I want to output my floats without the ending zeros.
Example: float 3.570000 should be outputted as 3.57
and float 3.00000 should be outputted as 3.0 (so here would be the exception!)
A more efficient and (in my opinion) clearer form of paxdiablo's morphNumericString(). Sorry not compiled or tested.
void morphNumericString( char *s )
{
char *p, *end, *decimal, *nonzero;
// Find the last decimal point and non zero character
end = p = strchr(s,'\0');
decimal = nonzero = NULL;
while( p > s )
{
p--;
if( !nonzero && *p!='0' )
{
nonzero = p;
}
if( !decimal && *p=='.' )
{
decimal = p;
break; // nonzero must also be non NULL, so stop early
}
}
// eg "4.3000" -> "4.3"
if( decimal && nonzero && nonzero>decimal )
*(nonzero+1) = '\0';
// eg if(decimal) "4.0000" -> "4.0"
// if(!decimal) "4" -> "4.0"
else
strcpy( decimal?decimal:end, ".0" );
}
This is not possible with standard printf semantics. In, the past, I've had to do this by outputting to a string (with something like "%.20f") then post-processing the string.
Something like this is probably what you're looking for:
#include <stdio.h>
void morphNumericString (char *s) {
char *p;
int count;
// Find decimal point, if any.
p = strchr (s,'.');
if (p == NULL) {
// No decimal, just add one fractional position.
strcat (s, ".0");
} else {
// Decimal, start stripping off trailing zeros.
while (s[strlen(s)-1] == '0') {
s[strlen(s)-1] = '\0';
}
// If all fractional positions were zero, add one.
if (s[strlen(s)-1] == '.') {
strcat (s, "0");
}
}
}
int main (int argc, char *argv[]) {
char str[100];
int i;
for (i = 1; i < argc; i++) {
strcpy (str, argv[i]);
morphNumericString (str);
printf ("[%s] -> [%s]\n", argv[i], str);
}
return 0;
}
The code runs through each of it's arguments, morphing each one in turn. The following transcript shows how it works:
pax> ./qq 3.750000 12 12.507 47.90 56.0000000 76.0 0
[3.750000] -> [3.75]
[12] -> [12.0]
[12.507] -> [12.507]
[47.90] -> [47.9]
[56.0000000] -> [56.0]
[76.0] -> [76.0]
[0] -> [0.0]
You should be aware however that, if using floats or doubles, you'll have to watch out for the normal floating point inaccuracies. On my system 3.57 is actually 3.5699999999999998401, which won't be truncated at all.
Of course, you can get around that problem by using a less-specific number of output digits in the sprintf, something less than the actual accuracy of the floating point point. For example, "%.10f" on my system outputs 3.5700000000 which will be truncated. Adding the following lines to main:
sprintf (str, "%.10f", 3.57);
morphNumericString (str);
printf ("[%.10f] -> [%s]\n", 3.57, str);
sprintf (str, "%.10f", 3.0);
morphNumericString (str);
printf ("[%.10f] -> [%s]\n", 3.0, str);
will result in the following added output:
[3.5700000000] -> [3.57]
[3.0000000000] -> [3.0]
as per your test data.
One other possibility (if your input range and precision can be controlled) is to use the g format specifier. This outputs in either f format where the precision is the maximum number of digits (rather than fixed number like f) or exponential format (e).
Basically, it prefers the non-exponential output format as long as all the information is shown. It will switch to exponential only if that will deliver more information. A simple example is the format string "%.4g" with 3.7 and .0000000004. The former would be printed as 3.7, the latter as 4e-10.
Update: for those more concerned about performance than robustness or readability, you could try the following (unnecessary in my opinion but to each their own):
void morphNumericString (char *s) {
char *p = strchr (s,'.');
if (p == NULL) {
strcat (s, ".0");
return;
}
p = &(p[strlen(p)-1]);
while ((p != s) && (*p == '0') && (*(p-1) != '.'))
*p-- = '\0';
}
It may be faster but, given the extreme optimisations I've seen modern compilers do, you can never be too sure. I tend to code for readability first and only worry about speed when it becomes an issue (YAGNI can apply equally to performance as well as functionality).
My approach is:
implement a function trim(char c) to eliminate the trail 'c', for examaple:
void trim(std::string &str, char c) {
size_t len = str.length();
const char *s = str.c_str();
const char *p = s + len - 1;
while (p != s && *p == c) {
-- p;
}
++ p;
size_t end = p - s;
str = str.substr(0, end);
}
char buf[32];
printf(buf, "%f", 0.01);
std::string s(buf);
trim(buf, '0');
void tidyFloatRepresentation(char *s)
{
size_t end = strlen (s);
size_t i = end - 1;
char *lastZero = NULL;
while (i && s[i] == '0') lastZero = &s[i--];
while (i && s[i] != '.') i--;
if (lastZero && s[i] == '.')
