I am writing to an LCD and have developed the drivers to accept a string input for the value. I have a variable called "cycles" that is a uint16. I need to convert that value to a string and write it to the display. My string write function prototype is as follows:
void lcd_string(uint8_t column, uint8_t page, const uint8_t *font_address, const char *str)
My way of displaying the value is to break the value into individual digits and write them individually to the display at the proper position.
I have the following code that works for what I want to do, but I would like to eliminate the long if/else if statement if possible.
loop = 0;
while (cycles_1 > 0) {
temp = cycles_1 % 10;
if (temp == 9) lcd_string(36 - (6 * loop),6,font_6x8_num,"9");
else if (temp == 8) lcd_string(36 - (6 * loop),6,font_6x8_num,"8");
else if (temp == 7) lcd_string(36 - (6 * loop),6,font_6x8_num,"7");
else if (temp == 6) lcd_string(36 - (6 * loop),6,font_6x8_num,"6");
else if (temp == 5) lcd_string(36 - (6 * loop),6,font_6x8_num,"5");
else if (temp == 4) lcd_string(36 - (6 * loop),6,font_6x8_num,"4");
else if (temp == 3) lcd_string(36 - (6 * loop),6,font_6x8_num,"3");
else if (temp == 2) lcd_string(36 - (6 * loop),6,font_6x8_num,"2");
else if (temp == 1) lcd_string(36 - (6 * loop),6,font_6x8_num,"1");
else if (temp == 0) lcd_string(36 - (6 * loop),6,font_6x8_num,"0");
cycles_1 /= 10;
loop++;
}
I tried the following but the string was not writing to the display.
loop = 0;
while (cycles_1 > 0) {
temp = cycles_1 % 10;
lcd_string(36 - (6 * loop),6,font_6x8_num,{0x30 + temp, '\0'});
cycles_1 /= 10;
loop++;
}
I figured adding 0x30 to the temp value would convert it to an ASCII number and then I would terminate it with a null termination character, but this doesn't seem to be working. Any suggestions of what I might try?
If you have a number named temp between 0 and 9, you can convert it to a 1-character C string with a null terminator using this code:
char str[2] = { '0' + temp };
There is no need to have a separate case for each digit.
As you tagged the question [avr], you are using GNU tools for AVR, which use AVR-LibC as their libc implementation. It provides utoa which converts an unsigned (which is uint16_t) to string. Notice that utoa is non-standard.
#include <stdlib.h>
#include <stdint.h>
...
uint16_t cycles_1 = 12345;
char buf[6];
utoa (cycles_1, buf, 10);
lcd_string (..., buf, ...);
In case cycles_1 is a 16-bit signed type, there is itoa.
utoa and itoa are implemented in assembly; they are fast and have a small foot-print. They don't use division or modulo.
If you really want to output each character individually, you can
char buf[2], var = 3;
buf[1] = '\0';
buf[0] = '1'; lcd_string (..., buf, ...);
buf[0] = '2'; lcd_string (..., buf, ...);
buf[0] = '0' + var; lcd_string (..., buf, ...);
...
or, using it with your code:
uint8_t loop = 0;
while (cycles_1 > 0)
{
char temp = cycles_1 % 10;
char str[2] = { '0' + temp, 0 };
lcd_string (36 - loop, 6, font_6x8_num, str);
cycles_1 /= 10;
loop += 6;
}
Notice however that this is slow and expensive, as it need division (and perhaps also modulo) for each digit.
I am trying to use StringBuilder to create the output that is being sent over the serial port for a log file. The output is stored in a byte array, and I am recursing through it.
ref class UART_G {
public:
static array<System::Byte>^ message = nullptr;
static uint8_t message_length = 0;
};
static void logSend ()
{
StringBuilder^ outputsb = gcnew StringBuilder();
outputsb->Append("Sent ");
for (uint8_t i = 0; i < UART_G::message_length; i ++)
{
unsigned char mychar = UART_G::message[i];
if (
(mychar >= ' ' && mychar <= 'Z') || //Includes 0-9, A-Z.
