I have no idea why it doesn't work. What is more I can't even say what are errors about ;/
Can any1 explain what are errors about?
The code is suposed to :
create a string with word like - mom.
then create 2d array to fill it by string. Free spaces fill with _.So mom box =
[m] [o]
[m] [_]
now fill next array with text that follows from colums. mom_ filled to new array will look like mmo_. Then I cout crypted text. I hope u understood whatI did there :D
here is code
//wal = kolumny=wiersze
#include <cstdlib>
#include <iostream>
#include <string>
#include <cstring>
using namespace std;
void pole(int &a,const int &l);
void tab(const char &s[],char &d[], char &f[],const int a);
int main(){
string code;
cin >> code;
int wall=1;
int d=code.length();
char tekst[d];
pole(wall,d);
strcpy(tekst,code);
char kw[wall][wall];
char szyfr[d];
tab(tekst,kw,szyfr,wall);
for (int i=0;i<d;i++)
cout << szyfr[i] << endl;
system("PAUSE");
return 0;
}
void pole(int &a,const int &l){
if (a*a < l)
pole(a+=1,l);
}
void tab(const char &s[],char &d[], char &f[],const int a){
int i=0;
for (int x=0;x<a;x++,i++){
for (int y=0;y<a;y++,i++){
if(s[i])
d[x][y]=s[i];
else d[x][y]=='_';
f[i]=d[x][y];
}
}
}
d[x][y] has no meaning in tab d is a single dimension array. You will have to pass the first dimension as argument and use it when indexing. Something like:
void tab(const char &s[],char* &d, char &f[],const int a, int d_num_cols){
int i=0;
for (int x=0;x<a;x++,i++){
for (int y=0;y<a;y++,i++){
if(s[i])
d[x*d_num_cols + y]=s[i];
else d[x*d_num_cols + y]=='_';
f[i]=d[x*d_num_cols + y];
}
}
}
Related
I'm sorting an array of names and my IDE is giving me "no matching function for call to 'strcpy'.
Here are the values I set up:
char Names [MaxNames] [MaxCharsPerName + 1];
const int MaxNames (20);
const int MaxCharsPerName (15);
Here is my function
void SortNames (const char Names[][MaxCharsPerName + 1], int NumNames)
{
int i;
int NumElements;
bool Sorted;
char Temp; // size 15?
NumElements = NumNames;
do {
Sorted = true;
NumElements--;
for (i = 0; i < NumNames; i++)
{
if(Names[i-1] > Names[i]){
strcpy(Temp, Names[i]);
strcpy(Names[i], Names[i+1]);
strcpy(Names[i+1], Temp);
}
}
} while (!Sorted);
Do I have to use a reference or something?
Oh and these are at the top:
include
using namespace std;
#include "Constants.h"
#include "Functions.h"
#include <string.h>
#include <stdio.h>
char Temp; // size 15?
Temp is a char and strcpy expects a char *(and compiler will give a compilation error in your case).
Use a char array instead-
char Temp[MaxCharsPerName + 1]; // any desired size but reserve place for '\0'
And then pass it to strcpy.
Sorting an array of names could be done much easier:
vector<string> names;
// ...
sort(begin(names), end(names), less<string>());
for (const auto& name : names) cout << name << '\n';
I wrote a simple C++ program to reverse a string. I store a string in character array. To reverse a string I am using same character array and temp variable to swap the characters of an array.
#include<iostream>
#include<string>
using namespace std;
void reverseChar(char* str);
char str[50],rstr[50];
int i,n;
int main()
{
cout<<"Please Enter the String: ";
cin.getline(str,50);
reverseChar(str);
cout<<str;
return 0;
}
void reverseChar(char* str)
{
for(i=0;i<sizeof(str)/2;i++)
{
char temp=str[i];
str[i]=str[sizeof(str)-i-1];
str[sizeof(str)-i-1]=temp;
}
}
Now this method is not working and, I am getting the NULL String as result after the program execution.
So I want to know why I can't equate character array, why wouldn't this program work. And what is the solution or trick that I can use to make the same program work?
sizeof(str) does not do what you expect.
Given a char *str, sizeof(str) will not give you the length of that string. Instead, it will give you the number of bytes that a pointer occupies. You are probably looking for strlen() instead.
If we fixed that, we would have:
for(i=0;i<strlen(str)/2;i++)
{
char temp=str[i];
str[i]=str[strlen(str)-i-1];
str[strlen(str)-i-1]=temp;
}
This is C++, use std::swap()
In C++, if you want to swap the contents of two variables, use std::swap instead of the temporary variable.
So instead of:
char temp=str[i];
str[i]=str[strlen(str)-i-1];
str[strlen(str)-i-1]=temp;
You would just write:
swap(str[i], str[sizeof(str) - i - 1]);
Note how much clearer that is.
You're using C++, just use std::reverse()
std::reverse(str, str + strlen(str));
Global variables
It's extremely poor practice to make variables global if they don't need to be. In particular, I'm referring to i about this.