{
if (lastZero == &s[i] + 1)
*(lastZero + 1) = '\0';
else
*lastZero = '\0';
}
else
{
strcpy (&s[end + 1], ".0");
}
}
This will fail when the input ends with a decimal point, but for the most part, it will truncate (or append) appropriately.
Here's another string modification function. I did test it.
It's longer than Bill's and Diablo's, but it handles trailing 9's as well as 0's and should perform well.
Leaves a single trailing zero after the dot. You'll have to truncate using sprintf to get proper trailing 9's.
void morphNumericString( char *s ) {
char *point = s;
while ( * point && * point != '.' ) ++ point;
char *last = strchr( point, 0 );
if ( point == last ) {
* point = '.';
++ last;
}
if ( point == last - 1 ) {
* last = '0';
} else {
-- last;
if ( * last == '0' ) {
while ( * last == '0' ) -- last;
} else if ( * last == '9' ) {
while ( * last == '9' || * last == '.' ) {
if ( * last == '9' ) * last = '0';
-- last;
}
( * last ) ++;
}
if ( last < point + 1 ) last = point + 1;
}
* ++ last = 0;
}
Edit: Yikes, this fails on input like 999.999. Left as an exercise to the reader ;v)
It might be easier to calculate the fractional part directly:
double value= -3.57;
double int_part;
double frac= modf(value, &int_part);
int64 int_part_as_int= int_part;
int significant_digits= 10;
int64 fractional_scale= pow(10., significant_digits);
int64 fraction_magnitude= fabs(frac)*fractional_scale + 0.5;
fractional_magnitude/fractional_scale will be the fraction rounded to significant_digits sig figs. Even with doubles, this is guaranteed not to overflow.
Formatting the fraction should be straightforward.
Related
#include<string>
#include<cstring>
class Solution {
void shift_left(char* c, const short unsigned int bits) {
const unsigned short int size = sizeof(c);
memmove(c, c+bits, size - bits);
memset(c+size-bits, 0, bits);
}
public:
string longestPalindrome(string s) {
char* output = new char[s.length()];
output[0] = s[0];
string res = "";
char* n = output;
auto e = s.begin() + 1;
while(e != s.end()) {
char letter = *e;
char* c = n;
(*++n) = letter;
if((letter != *c) && (c == &output[0] || letter != (*--c)) ) {
++e;
continue;
}
while((++e) != s.end() && c != &output[0]) {
if((letter = *e) != (*--c)) {
const unsigned short int bits = c - output + 1;
shift_left(output, bits);
n -= bits;
break;
}
(*++n) = letter;
}
string temp(output);
res = temp.length() > res.length()? temp : res;
shift_left(output, 1);
--n;
}
return res;
}
};
input string longestPalindrome("babad");
the program works fine and prints out "bab" as the longest palindrome but there's a heap overflow somewhere. Error like this appears:
Read of size 6 at ...memory address... thread T0
"babad" is size 5 and after going over this for an hour. I don't see the point where the iteration ever exceeds 5
There is 3 pointers here that iterate.
e as the element of string s.
n which is the pointer to the next char of output.
and c which is a copy of n and decrements until it reaches the address of &output[0].
maybe it's something with the memmove or memset since I've never used it before.
I'm completely lost
TL;DR : mixture of char* and std::string are not really good idea if you don't understand how exactly it works.
If you want to length of string you cant do this const unsigned short int size = sizeof(c); (sizeof will return size of pointer (which is commonly 4 on 32-bit machine and 8 on 64-bit machine). You must do this instead: const size_t size = strlen(c);
Address sanitizers is right that you (indirectly) are trying to get an memory which not belongs to you.
How does constructor of string from char* works?
Answer: char* is considered as c-style string, which means that it must be null '\0' terminated.
More details: constructor of string from char* calls strlen-like function which looks like about this:
https://en.cppreference.com/w/cpp/string/byte/strlen
int strlen(char *begin){
int k = 0;
while (*begin != '\0'){
++k;
++begin;
}
return k;
}
If c-style char* string does not contain '\0' it cause accessing memory which doesn't belongs to you.
How to fix?
Answer (two options):
not use mixture of char* and std::string
char* output = new char[s.length()]; replace with char* output = new char[s.length() + 1]; memset(output, 0, s.length() + 1);
Also you must delete all memory which you newed. So add delete[] output; before return res;
I am a little bit stuck and cant find out what is wrong here.
