6
votes
Accepted
Pset3 Helpers.c Frequency Not Working
For "A4", this returns 440 for me.
Note that C's ^ operator is bitwise XOR, not power. Also, in C, 1/12 equals 0 (integer division truncates, you could avoid this by using floating point values like ...
- 21k
3
votes
Accepted
Problem with pow() function, int frequency, helpers.c, pset3, music
(pow(2, index / 12)
the issue is that both index and 12 are integers, so with integer division, the result will always be a whole number.
Try using 12.0 instead of 12 to force a floating point ...
- 18.7k
2
votes
Accepted
Music frequency accidental and note issue
When you write 9/12, that's an integer division. Integer division results in an integer again, truncating the result. You want a floating point division instead. For that, make at least one of the ...
- 21k
1
vote
Need help with Frequency in pset3. Been stuck for a while
IMPORTANT NOTE
This exercise is not included in CS50x 2019. The courseware changed at the beginning of 2019; music has been retired. Which is not to say you shouldn't persevere. But as far as I know, ...
- 28.6k
1
vote
Accepted
Pset3 Frequency Algorithm
Your semitones should be relative to A of the same octave. But C of the same octave is -9 semitones, not +3. That one would be C of the next octave, explaining the double frequency the test found.
- 21k
1
vote
signed integer overflow
The issue with this was that I have signed char to int and havent changed anything.
I asummed that if char is 4 that the int will be also 4 but it is not (its 52) so that's why my loop was giving so ...
- 19
1
vote
Accepted
pset3 load of null pointer of type 'char'
You can use note as a char * or as a char[], because it is a string.
So you can use it as an array of char's, or as a pointer to a char.
But you are using it as a pointer to a char[] ...
You should ...
- 1,009
1
vote
Accepted
Why are these if-else statements not letting my program compile?
Look at the following code:
if(octave == 4)
{
if(note[0] == 'A' && accidental == '#')
answer = (440 * pow((double)2, (double)1/12));
return answer;
else if(note[0] == ...
- 69.3k
1
vote
Accepted
PSET3 - Frequency giving wrong octave result
note[strlen(note) - 1] is the ASCII value of the digit. You would need to subtract the character '0' (that's same as subtracting 48) to find the value you expect.
So you could use note[strlen(note) - ...
- 21k
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