(mychar >= '^' && mychar <= '~') || //Includes a-z.
(mychar >= 128 && mychar <= 254)) //I think these are okay.
{
outputsb->Append(L""+mychar);
}
else
{
outputsb->Append("[");
outputsb->Append(mychar);
outputsb->Append("]");
}
}
log_line(outputsb->ToString());
}
I want all plain text characters (eg A, :) to be sent as text, while functional characters (eg BEL, NEWLINE) will be sent like [7][13].
What is happening is that the StringBuilder, in all cases, is outputting the character as a number. For example, A is being sent out as 65.
For example, if I have the string 'APPLE' and a newline in my byte array, I want to see:
Sent APPLE[13]
Instead, I see:
Sent 6580807669[13]
I have tried every way imaginable to get it to display the character properly, including type-casting, concatenating it to a string, changing the variable type, etc... I would really appreciate if anyone knows how to do this. My log files are largely unreadable without this function.
You're getting the ASCII values because the compiler is choosing one of the Append overloads that takes an integer of some sort. To fix this, you could do a explicit cast to System::Char, to force the correct overload.
However, that won't necessarily give the proper results for 128-255. You could cast a value in that range from Byte to Char, and it'll give something, but not necessarily what you expect. First off, 0x80 through 0x9F are control characters, and whereever you're getting the bytes from might not intend the same representation for 0xA0 through 0xFF as Unicode has.
In my opinion, the best solution would be to use the "[value]" syntax that you're using for the other control characters for 0x80 through 0xFF as well. However, if you do want to convert those to characters, I'd use Encoding::Default, not Encoding::ASCII. ASCII only defines 0x00 through 0x7F, 0x80 and higher will come out as "?". Encoding::Default is whatever code page is defined for the language you have selected in Windows.
Combine all that, and here's what you'd end up with:
for (uint8_t i = 0; i < UART_G::message_length; i ++)
{
unsigned char mychar = UART_G::message[i];
if (mychar >= ' ' && mychar <= '~' && mychar != '[' && mychar != ']')
{
// Use the character directly for all ASCII printable characters,
// except '[' and ']', because those have a special meaning, below.
outputsb->Append((System::Char)(mychar));
}
else if (mychar >= 128)
{
// Non-ASCII characters, use the default encoding to convert to Unicode.
outputsb->Append(Encoding::Default->GetChars(UART_G::message, i, 1));
}
else
{
// Unprintable characters, use the byte value in brackets.
// Also do this for bracket characters, so there's no ambiguity
// what a bracket means in the logs.
outputsb->Append("[");
outputsb->Append((unsigned int)mychar);
outputsb->Append("]");
}
}
You are recieveing ascii value of the string .
See the Ascii chart
65 = A
80 = P
80 = P
76 = L
69 = E
Just write a function that converts the ascii value to string
Here is the code I came up with which resolved the issue:
static void logSend ()
{
StringBuilder^ outputsb = gcnew StringBuilder();
ASCIIEncoding^ ascii = gcnew ASCIIEncoding;
outputsb->Append("Sent ");
for (uint8_t i = 0; i < UART_G::message_length; i ++)
{
unsigned char mychar = UART_G::message[i];
if (
(mychar >= ' ' && mychar <= 'Z') || //Includes 0-9, A-Z.
(mychar >= '^' && mychar <= '~') || //Includes a-z.
(mychar >= 128 && mychar <= 254)) //I think these are okay.
{
outputsb->Append(ascii->GetString(UART_G::message, i, 1));
}
else
{
outputsb->Append("[");
outputsb->Append(mychar);
outputsb->Append("]");
}
}
log_line(outputsb->ToString());
}
I still appreciate any alternatives which are more efficient or simpler to read.
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Definition:
A palindrome is a word, phrase, number or other sequence of units that has the property of reading the same in either direction
How to check if the given string is a palindrome?
This was one of the FAIQ [Frequently Asked Interview Question] a while ago but that mostly using C.
Looking for solutions in any and all languages possible.