I have an assignment to enter a sentence into char array and if there are duplicate and followed words(example : same same , diff diff. but not : same word same.) they should be removed.
here is the function I wrote:
void Same(char arr[], char temp[]){
int i = 0, j = 0, f = 0, *p, k = 0, counter = 0;
for (i = 0; i < strlen(arr); i++){
while (arr[i] != ' ' && i < strlen(arr)){
temp[k] = arr[i];
i++;
k++;
counter++;
}
temp[k] = '\0';
k = 0;
p = strstr((arr + i), (temp + j));
if (p != NULL && (*p == arr[i])){
for (f = 0; f < strlen(p); f++){
*p = '*';
p++;
}
f = 0;
}
j = counter;
}
}
strtok is a handy function to grab the next word from a list (strsep is a better one, but is less likely to be available on your system). Using strtok, an approach like the following might work, at least for simple examples...
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAXPHRASELEN 1000
#define MAXTOKLEN 100
int main(int argc, char ** argv)
{
// Here is the sentence we are looking at
char * tmp = "This is a test and and another test";
// We will copy it to this variable
char phrase[MAXPHRASELEN+1];
strcpy(phrase, tmp);
// And will put the altered text in this variable
char new_phrase[MAXPHRASELEN+1];
// This will be the last word we looked at
char * lasttok = malloc(MAXTOKLEN+1);
// This will be the current word
char * tok = malloc(MAXTOKLEN+1);
// Both words are initially empty
new_phrase[0] = '\0';
lasttok[0] = '\0';
// Get the first word
lasttok = strtok(phrase, " ");
// If there is a word...
if (lasttok != NULL) {
// Put it in the altered text and add a space
strcat(new_phrase, lasttok);
strcat(new_phrase, " ");
// As long as there is a next word
while ( (tok = strtok(NULL, " ")) != NULL ) {
// See if it is the same as the last word
if (strcmp(tok,lasttok) != 0) {
// If it isn't, copy it to the altered text
strcat(new_phrase, tok);
// and add a space
strcat(new_phrase, " ");
// The current word becomes the last word
lasttok = tok;
}
}
}
// Print the lot
printf("%s\n", new_phrase);
}
If you really must write your own routine for grabbing the individual words, you could do worse than emulate strtok. It maintains a pointer to the beginning of current word in the string and puts a null character at the next separator (space character). When called again, it just moves the pointer to the character past the null, and puts another null after the next separator. Most string functions, when passed the pointer, will see the null as the end of the string and so just deal with the current word.
Minus comments, headers, and initialisation, it looks less threatening...
lasttok = strtok(phrase, " ");
if (lasttok != NULL) {
strcat(new_phrase, lasttok);
strcat(new_phrase, " ");
while ( (tok = strtok(NULL, " ")) != NULL ) {
if (strcmp(tok,lasttok) != 0) {
strcat(new_phrase, tok);
strcat(new_phrase, " ");
lasttok = tok;
}
}
}
printf("%s\n", new_phrase);
I have the following code that solves a small image using Tesseract.
char *answer = tess_api.GetUTF8Text();
I know beforehand that the result will always start with the character '+' and it's one word so I want to get rid of any junk it finds.
I get the result as "G+ABC S\n\n" and I need only +ABC. So basically I need to ignore anything before + and everything after the first space. I was thinking I should use rindex to find the position of + and spaces.
std::string ParseString(const std::string& s)
{
size_t plus = s.find_first_of('+');
size_t space = s.find_first_of(" \n", plus);
return s.substr(plus, space-plus);
}
int main()
{
std::cout << ParseString("G+ABC S\n\n").c_str() << std::endl;
std::cout << ParseString("G +ABC\ne\n").c_str() << std::endl;
return 0;
}
Gives
+ABC
+ABC
If you really can't use strings then something like this might do
char *ParseString2(char *s)
{
int plus,end;
for (plus = 0 ; s[plus] != '+' ; ++plus){}
for (end = plus ; s[end] != ' ' && s[end] != '\n' ; ++end){}
char *result = new char[end - plus + 1];
memcpy(result, s + plus, end - plus);
result[end - plus] = 0;
return result;
}
You can use:
// just scan "answer" to find out where to start and where to end
int indexStart = // find the index of '+'
int indexEnd = // find the index before space
int length = indexEnd-indexStart+1;
char *dataYouWant = (char *) malloc(length+1); // result will be stored here
memcpy( dataYouWant, &answer[indexStart], length );
// for example answer = "G+ABC S\n\n"
dataYouWant[length] = '\0'; // dataYouWant will be "+ABC"
You can check out Strings in c, how to get subString for other alternatives.
