I am having a problem reading in a portion of a text file using fscanf. The difficulty is only occurring with the characters right before the end of the line.
The text file I am reading in is of the format:
38 59 26.21 N
76 56 24.20 W
...
What I need the code to do is read in the values, convert them into radians and then determine whether or not the sign needs to be adjusted (if it is "W" it needs to be negative).
for ( int i = 0; i < 10; i ++) {
// accepts one line of coordinates at a time from the text file
fscanf_s(f, "%lf %lf %lf %s\n", °, &min, &sec, coordinate);
printf("%s\n", coordinate);
// converting into radians
val = (deg*(pi/180)) + ((min/60) * (pi/180)) + ((sec/3600) * (pi/180));
// checking coordinate to decide whether or not to change the sign
if ((strcmp(coordinate, "S") == 0) || (strcmp(coordinate, "W") == 0))
{
val = val*-1;
}
// are dealing with
if (EVEN(i))
{
lat[count1][0] = val;
count1++;
}
else
{
lon[count2][0] = val;
count2++;
}
}
So far this code works perfectly in reading in the floating point numbers in the text file and converting them to radians, however it is not storing the characters at the end of the line for some reason.
I have looked all over and cannot figure out why this is the case.
Thank you for your help!
Related
I'm having some trouble trying to build a pyramid out of "#" for mario.c in problem set 1. My program seems to be having some trouble with handling input.
I get these errors:
:( handles a height of 1 correctly
expected ""#"", not "" #"\n"##""
:( handles a height of 2 correctly
expected "" #"\n"##"", not "" #"\n" ##"\n..."
:( handles a height of 8 correctly
expected "" #"\n" ...", not "" #"\n"..."
:( rejects a height of 9, and then accepts a height of 2
expected "" #"\n"##"", not "" #"\n" ##"\n..."
Here is my code:
#include <cs50.h>
#include <stdio.h>
int main(void)
{
int height;
do
{
height = get_int("How many rows for the pyramid?\n");
}
while (height < 1 || height > 8); //This loops until the user input 1 - 8 rows
for (int row = 0; row <= height; row++) //This says to print a new row until it reaches the user inputted height
{
for (int spaces = (height - (row + 1)); spaces >= 0; spaces--) //This formula prints less spaces as more rows are printed
{
printf(" ");
}
for (int hashes = 1; hashes <= (row + 1); hashes++) //This formula prints more hashes as more rows are printed
{
printf("#");
}
printf("\n");
}
}
Any help is appreciated!!
Your code prints an extra line of hashtags creating a pyramid which is 1 greater in height than the user input. You can see this in the error message you get:
:( handles a height of 1 correctly expected ""#"", not "" #"\n"##""
Hence, the output of your program is #\n## which equates to
#
##
The required output for a height of 1 is:
#
To fix your issue, you have to iterate through your code one lesser time than you do currently for any number. Hence, perhaps consider changing the condition: row <= height; in your for loop.
Consider changing the comparison to just <. By doing this, the code inside the for loop will run one lesser time.
Very new to C#, so this could be a silly question.
I am working with alot of UInt64's. These are expressed as hex right? If we look at its binary representation, can we return such an array that if we apply the 'or' operation to, we will arrive back at the original UInt64?
For example, let's say
x = 1011
Then, I am looking for an efficient way to arrive at,
f(x) = {1000, 0010, 0001}
Where these numbers are in hex, rather than binary. Sorry, I am new to hex too.
I have a method already, but it feels inefficient. I first convert to a binary string, and loop over that string to find each '1'. I then add the corresponding binary number to an array.
Any thoughts?
Here is a better example. I have a hexadecimal number x, in the form of,
UInt64 x = 0x00000000000000FF
Where the binary representation of x is
0000000000000000000000000000000000000000000000000000000011111111
I wish to find an array consisting of hexadecimal numbers (UInt64??) such that the or operation applied to all members of that array would result in x again. For example,
f(x) = {0x0000000000000080, // 00000....10000000
0x0000000000000040, // 00000....01000000
0x0000000000000020, // 00000....00100000
0x0000000000000010, // 00000....00010000
0x0000000000000008, // 00000....00001000
0x0000000000000004, // 00000....00000100
0x0000000000000002, // 00000....00000010
0x0000000000000001 // 00000....00000001
}
I think the question comes down to finding an efficient way to find the index of the '1's in the binary expansion...