PHP sample:
$string = "A man, a plan, a canal, Panama";
function is_palindrome($string)
{
$a = strtolower(preg_replace("/[^A-Za-z0-9]/","",$string));
return $a==strrev($a);
}
Removes any non-alphanumeric characters (spaces, commas, exclamation points, etc.) to allow for full sentences as above, as well as simple words.
Windows XP (might also work on 2000) or later BATCH script:
#echo off
call :is_palindrome %1
if %ERRORLEVEL% == 0 (
echo %1 is a palindrome
) else (
echo %1 is NOT a palindrome
)
exit /B 0
:is_palindrome
set word=%~1
set reverse=
call :reverse_chars "%word%"
set return=1
if "$%word%" == "$%reverse%" (
set return=0
)
exit /B %return%
:reverse_chars
set chars=%~1
set reverse=%chars:~0,1%%reverse%
set chars=%chars:~1%
if "$%chars%" == "$" (
exit /B 0
) else (
call :reverse_chars "%chars%"
)
exit /B 0
Language agnostic meta-code then...
rev = StringReverse(originalString)
return ( rev == originalString );
C# in-place algorithm. Any preprocessing, like case insensitivity or stripping of whitespace and punctuation should be done before passing to this function.
boolean IsPalindrome(string s) {
for (int i = 0; i < s.Length / 2; i++)
{
if (s[i] != s[s.Length - 1 - i]) return false;
}
return true;
}
Edit: removed unnecessary "+1" in loop condition and spent the saved comparison on removing the redundant Length comparison. Thanks to the commenters!
C#: LINQ
var str = "a b a";
var test = Enumerable.SequenceEqual(str.ToCharArray(),
str.ToCharArray().Reverse());
A more Ruby-style rewrite of Hal's Ruby version:
class String
def palindrome?
(test = gsub(/[^A-Za-z]/, '').downcase) == test.reverse
end
end
Now you can call palindrome? on any string.
Unoptimized Python:
>>> def is_palindrome(s):
... return s == s[::-1]
Java solution:
public class QuickTest {
public static void main(String[] args) {
check("AmanaplanacanalPanama".toLowerCase());
check("Hello World".toLowerCase());
}
public static void check(String aString) {
System.out.print(aString + ": ");
char[] chars = aString.toCharArray();
for (int i = 0, j = (chars.length - 1); i < (chars.length / 2); i++, j--) {
if (chars[i] != chars[j]) {
System.out.println("Not a palindrome!");
return;
}
}
System.out.println("Found a palindrome!");
}
}
Using a good data structure usually helps impress the professor:
Push half the chars onto a stack (Length / 2).
Pop and compare each char until the first unmatch.
If the stack has zero elements: palindrome.
*in the case of a string with an odd Length, throw out the middle char.
C in the house. (not sure if you didn't want a C example here)
bool IsPalindrome(char *s)
{
int i,d;
int length = strlen(s);
char cf, cb;
for(i=0, d=length-1 ; i < length && d >= 0 ; i++ , d--)
{
while(cf= toupper(s[i]), (cf < 'A' || cf >'Z') && i < length-1)i++;
while(cb= toupper(s[d]), (cb < 'A' || cb >'Z') && d > 0 )d--;
if(cf != cb && cf >= 'A' && cf <= 'Z' && cb >= 'A' && cb <='Z')
return false;
}
return true;
}
That will return true for "racecar", "Racecar", "race car", "racecar ", and "RaCe cAr". It would be easy to modify to include symbols or spaces as well, but I figure it's more useful to only count letters(and ignore case). This works for all palindromes I've found in the answers here, and I've been unable to trick it into false negatives/positives.
Also, if you don't like bool in a "C" program, it could obviously return int, with return 1 and return 0 for true and false respectively.