P.S. suggestion: use string instead in C++, it will be much easier (check out #DavidSykes's answer).
Recently I was asked in an interview to convert the string "aabbbccccddddd" to "a2b3c4d5". The goal is to replace each repeated character with a single occurrence and a repeat count. Here 'a' is repeated twice in the input, so we have to write it as 'a2' in the output. Also I need to write a function to reverse the format back to the original one (e.g. from the string "a2b3c4d5" to "aabbbccccddddd"). I was free to use either C or C++. I wrote the below code, but the interviewer seemed to be not very happy with this. He asked me to try a smarter way than this.
In the below code, I used formatstring() to eliminate repeated chars by just adding the repeated count and used reverseformatstring() to convert back to the original string.
void formatstring(char* target, const char* source) {
int charRepeatCount = 1;
bool isFirstChar = true;
while (*source != '\0') {
if (isFirstChar) {
// Always add the first character to the target
isFirstChar = false;
*target = *source;
source++; target++;
} else {
// Compare the current char with previous one,
// increment repeat count
if (*source == *(source-1)) {
charRepeatCount++;
source++;
} else {
if (charRepeatCount > 1) {
// Convert repeat count to string, append to the target
char repeatStr[10];
_snprintf(repeatStr, 10, "%i", charRepeatCount);
int repeatCount = strlen(repeatStr);
for (int i = 0; i < repeatCount; i++) {
*target = repeatStr[i];
target++;
}
charRepeatCount = 1; // Reset repeat count
}
*target = *source;
source++; target++;
}
}
}
if (charRepeatCount > 1) {
// Convert repeat count to string, append it to the target
char repeatStr[10];
_snprintf(repeatStr, 10, "%i", charRepeatCount);
int repeatCount = strlen(repeatStr);
for (int i = 0; i < repeatCount; i++) {
*target = repeatStr[i];
target++;
}
}
*target = '\0';
}
void reverseformatstring(char* target, const char* source) {
int charRepeatCount = 0;
bool isFirstChar = true;
while (*source != '\0') {
if (isFirstChar) {
// Always add the first character to the target
isFirstChar = false;
*target = *source;
source++; target++;
} else {
// If current char is alpha, add it to the target
if (isalpha(*source)) {
*target = *source;
target++; source++;
} else {
// Get repeat count of previous character
while (isdigit(*source)) {
int currentDigit = (*source) - '0';
charRepeatCount = (charRepeatCount == 0) ?
currentDigit : (charRepeatCount * 10 + currentDigit);
source++;
}
// Decrement repeat count as we have already written
// the first unique char to the target
charRepeatCount--;
// Repeat the last char for this count
while (charRepeatCount > 0) {
*target = *(target - 1);
target++;
charRepeatCount--;
}
}
}
}
*target = '\0';
}
I didn't find any issues with above code. Is there any other better way of doing this?
The approach/algorithm is fine, perhaps you could refine and shrink the code a bit (by doing something simpler, there's no need to solve this in an overly complex way). And choose an indentation style that actually makes sense.
A C solution:
void print_transform(const char *input)
{
for (const char *s = input; *s;) {
char current = *s;
size_t count = 1;
while (*++s == current) {
count++;
}
if (count > 1) {
printf("%c%zu", current, count);
} else {
putc(current, stdout);
}
}
putc('\n', stdout);
}
(This can be easily modified so that it returns the transformed string instead, or writes it to a long enough buffer.)
A C++ solution:
std::string transform(const std::string &input)
{
std::stringstream ss;
std::string::const_iterator it = input.begin();
while (it != input.end()) {
char current = *it;
std::size_t count = 1;
while (++it != input.end() && *it == current) {
count++;
}
if (count > 1) {
ss << current << count;
} else {
ss << current;
}
}
return ss.str();
}
Since several others have suggested very reasonable alternatives, I'd like to offer some opinions on what I think is your underlying question: "He asked me to try a smarter way than this.... Is there any other better way of doing this?"
When I interview a developer, I'm looking for signals that tell me how she approaches a problem:
Most important, as H2CO3 noted, is correctness: will the code work? I'm usually happy to overlook small syntax errors (forgotten semicolons, mismatched parens or braces, and so on) if the algorithm is sensible.
Proper use of the language, especially if the candidate claims expertise or has had extensive experience. Does he understand and use idioms appropriately to write straightforward, uncomplicated code?
Can she explain her train of thought as she formulates her solution? Is it logical and coherent, or is it a shotgun approach? Is she able and willing to communicate well?