public static UInt64[] findOccupiedSquares(UInt64 pieces){
UInt64[] toReturn = new UInt64[BitOperations.PopCount(pieces)];
if (BitOperations.PopCount(pieces) == 1){
toReturn[0] = pieces;
}
else{
int i = 0;
int index = 0;
while (pieces != 0){
i += 1;
pieces = pieces >> 1;
if (BitOperations.TrailingZeroCount(pieces) == 0){ // One
int rank = (int)(i / 8);
int file = i - (rank * 8);
toReturn[index] = LUTable.MaskRank[rank] & LUTable.MaskFile[file];
index += 1;
}
}
}
return toReturn;
}
Your question still confuses me as you seem to be mixing the concepts of numbers and number representations. i.e. There is an integer and then there is a hexadecimal representation of that integer.
You can very simply break any integer into its base-2 components.
ulong input = 16094009876; // example input
ulong x = 1;
var bits = new List<ulong>();
do
{
if ((input & x) == x)
{
bits.Add(x);
}
x <<= 1;
} while (x != 0);
bits is now a list of integers which each represent one of the binary 1 bits within the input. This can be verified by adding (or ORing - same thing) all the values. So this expression is true:
bits.Aggregate((a, b) => a | b) == input
If you want hexadecimal representations of those integers in the list, you can simply use ToString():
var hexBits = bits.Select(b => b.ToString("X16"));
If you want the binary representations of the integers, you can use Convert:
var binaryBits = bits.Select(b => Convert.ToString((long)b, 2).PadLeft(64, '0'));
I found a script that converts binary to string but how can I input a string and get the binary representation? so say I put in "P" I want it to output 01010000 as a string.
I have this but it is not what I am trying to do - it converts a string containing a binary number into a real value of that number:
///string_to_binary(string)
var str = argument0;
var output = "";
for(var i = 0; i < string_length(str); i++){
if(string_char_at(str, i + 1) == "0"){
output += "0";
}
else{
output += "1";
}
}
return real(output);
Tip: search for GML or other language term, these questions answered many times. Also please check your tag as it is the IDE tag, not language tag.
Im not familiar with GML myself, but a quick search showed this:
At least semi-official method for exactly this: http://www.gmlscripts.com/script/bytes_to_bin
/// bytes_to_bin(str)
//
// Returns a string of binary digits, 1 bit each.
//
// str raw bytes, 8 bits each, string
//
/// GMLscripts.com/license
{
var str, bin, p, byte;
str = argument0;
bin = "";
p = string_length(str);
repeat (p) {
byte = ord(string_char_at(str,p));
repeat (8) {
if (byte & 1) bin = "1" + bin else bin = "0" + bin;
byte = byte >> 1;
}
p -= 1;
}
return bin;
}
GML forum (has several examples) https://www.reddit.com/r/gamemaker/comments/4opzhu/how_could_i_convert_a_string_to_binary/
///string_to_binary(string)
var str = argument0;
var output = "";
for(var i = 0; i < string_length(str); i++){
if(string_char_at(str, i + 1) == "0"){
output += "0";
}
else{
output += "1";
}
}
return real(output);
And other language examples:
C++ Fastest way to Convert String to Binary?
#include <string>
#include <bitset>
#include <iostream>
using namespace std;
int main(){
string myString = "Hello World";
for (std::size_t i = 0; i < myString.size(); ++i)
{
cout << bitset<8>(myString.c_str()[i]) << endl;
}
}
Java: Convert A String (like testing123) To Binary In Java
String s = "foo";
byte[] bytes = s.getBytes();
StringBuilder binary = new StringBuilder();
for (byte b : bytes)
{
int val = b;
for (int i = 0; i < 8; i++)
{
binary.append((val & 128) == 0 ? 0 : 1);
val <<= 1;
}
binary.append(' ');
}
System.out.println("'" + s + "' to binary: " + binary);
JS: How to convert text to binary code in JavaScript?
function convert() {
var output = document.getElementById("ti2");
var input = document.getElementById("ti1").value;
output.value = "";
for (var i = 0; i < input.length; i++) {
output.value += input[i].charCodeAt(0).toString(2) + " ";
}
}
I was looking around for a simple GML script to convert a decimal to binary and return the bits in an array. I didn't find anything for my need and to my liking so I rolled my own. Short and sweet.
The first param is the decimal number (string or decimal) and the second param is the bit length.
