (platform dependent) methods using lookup tables. Rustically this could look like this:
template <typename InputIterator>
void substitute( InputIterator first, InputIterator last,
char const* values )
{
char lookup[UCHAR_MAX+1];
for (int c = 0; c != sizeof lookup; ++c)
if (std::isupper(c))
lookup[c] = values[c - 'A'];
else if (std::islower(c))
lookup[c] = values[c - 'a' + 26];
else
lookup[c] = c;
// Further adjustments to the table can be made here
std::transform( first, last, first,
[&] (unsigned char c) {return lookup[c];} );
}
Demo. It depends on stuff that normal character sets should easily suffice. The table can also be created elsewhere and passed to transform via the unwrapped one-liner. This is especially efficient for large string transformations as no branching is necessary in the implicit loop in transform.

Answer 2

So, you want to replace every char in your string with a char from keyboard string using original char's position in alphabet?
I would consider using a map alphabet char to keyboard char. Then iterate through string and replace its characters.
#include <string>
#include <algorithm>
#include <map>
int main(int argc, char* argv[])
{
std::map<char, char> alphabetToKeyboard;
alphabetToKeyboard['A'] = 'A';
alphabetToKeyboard['B'] = 'Z';
alphabetToKeyboard['C'] = 'E';
std::string str = "ABC";
std::transform(str.begin(), str.end(), str.begin(), [&alphabetToKeyboard](char c) {
return alphabetToKeyboard[c];
});
return 0;
}