Does he account for edge cases? And if so, does the intrinsic algorithm handle them, or is everything a special case? Although I'm happiest if the initial algorithm "just works" for all cases, I think it's perfectly acceptable to start with a verbose approach that covers all cases (or simply to add a "TODO" comment, noting that more work needs to be done), and then simplifying later, when it may be easier to notice patterns or duplicated code.
Does she consider error-handling? Usually, if a candidate starts by asking whether she can assume the input is valid, or with a comment like, "If this were production code, I'd check for x, y, and z problems," I'll ask what she would do, then suggest she focus on a working algorithm for now and (maybe) come back to that later. But I'm disappointed if a candidate doesn't mention it.
Testing, testing, testing! How will the candidate verify his code works? Does he walk through the code and suggest test cases, or do I need to remind him? Are the test cases sensible? Will they cover the edge cases?
Optimization: as a final step, after everything works and has been validated, I'll sometimes ask the candidate if she can improve her code. Bonus points if she suggests it without my prodding; negative points if she spends a lot of effort worrying about it before the code even works.
Applying these ideas to the code you wrote, I'd make these observations:
Using const appropriately is a plus, as it shows familiarity with the language. During an interview I'd probably ask a question or two about why/when to use it.
The proper use of char pointers throughout the code is a good sign. I tend to be pedantic about making the data types explicit within comparisons, particularly during interviews, so I'm happy to see, e.g.
while (*source != '\0') rather than the (common, correct, but IMO less careful) while(*source).
isFirstChar is a bit of a red flag, based on my "edge cases" point. When you declare a boolean to keep track of the code's state, there's often a way of re-framing the problem to handle the condition intrinsically. In this case, you can use charRepeatCount to decide if this is the first character in a possible series, so you won't need to test explicitly for the first character in the string.
By the same token, repeated code can also be a sign that an algorithm can be simplified. One improvement would be to move the conversion of charRepeatCount to a separate function. See below for an even better solution.
It's funny, but I've found that candidates rarely add comments to their code during interviews. Kudos for helpful ones, negative points for those of the ilk "Increment the counter" that add verbosity without information. It's generally accepted that, unless you're doing something weird (in which case you should reconsider what you've written), you should assume the person who reads your code is familiar with the programming language. So comments should explain your thought process, not translate the code back to English.
Excessive levels of nested conditionals or loops can also be a warning. You can eliminate one level of nesting by comparing each character to the next one instead of the previous one. This works even for the last character in the string, because it will be compared to the terminating null character, which won't match and can be treated like any other character.
There are simpler ways to convert charRepeatCount from an int to a string. For example, _snprintf() returns the number of bytes it "prints" to the string, so you can use
target += _snprintf(target, 10, "%i", charRepeatCount);
In the reversing function, you've used the ternary operator perfectly ... but it's not necessary to special-case the zero value: the math is the same regardless of its value. Again, there are also standard utility functions like atoi() that will convert the leading digits of a string into an integer for you.
Experienced developers will often include the increment or decrement operation as part of the condition in a loop, rather than as a separate statement at the bottom: while(charRepeatCount-- > 0). I'd raise an eyebrow but give you a point or two for humor and personality if you wrote this using the slide operator: while (charRepeatCount --> 0). But only if you'd promise not to use it in production.
Good luck with your interviewing!
I think your code is too complex for the task. Here's my approach (using C):
#include <ctype.h>
#include <stdio.h>
void format_str(char *target, char *source) {
int count;
char last;
while (*source != '\0') {
*target = *source;
last = *target;
target++;
source++;
for (count = 1; *source == last; source++, count++)
; /* Intentionally left blank */
if (count > 1)
target += sprintf(target, "%d", count);
}
*target = '\0';
}
void convert_back(char *target, char *source) {
char last;
int val;
while (*source != '\0') {
if (!isdigit((unsigned char) *source)) {
last = *source;
*target = last;
target++;
source++;
}
else {
for (val = 0; isdigit((unsigned char) *source); val = val*10 + *source - '0', source++)
; /* Intentionally left blank */
while (--val) {
*target = last;
target++;
}
}
}
*target = '\0';
}
format_str compresses the string, and convert_back uncompresses it.
Your code "works", but it doesn't adhere to some common patterns used in C++. You should have:
used std::string instead of plain char* array(s)
pass that string as const reference to avoid modification, since you write the result somewhere else;
use C++11 features such as ranged based for loops and lambdas as well.
I think the interviewer's purpose was to test your ability to deal with the C++11 standard, since the algorithm itself was pretty trivial.