// dec_to_bin(num, len);
// argument0, decimal string
// argument1, integer
var num = real(argument0);
var len = argument1;
var bin = array_create(len, 0);
for (var i = len - 1; i >= 0; --i) {
bin[i] = floor(num % 2);
num -= num / 2;
}
return bin;
Usage:
dec_to_bin("48", 10);
Output:
{ { 0,0,0,0,1,1,0,0,0,0 }, }
i think the binary you mean is the one that computers use, if thats the case, just use the common binary and add a kind of identification.
binary is actually simple, instead of what most people think.
every digit represents the previous number *2 (2¹, 2², 2³...) so we get:
1, 2, 4, 8, 16, 32, 64, 128, 256, 512...
flip it and get:
...512, 256, 128, 64, 32, 16, 8, 4, 2, 1
every digit is "activated" with 1's, plus all the activated number ant thats the value.
ok, so binary is basically another number system, its not like codes or something. Then how are letters and other characters calculated?
they arent ;-;
we just represent then as their order on their alphabets, so:
a=1
b=2
c=3
...
this means that "b" in binary would be "10", but "2" is also "10". So thats where computer's binary enter.
they just add a identification before the actual number, so:
letter_10 = b
number_10 = 2
signal_10 = "
wait, but if thats binary there cant be letter on it, instead another 0's and 1's are used, so:
011_10 = b
0011_10 = 2
001_10 = "
computers also cant know where the number starts and ends, so you have to always use the same amount of numbers, which is 8. now we get:
011_00010 = b
0011_0010 = 2
001_00010 = "
then remove the "_" cuz again, computers will only use 0's and 1's. and done!
so what i mean is, just use the code you had and add 00110000 to the value, or if you want to translate these numbers to letters as i wanted just add 01100000
in that case where you have the letter and wants the binary, first convert the letter to its number, for it just knows that the letters dont start at 1, capitalized letters starts at 64 and the the non-capitalized at 96.
ord("p")=112
112-96=16
16 in binary is 10000
10000 + 01100000 = 01110000
"p" in binary is 01110000
ord("P")=80
80-64=16
16 in binary is 10000
10000 + 01000000 = 01010000
"P" in binary is 01010000
thats just a explanation of what the code should do, actually im looking for a simple way to turn binary cuz i cant understand much of the code you showed.
(011)
1000 1111 10000 101 1001 1000 101 1100 10000 101 100
I am trying to count two binary numbers from string. The maximum number of counting digits have to be 253. Short numbers works, but when I add there some longer numbers, the output is wrong. The example of bad result is "10100101010000111111" with "000011010110000101100010010011101010001101011100000000111000000000001000100101101111101000111001000101011010010111000110".
#include <iostream>
#include <stdlib.h>
using namespace std;
bool isBinary(string b1,string b2);
int main()
{
string b1,b2;
long binary1,binary2;
int i = 0, remainder = 0, sum[254];
cout<<"Get two binary numbers:"<<endl;
cin>>b1>>b2;
binary1=atol(b1.c_str());
binary2=atol(b2.c_str());
if(isBinary(b1,b2)==true){
while (binary1 != 0 || binary2 != 0){
sum[i++] =(binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder =(binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0){
sum[i++] = remainder;
}
--i;
cout<<"Result: ";
while (i >= 0){
cout<<sum[i--];
}
cout<<endl;
}else cout<<"Wrong input"<<endl;
return 0;
}
bool isBinary(string b1,string b2){
bool rozhodnuti1,rozhodnuti2;
for (int i = 0; i < b1.length();i++) {
if (b1[i]!='0' && b1[i]!='1') {
rozhodnuti1=false;
break;
}else rozhodnuti1=true;
}
for (int k = 0; k < b2.length();k++) {
if (b2[k]!='0' && b2[k]!='1') {
rozhodnuti2=false;
break;
}else rozhodnuti2=true;
}
if(rozhodnuti1==false || rozhodnuti2==false){ return false;}
else{ return true;}
}
One of the problems might be here: sum[i++]
This expression, as it is, first returns the value of i and then increases it by one.
Did you do it on purporse?
Change it to ++i.
It'd help if you could also post the "bad" output, so that we can try to move backward through the code starting from it.
EDIT 2015-11-7_17:10
Just to be sure everything was correct, I've added a cout to check what binary1 and binary2 contain after you assing them the result of the atol function: they contain the integer numbers 547284487 and 18333230, which obviously dont represent the correct binary-to-integer transposition of the two 01 strings you presented in your post.
Probably they somehow exceed the capacity of atol.
Also, the result of your "math" operations bring to an even stranger result, which is 6011111101, which obviously doesnt make any sense.