Here's a python way. Note: this isn't really that "pythonic" but it demonstrates the algorithm.
def IsPalindromeString(n):
myLen = len(n)
i = 0
while i <= myLen/2:
if n[i] != n[myLen-1-i]:
return False
i += 1
return True
Delphi
function IsPalindrome(const s: string): boolean;
var
i, j: integer;
begin
Result := false;
j := Length(s);
for i := 1 to Length(s) div 2 do begin
if s[i] <> s[j] then
Exit;
Dec(j);
end;
Result := true;
end;
I'm seeing a lot of incorrect answers here. Any correct solution needs to ignore whitespace and punctuation (and any non-alphabetic characters actually) and needs to be case insensitive.
A few good example test cases are:
"A man, a plan, a canal, Panama."
"A Toyota's a Toyota."
"A"
""
As well as some non-palindromes.
Example solution in C# (note: empty and null strings are considered palindromes in this design, if this is not desired it's easy to change):
public static bool IsPalindrome(string palindromeCandidate)
{
if (string.IsNullOrEmpty(palindromeCandidate))
{
return true;
}
Regex nonAlphaChars = new Regex("[^a-z0-9]");
string alphaOnlyCandidate = nonAlphaChars.Replace(palindromeCandidate.ToLower(), "");
if (string.IsNullOrEmpty(alphaOnlyCandidate))
{
return true;
}
int leftIndex = 0;
int rightIndex = alphaOnlyCandidate.Length - 1;
while (rightIndex > leftIndex)
{
if (alphaOnlyCandidate[leftIndex] != alphaOnlyCandidate[rightIndex])
{
return false;
}
leftIndex++;
rightIndex--;
}
return true;
}
EDIT: from the comments:
bool palindrome(std::string const& s)
{
return std::equal(s.begin(), s.end(), s.rbegin());
}
The c++ way.
My naive implementation using the elegant iterators. In reality, you would probably check
and stop once your forward iterator has past the halfway mark to your string.
#include <string>
#include <iostream>
using namespace std;
bool palindrome(string foo)
{
string::iterator front;
string::reverse_iterator back;
bool is_palindrome = true;
for(front = foo.begin(), back = foo.rbegin();
is_palindrome && front!= foo.end() && back != foo.rend();
++front, ++back
)
{
if(*front != *back)
is_palindrome = false;
}
return is_palindrome;
}
int main()
{
string a = "hi there", b = "laval";
cout << "String a: \"" << a << "\" is " << ((palindrome(a))? "" : "not ") << "a palindrome." <<endl;
cout << "String b: \"" << b << "\" is " << ((palindrome(b))? "" : "not ") << "a palindrome." <<endl;
}
boolean isPalindrome(String str1) {
//first strip out punctuation and spaces
String stripped = str1.replaceAll("[^a-zA-Z0-9]", "");
return stripped.equalsIgnoreCase((new StringBuilder(stripped)).reverse().toString());
}
Java version
Here's my solution, without using a strrev. Written in C#, but it will work in any language that has a string length function.
private static bool Pal(string s) {
for (int i = 0; i < s.Length; i++) {
if (s[i] != s[s.Length - 1 - i]) {
return false;
}
}
return true;
}
Here's my solution in c#
static bool isPalindrome(string s)
{
string allowedChars = "abcdefghijklmnopqrstuvwxyz"+
"1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string compareString = String.Empty;
string rev = string.Empty;
for (int i = 0; i <= s.Length - 1; i++)
{
char c = s[i];
if (allowedChars.IndexOf(c) > -1)
{
compareString += c;
}
}
for (int i = compareString.Length - 1; i >= 0; i--)
{
char c = compareString[i];
rev += c;
}
return rev.Equals(compareString,
StringComparison.CurrentCultureIgnoreCase);
}
Here's a Python version that deals with different cases, punctuation and whitespace.
import string
def is_palindrome(palindrome):
letters = palindrome.translate(string.maketrans("",""),
string.whitespace + string.punctuation).lower()
return letters == letters[::-1]
Edit: Shamelessly stole from Blair Conrad's neater answer to remove the slightly clumsy list processing from my previous version.