Perhaps the interviewer wanted to test your knowledge of existing standard library tools. Here's how my take could look in C++:
#include <string>
#include <sstream>
#include <algorithm>
#include <iostream>
typedef std::string::const_iterator Iter;
std::string foo(Iter first, Iter last)
{
Iter it = first;
std::ostringstream result;
while (it != last) {
it = std::find_if(it, last, [=](char c){ return c != *it; });
result << *first << (it - first);
first = it;
}
return result.str();
}
int main()
{
std::string s = "aaabbbbbbccddde";
std::cout << foo(s.begin(), s.end());
}
An extra check is needed for empty input.
try this
std::string str="aabbbccccddddd";
for(int i=0;i<255;i++)
{
int c=0;
for(int j=0;j<str.length();j++)
{
if(str[j] == i)
c++;
}
if(c>0)
printf("%c%d",i,c);
}
My naive approach:
void pack( char const * SrcStr, char * DstBuf ) {
char const * Src_Ptr = SrcStr;
char * Dst_Ptr = DstBuf;
char c = 0;
int RepeatCount = 1;
while( '\0' != *Src_Ptr ) {
c = *Dst_Ptr = *Src_Ptr;
++Src_Ptr; ++Dst_Ptr;
for( RepeatCount = 1; *Src_Ptr == c; ++RepeatCount ) {
++Src_Ptr;
}
if( RepeatCount > 1 ) {
Dst_Ptr += sprintf( Dst_Ptr, "%i", RepeatCount );
RepeatCount = 1;
}
}
*Dst_Ptr = '\0';
};
void unpack( char const * SrcStr, char * DstBuf ) {
char const * Src_Ptr = SrcStr;
char * Dst_Ptr = DstBuf;
char c = 0;
while( '\0' != *Src_Ptr ) {
if( !isdigit( *Src_Ptr ) ) {
c = *Dst_Ptr = *Src_Ptr;
++Src_Ptr; ++Dst_Ptr;
} else {
int repeat_count = strtol( Src_Ptr, (char**)&Src_Ptr, 10 );
memset( Dst_Ptr, c, repeat_count - 1 );
Dst_Ptr += repeat_count - 1;
}
}
*Dst_Ptr = '\0';
};
But if interviewer asks for error-handling than solution turns to be much more complex (and ugly). My portable approach:
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
// for MSVC
#ifdef _WIN32
#define snprintf sprintf_s
#endif
int pack( char const * SrcStr, char * DstBuf, size_t DstBuf_Size ) {
int Err = 0;
char const * Src_Ptr = SrcStr;
char * Dst_Ptr = DstBuf;
size_t SrcBuf_Size = strlen( SrcStr ) + 1;
char const * SrcBuf_End = SrcStr + SrcBuf_Size;
char const * DstBuf_End = DstBuf + DstBuf_Size;
char c = 0;
int RepeatCount = 1;
// don't forget about buffers intercrossing
if( !SrcStr || !DstBuf || 0 == DstBuf_Size \
|| (DstBuf < SrcBuf_End && DstBuf_End > SrcStr) ) {
return 1;
}
// source string must contain no digits
// check for destination buffer overflow
while( '\0' != *Src_Ptr && Dst_Ptr < DstBuf_End - 1 \
&& !isdigit( *Src_Ptr ) && !Err ) {
c = *Dst_Ptr = *Src_Ptr;
++Src_Ptr; ++Dst_Ptr;
for( RepeatCount = 1; *Src_Ptr == c; ++RepeatCount ) {
++Src_Ptr;
}
if( RepeatCount > 1 ) {
int res = snprintf( Dst_Ptr, DstBuf_End - Dst_Ptr - 1, "%i" \
, RepeatCount );
if( res < 0 ) {
Err = 1;
} else {
Dst_Ptr += res;
RepeatCount = 1;
}
}
}
*Dst_Ptr = '\0';
return Err;
};
int unpack( char const * SrcStr, char * DstBuf, size_t DstBuf_Size ) {
int Err = 0;
char const * Src_Ptr = SrcStr;
char * Dst_Ptr = DstBuf;
size_t SrcBuf_Size = strlen( SrcStr ) + 1;
char const * SrcBuf_End = SrcStr + SrcBuf_Size;
char const * DstBuf_End = DstBuf + DstBuf_Size;
char c = 0;
// don't forget about buffers intercrossing
// first character of source string must be non-digit
if( !SrcStr || !DstBuf || 0 == DstBuf_Size \
|| (DstBuf < SrcBuf_End && DstBuf_End > SrcStr) || isdigit( SrcStr[0] ) ) {
return 1;
}
// check for destination buffer overflow
while( '\0' != *Src_Ptr && Dst_Ptr < DstBuf_End - 1 && !Err ) {
if( !isdigit( *Src_Ptr ) ) {
c = *Dst_Ptr = *Src_Ptr;
++Src_Ptr; ++Dst_Ptr;
} else {
int repeat_count = strtol( Src_Ptr, (char**)&Src_Ptr, 10 );
if( !repeat_count || repeat_count - 1 > DstBuf_End - Dst_Ptr - 1 ) {
Err = 1;
} else {
memset( Dst_Ptr, c, repeat_count - 1 );
Dst_Ptr += repeat_count - 1;
}
}
}
*Dst_Ptr = '\0';
return Err;
};
int main() {
char str[] = "aabbbccccddddd";
char buf1[128] = {0};
char buf2[128] = {0};
pack( str, buf1, 128 );
printf( "pack: %s -> %s\n", str, buf1 );
unpack( buf1, buf2, 128 );
printf( "unpack: %s -> %s\n", buf1, buf2 );
return 0;
}
Test: http://ideone.com/Y7FNE3. Also works in MSVC.