What do you mean, exactly, when you say you want to count these two numbers? Maybe you want to make a sum? I guess that's it.
But then, again, what you got there is two signed integer numbers and not two binaries, which means those %10 and %2 operations are (probably) misused.
EDIT 2015-11-07_17:20
I've tried to use your program with small binary strings and it actually works; with small binary strings.
It's a fact(?), at this point, that atol cant handle numerical strings that long.
My suggestion: use char arrays instead of strings and replace 0 and 1 characters with numerical values (if (bin1[i]){bin1[i]=1;}else{bin1[i]=0}) with which you'll be able to perform all the math operations you want (you've already written a working sum function, after all).
Once done with the math, you can just convert the char array back to actual characters for 0 and 1 and cout it on the screen.
EDIT 2015-11-07_17:30
Tested atol on my own: it correctly converts only strings that are up to 10 characters long.
Anything beyond the 10th character makes the function go crazy.
I was doing a programmed that require me to prepare two separate input file which was name group1.txt and group2.txt and then create a programmed that finds the average score for each group
I wrote this but it cant find my file for reasons I do not know
some help maybe thanks !
Here is the code:
#include <iostream>
#include <fstream>
#include <string>
#include <stdlib.h>
using namespace std;
int main()
{
string line;
const int num_lines = 10; //change numline to any number you like, its to set the size of the array
string sub_code[num_lines];
float avrgs[num_lines];
string sub_code2[num_lines];
float avrgs2[num_lines];
ifstream myfile ("group1.txt");
int sum=0,score=0,j=1,i=0,i2=0;
double ave=0.0;
if (myfile.is_open())
{
while ( myfile>>line )
{
//cout<<line<<endl;
sub_code[i] = line;
while ( myfile>>score && score <100 && score >= 0)
{
sum += score;
j++;
}
ave = sum/j;
j=1;
sum = 0;
avrgs[i]=ave;
i++;
}
}
else
{cout << "Unable to open file"<<endl;}
myfile.close();
//this is for the secode line
ifstream myfile2 ("group2.txt");
if (myfile2.is_open())
{
while ( myfile2>>line )
{
//cout<<line<<endl;
sub_code2[i2] = line; //add all the subject code into the array to store sub codes
while ( myfile2>>score && score <100 && score >= 0) //well basically the score should be between 0 - 100,
{ //so -999 wont be read. Can change to while ( myfile2>>score && score!=-999)
sum += score; //read each grade and add it to sum
j++; //just to know how many grades are there so that division can be done
}
ave = sum/j; //find the average.
j=1; //set j back to 1, cause j is used to count the number of marks.
sum = 0; //since its sum+=score, we need to set sum back to 0, or else it will be adding on to the old marks
avrgs2[i2]=ave; //add that calculated ave into the array for average
i2++; //i2 is to basically know how many entries are in the file for grp2
}
}
else
{cout << "Unable to open file"<<endl;}
myfile2.close();
int gr1,gr2;
//outputing the averages and so on...
for (gr1=0;gr1<i;gr1++)
{
for(gr2=0;gr2<i2;gr2++)
{
if(sub_code[gr1]==sub_code2[gr2]) // compare subject id before displaying
{
cout<<sub_code[gr1]<<"\t"<<" 1 "<<avrgs[gr1]<<endl;
cout<<sub_code2[gr2]<<"\t"<<" 2 "<<avrgs2[gr2]<<endl;
break;
}
}
}
//
cout<<endl;
double grpave1=0,grpave2=0;
//to find the average of each group
for (gr1=0;gr1<i;gr1++)
{
grpave1+=avrgs[gr1];
}
for(gr2=0;gr2<i2;gr2++)
{
grpave2+=avrgs2[gr2];
}
grpave1=grpave1/i;
grpave2=grpave2/i2;
cout<<"Average for group 1:"<<grpave1<<endl;
cout<<"Average for group 2:"<<grpave2<<endl;
system("pause");
return 0;
}
Just need to know how to get my file ! I have put in Desktop, MY document, Projects, With the C++ file and I have no idea how !! Somemore I need to have a soft copy which I dun know whether it will be able to find my file in there !
The way you specify the file name, as "group1.txt" the file will only be found by the program if it's in the current working directory of the compiled program.
There are two ways to solve this problem:
actually copy the file to the working directory of the program. In many cases that will be the directory the executable resides in.
use an absolute filename, like ("c:\Users\youruser\Desktop\group1.txt" for a file stored on the desktop on a Win7 system.