C++
std::string a = "god";
std::string b = "lol";
std::cout << (std::string(a.rbegin(), a.rend()) == a) << " "
<< (std::string(b.rbegin(), b.rend()) == b);
Bash
function ispalin { [ "$( echo -n $1 | tac -rs . )" = "$1" ]; }
echo "$(ispalin god && echo yes || echo no), $(ispalin lol && echo yes || echo no)"
Gnu Awk
/* obvious solution */
function ispalin(cand, i) {
for(i=0; i<length(cand)/2; i++)
if(substr(cand, length(cand)-i, 1) != substr(cand, i+1, 1))
return 0;
return 1;
}
/* not so obvious solution. cough cough */
{
orig = $0;
while($0) {
stuff = stuff gensub(/^.*(.)$/, "\\1", 1);
$0 = gensub(/^(.*).$/, "\\1", 1);
}
print (stuff == orig);
}
Haskell
Some brain dead way doing it in Haskell
ispalin :: [Char] -> Bool
ispalin a = a == (let xi (y:my) = (xi my) ++ [y]; xi [] = [] in \x -> xi x) a
Plain English
"Just reverse the string and if it is the same as before, it's a palindrome"
Ruby:
class String
def is_palindrome?
letters_only = gsub(/\W/,'').downcase
letters_only == letters_only.reverse
end
end
puts 'abc'.is_palindrome? # => false
puts 'aba'.is_palindrome? # => true
puts "Madam, I'm Adam.".is_palindrome? # => true
An obfuscated C version:
int IsPalindrome (char *s)
{
char*a,*b,c=0;
for(a=b=s;a<=b;c=(c?c==1?c=(*a&~32)-65>25u?*++a,1:2:c==2?(*--b&~32)-65<26u?3:2:c==3?(*b-65&~32)-(*a-65&~32)?*(b=s=0,a),4:*++a,1:0:*++b?0:1));
return s!=0;
}
This Java code should work inside a boolean method:
Note: You only need to check the first half of the characters with the back half, otherwise you are overlapping and doubling the amount of checks that need to be made.
private static boolean doPal(String test) {
for(int i = 0; i < test.length() / 2; i++) {
if(test.charAt(i) != test.charAt(test.length() - 1 - i)) {
return false;
}
}
return true;
}
Another C++ one. Optimized for speed and size.
bool is_palindrome(const std::string& candidate) {
for(std::string::const_iterator left = candidate.begin(), right = candidate.end(); left < --right ; ++left)
if (*left != *right)
return false;
return true;
}
Lisp:
(defun palindrome(x) (string= x (reverse x)))
Three versions in Smalltalk, from dumbest to correct.
In Smalltalk, = is the comparison operator:
isPalindrome: aString
"Dumbest."
^ aString reverse = aString
The message #translateToLowercase returns the string as lowercase:
isPalindrome: aString
"Case insensitive"
|lowercase|
lowercase := aString translateToLowercase.
^ lowercase reverse = lowercase
And in Smalltalk, strings are part of the Collection framework, you can use the message #select:thenCollect:, so here's the last version:
isPalindrome: aString
"Case insensitive and keeping only alphabetic chars
(blanks & punctuation insensitive)."
|lowercaseLetters|
lowercaseLetters := aString
select: [:char | char isAlphabetic]
thenCollect: [:char | char asLowercase].
^ lowercaseLetters reverse = lowercaseLetters
Note that in the above C++ solutions, there was some problems.
One solution was inefficient because it passed an std::string by copy, and because it iterated over all the chars, instead of comparing only half the chars. Then, even when discovering the string was not a palindrome, it continued the loop, waiting its end before reporting "false".
The other was better, with a very small function, whose problem was that it was not able to test anything else than std::string. In C++, it is easy to extend an algorithm to a whole bunch of similar objects. By templating its std::string into "T", it would have worked on both std::string, std::wstring, std::vector and std::deque. But without major modification because of the use of the operator <, the std::list was out of its scope.
My own solutions try to show that a C++ solution won't stop at working on the exact current type, but will strive to work an anything that behaves the same way, no matter the type. For example, I could apply my palindrome tests on std::string, on vector of int or on list of "Anything" as long as Anything was comparable through its operator = (build in types, as well as classes).