Try to make do with less boilerplate:
#include <iostream>
#include <iterator>
#include <sstream>
using namespace std;
template<typename in_iter,class ostream>
void torle(in_iter i, ostream &&o)
{
while (char c = *i++) {
size_t n = 1;
while ( *i == c )
++n, ++i;
o<<c<<n;
}
}
template<class istream, typename out_iter>
void fromrle(istream &&i, out_iter o)
{
char c; size_t n;
while (i>>c>>n)
while (n--) *o++=c;
}
int main()
{
typedef ostream_iterator<char> to;
string line; stringstream converted;
while (getline(cin,line)) {
torle(begin(line),converted);
cout<<converted.str()<<'\n';
fromrle(converted,ostream_iterator<char>(cout));
cout<<'\n';
}
}
I made this program just out of interest and wanted to make it better. My problem is that I want to make a nested for-loop to carry out the iterations but I can't get my head around it, I have tried many times but my head is melting. Any help would be greatly appreciated. Also for some reason on windows and openSuse (from what I have seen) the program prints out some random characters after the expected output, a solution to this would be a great bonus. Thanks !
Sorry I didn't make it clearer, the point of the code is to be able to theoretically generate every combination of letters from AAAAAAAA to ZZZZZZZZ.
1) No it's not homework
#include <iostream>
using namespace std;
int main()
{
char pass [] = {'A','A','A','A','A','A','A','A'};
while(pass[0] != '[')
{
pass[7]++;
if(pass[7]=='[')
{
pass[6]++;
pass[7] = 'A';
}
if(pass[6] == '[')
{
pass[6] = 'A';
pass[5]++;
}
if(pass[5] == '[')
{
pass[5] = 'A';
pass[4]++;
}
if(pass[4] == '[')
{
pass[4] = 'A';
pass[3]++;
}
if(pass[3] == '[')
{
pass[3] = 'A';
pass[2]++;
}
if(pass[2] == '[')
{
pass[2] = 'A';
pass[1]++;
}
if(pass[1] == '[')
{
pass[1] = 'A';
pass[0]++;
}
cout << pass << endl;
}
return 0;
}
Maybe like this:
const char char_first = 'A';
const char char_last = '[';
const unsigned int passlen = 8;
while (pass[0] != char_last)
{
++pass[passlen - 1];
for (unsigned int i = passlen - 1; i != 0; --i)
{
if (pass[i] == char_last)
{
++pass[i - 1]; // OK, i is always > 0
pass[i] = char_first;
}
}
}
For printing, include <string> and say:
std::cout << std::string(pass, passlen) << std::endl;
I took the liberty of making a few of the magic numbers into constants. If you're ever going to refactor this into a separate function, you'll see the merit of this.
Since (to output it) you use pass as a C string, it should be null terminated. Since it is not, garbage is printed. So you could define it as:
char pass [] = {'A','A','A','A','A','A','A','A','\0'};
or simpler
char pass[] = "AAAAAAAAA";
I'd forget about carrying on my own and just convert to/from numbers. What you're doing here is basically printing a numbers whose digits range from 'A' to ']', mappable to 0-28 via the magic of ASCII (why no ^ in passwords?)
Printing the number of anything then really boils down to
#include <iostream>
#include <cmath>
using namespace std;
std::string format(long num, int ndigits) {
if(ndigits == 0) {
return "";
} else {
char digit = 'A' + num % 28;
return format(num / 28, ndigits - 1) + digit;
}
}
int main()
{
for(int i = 0 ; i < powl(28,8) ; ++i) {
cout << format(i, 8) << endl;
}
}
You may still want to work in a char array instead of producing a billion temporary strings if you're serious about the loop, but the principle stays the same.
