Thank you! That is a good point. I hope this article was helpful. Your comment helped me understand a better way to solve this problem set, and i’m sure it will help other programmers who read the comments. Happy new year to you!

1) this segment:

given_octave = atof(&note[octave_pos]);
exponent = fabs(given_octave – STANDARD_OCTAVE);

if ( given_octave > STANDARD_OCTAVE)
frequency = STANDARD_HZ * pow(2, exponent);

if ( given_octave <= STANDARD_OCTAVE)
frequency = STANDARD_HZ / pow(2, exponent);

is absolutely fine without the "if" loop because there is absolutely no need to calculate the absolute value of the exponent. You are fine like this:

given_octave = atof(&note[octave_pos]);
exponent = (given_octave – 4);
frequency = 440.00 * pow(2, exponent);

2) The following segment

int semitones = 0;
char given_letter = note[0];
char white_keys[] = {'C', '#', 'D', '#', 'E', 'F', '#', 'G', '#', 'A', '#', 'B'};
int a_index = 9;
if (given_letter != 'B')
{
while(white_keys[a_index] != given_letter)
{
semitones–;
a_index–;
}
}
else
{
while(white_keys[a_index] != given_letter)
{
semitones++;
a_index++;
}
}

could obviously be simplified a little bit to exclude the 'if' loop

int semitones = 2;
char given_letter = note[0];
char white_keys[] = {'C', '#', 'D', '#', 'E', 'F', '#', 'G', '#', 'A', '#', 'B'};
int a_index = 11;

while(white_keys[a_index] != given_letter)
{
semitones–;
a_index–;
}