Note that the template can even be extended with an optional type that can be used to compare the data. For example, if you want to compare in a case insensitive way, or even compare similar characters (like è, é, ë, ê and e).
Like king Leonidas would have said: "Templates ? This is C++ !!!"
So, in C++, there are at least 3 major ways to do it, each one leading to the other:
Solution A: In a c-like way
The problem is that until C++0X, we can't consider the std::string array of chars as contiguous, so we must "cheat" and retrieve the c_str() property. As we are using it in a read-only fashion, it should be ok...
bool isPalindromeA(const std::string & p_strText)
{
if(p_strText.length() < 2) return true ;
const char * pStart = p_strText.c_str() ;
const char * pEnd = pStart + p_strText.length() - 1 ;
for(; pStart < pEnd; ++pStart, --pEnd)
{
if(*pStart != *pEnd)
{
return false ;
}
}
return true ;
}
Solution B: A more "C++" version
Now, we'll try to apply the same solution, but to any C++ container with random access to its items through operator []. For example, any std::basic_string, std::vector, std::deque, etc. Operator [] is constant access for those containers, so we won't lose undue speed.
template <typename T>
bool isPalindromeB(const T & p_aText)
{
if(p_aText.empty()) return true ;
typename T::size_type iStart = 0 ;
typename T::size_type iEnd = p_aText.size() - 1 ;
for(; iStart < iEnd; ++iStart, --iEnd)
{
if(p_aText[iStart] != p_aText[iEnd])
{
return false ;
}
}
return true ;
}
Solution C: Template powah !
It will work with almost any unordered STL-like container with bidirectional iterators
For example, any std::basic_string, std::vector, std::deque, std::list, etc.
So, this function can be applied on all STL-like containers with the following conditions:
1 - T is a container with bidirectional iterator
2 - T's iterator points to a comparable type (through operator =)
template <typename T>
bool isPalindromeC(const T & p_aText)
{
if(p_aText.empty()) return true ;
typename T::const_iterator pStart = p_aText.begin() ;
typename T::const_iterator pEnd = p_aText.end() ;
--pEnd ;
while(true)
{
if(*pStart != *pEnd)
{
return false ;
}
if((pStart == pEnd) || (++pStart == pEnd))
{
return true ;
}
--pEnd ;
}
}
A simple Java solution:
public boolean isPalindrome(String testString) {
StringBuffer sb = new StringBuffer(testString);
String reverseString = sb.reverse().toString();
if(testString.equalsIgnoreCase(reverseString)) {
return true;
else {
return false;
}
}
Many ways to do it. I guess the key is to do it in the most efficient way possible (without looping the string). I would do it as a char array which can be reversed easily (using C#).
string mystring = "abracadabra";
char[] str = mystring.ToCharArray();
Array.Reverse(str);
string revstring = new string(str);
if (mystring.equals(revstring))
{
Console.WriteLine("String is a Palindrome");
}
In Ruby, converting to lowercase and stripping everything not alphabetic:
def isPalindrome( string )
( test = string.downcase.gsub( /[^a-z]/, '' ) ) == test.reverse
end
But that feels like cheating, right? No pointers or anything! So here's a C version too, but without the lowercase and character stripping goodness:
#include <stdio.h>
int isPalindrome( char * string )
{
char * i = string;
char * p = string;
while ( *++i ); while ( i > p && *p++ == *--i );
return i <= p && *i++ == *--p;
}
int main( int argc, char **argv )
{
if ( argc != 2 )
{
fprintf( stderr, "Usage: %s <word>\n", argv[0] );
return -1;
}
fprintf( stdout, "%s\n", isPalindrome( argv[1] ) ? "yes" : "no" );
return 0;
}
Well, that was fun - do I get the job ;^)
Using Java, using Apache Commons String Utils:
public boolean isPalindrome(String phrase) {
phrase = phrase.toLowerCase().replaceAll("[^a-z]", "");
return StringUtils.reverse(phrase).equals(phrase);
}