First try to find the common parts in the expressions looking like
if(pass[7]=='[')
{
pass[6]++;
pass[7] = 'A';
}
You should think along a line like "There's always the same number here, and a one-lower number there". Then, you replace that notion of a number with a variable and find out which range the variable has. KerrekSB gave you a solution, try to arrive at similar code from your own reasoning.
You just have to play a bit with your while and make it fit a for-loop.
while(pass[0] != '[') becomes for (i=0; pass[0] != '['; i++)
then you can replace all ifs with only one:
if(pass[i+1] == '[')
{
pass[i+1] = 'A';
pass[i]++;
}
How did we come to that conclusion? Well if you check all your if-statements all that changes between them is the indices. You can see clearly that pattern so you just replace the indices with a variable.
For starters, this is definitely not a case for a nested loop. In fact,
your entire code boils down to:
pass = initialPattern();
while ( isValidPattern( pass ) ) {
nextPattern( pass );
std::cout << pass << std::endl;
}
(But I wonder if you don't really mean to do the output before the
increment.)
Now all you have to do is define the type of pass and relevant
functions; you might even consider
putting everything in a class, since all of the functions operate on the
same data instance.
Judging from your code, pass should be an std::string with 8
characters; the initialization could be written:
std::string pass( 8, 'A' );
isValidPattern apparently only looks at the first character. (I'm not
sure that's correct, but that's what your code does.) Something like:
bool
isValidPattern( std::string const& pattern )
{
return pattern[0] != '[';
}
according to your code, but something like:
struct NotIsUpper
{
bool operator()( char ch ) const
{
return ! ::isupper( static_cast<unsigned char>( ch ) );
}
};
bool
isValidPattern( std::string const& pattern )
{
return pattern.size() == 8
&& std::find_if( pattern.begin(), pattern.end(), NotIsUpper() )
== pattern.end();
}
would seem more appropriate. (Of course, if you're doing any sort of
coding with text, you'd already have NotIsUpper and its siblings in
your tool kit.)
Finally, nextPattern seems to be nothing more than a multi-digit
increment, where the data is stored in big-endian order. So the
following (classical) algorithm would seem appropriate:
void
nextPattern( std::string& pattern )
{
static char const firstDigit = 'A';
static char const lastDigit = 'Z';
static std::string const invalidPattern( 1, '[' );
std::string::reverse_iterator current = pattern.rbegin();
std::string::reverse_iterator end = pattern.rend();
while ( current != end && *current == lastDigit ) {
*current = firstDigit;
++ current;
}
if ( current != end ) {
++ *current;
} else {
pattern = invalidPattern;
}
}
Formally, there is no guarantee in the standard that the letters will
be encoded in sequential ascending order, so for maximum portability,
you probably should in fact use an std::vector<int> with values in the
range [0, 26), and map those to letters just befor output. This
would be trivial if you put all of these operations in a class, since
the internal representation wouldn't be visible to the client code.
Something like:
class PatternGenerator
{
std::vector<int> myData;
public:
explicit PatternGenerator()
: myData( 8, 0 )
{
}
void next()
{
static int const lastDigit = 26;
std::vector<int>::reverse_iterator current = pattern.rbegin();
std::vector<int>::reverse_iterator end = pattern.rend();
while ( current != end && *current == lastDigit - 1 ) {
*current = 0;
++ current;
}
if ( current != end ) {
++ *current;
} else {
myData.front() = lastDigit;
}
}
bool isValid() const
{
return myData.front() < lastDigit;
}
friend std::ostream& operator<<(
std::ostream& dest, PatternGenerator const& obj )
{
static char const characterMap[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for ( std::vector<int>::iterator current = obj.myData.current();
current != obj.myData.end():
++ current ) {
dest << characterMap[*current];
}
return dest;
}
};
(Note that things like isValid become simpler, because they can depend on the class invariants.)
Given this, all you have to write is:
int
main()
{
PatternGenerator pass;
while ( pass.isValid() ) {
std::cout << pass << std::endl;
pass.next();
}
return 0;
}
To do nested loops, you need to turn it inside-out.
You've written the code thinking as follows: go through all the possibilities for the last symbol, then change the second-last once and go back, etc. That's like counting up from 1, getting to 10 and putting a 1 in the tens column, etc.
Nested loops work the other way: go through the possibilities for the first symbol, allowing the inner loops to take care of possibilities for the other symbols each time. i.e., "list all those numbers, in order, that start with 0 in the millions place, then the ones that start with 1 etc.". In the outermost loop, you just set that value for the first digit, and the nested loops take care of the rest.