Executive Summary
If I was to write this function, it would look like one of the two following implementations:
void reverseChar(char* str) {
const size_t len = strlen(str);
for(size_t i=0; i<len/2; i++)
swap(str[i], str[len-i-1]);
}
void reverseChar(char* str) {
std::reverse(str, str + strlen(str));
}
When tested, both of these produce dlrow olleh on an input of hello world.
The problem is that within your function, str is not an array but a pointer. So sizeof will get you the size of the pointer, not the length of the array it points to. Also, even if it gave you the size of the array, that is not the length of the string. For this, better use strlen.
To avoid multiple calls to strlen, give the function another parameter, which tells the length:
void reverseChar(char* str, int len)
{
for(i=0; i<len/2; i++)
{
char temp=str[i];
str[i]=str[len-i-1];
str[len-i-1]=temp;
}
}
and call it with
reverseChar(str, strlen(str))
Another improvement, as mentioned in the comments, is to use std::swap in the loop body:
void reverseChar(char* str, int len)
{
for(i=0; i<len/2; i++)
{
std::swap(str[i], str[len-i-1]);
}
}
Also, there is std::reverse which does almost exactly that.
//reverse a string
#include<iostream>
using namespace std;
int strlen(char * str) {
int len = 0;
while (*str != '\0') {
len++;
str++;
}
return len;
}
void reverse(char* str, int len) {
for(int i=0; i<len/2; i++) {
char temp=str[i];
str[i]=str[len-i-1];
str[len-i-1]=temp;
}
}
int main() {
char str[100];
cin.getline(str,100);
reverse(str, strlen(str));
cout<<str<<endl;
getchar();
return 0;
}
If I were you, I would just write it like so:
int main()
{
string str;
cout << "Enter a string: " << endl;
getline(cin, str);
for (int x = str.length() - 1; x > -1; x--)
{
cout << str[x];
}
return 0;
}
This is a very simple way to do it and works great.
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
char str[80];
cout << "Enter a string bro: \n";
gets_s(str);
for (int i = strlen(str) - 1; i > -1; i--)
{
cout << str[i];
}
}
I have to admit, i have no idea how to use pointers, but I tried non the less. the problem with my program is that it shows the string in reverse, except for what was the first letter being missing and the entire string is moved one space forward with the first element being blank.
for example it show " olle" when typing "hello".
#include <iostream>
#include <string>
using namespace std;
string reverse(string word);
int main()
{
char Cstring[50];
cout<<"enter a word: ";
cin>>Cstring;
string results = reverse(Cstring);
cout <<results;
}
string reverse(string word)
{
char *front;
char *rear;
for (int i=0;i< (word.length()/2);i++)
{
front[0]=word[i];
rear[0]=word[word.length()-i];
word[i]=*rear;
word[word.length()-i]=*front;
}
return word;
}
The new code works perfectly. changed the strings to cstrings. the question technicaly asked for cstrings but i find strings easier so i work with strings then make the necesary changes to make it c string. figured out ho to initialize the rear and front as well.
#include <iostream>
#include <cstring>
using namespace std;
string reverse(char word[20]);
int main()
{
char Cstring[20];
cout<<"enter a word: ";
cin>>Cstring;
string results = reverse(Cstring);
cout <<results;
}
string reverse(char word[20])
{
char a='a';
char b='b';
char *front=&a;
char *rear=&b;
for (int i=0;i< (strlen(word)/2);i++)
{
front[0]=word[i];
rear[0]=word[strlen(word)-1-i];
word[i]=*rear;
word[strlen(word)-1-i]=*front;
}
return word;
}
char *front;
char *rear;
then later
front[0]=word[i];
rear[0]=word[strlen(word)-1-i];
Not good. Dereferencing uninitialized pointers invokes undefined behavior.
Apart from that, your code is overly complicated, it calls strlen() during each iteration (and even multiple times), which is superfluous, and the swap logic is also unnecessarily complex. Try using two pointers instead and your code will be much cleaner:
void rev_string(char *str)
{
char *p = str, *s = str + strlen(str) - 1;
while (p < s) {
char tmp = *p;
*p++ = *s;
*s-- = tmp;
}
}
The thing is, however, that in C++ there's rarely a good reason for using raw pointers. How about using std::reverse() instead?
string s = "foobar";
std::reverse(s.begin(), s.end());
inline void swap(char* a, char* b)
{
char tmp = *a;
*a = *b;
*b = tmp;
}
inline void reverse_string(char* pstart, char* pend)
{
while(pstart < pend)
{
swap(pstart++, pend--);
}
}
int main()
{
char pstring[] = "asfasd Lucy Beverman";
auto pstart = std::begin(pstring);
auto pend = std::end(pstring);
pend -= 2; // end points 1 past the null character, so have to go back 2
std::cout << pstring << std::endl;
reverse_string(pstart, pend);
std::cout << pstring << std::endl;
return 0;
}
you can also do it like this:
#include <iostream>
#include <cstring>
using namespace std;
string reverse(char word[20]);
int main()
{
char Cstring[20];
cout<<"enter a word: ";
cin>>Cstring;
string results = reverse(Cstring);
cout <<results;
}
string reverse(char word[20])
{
char a='a';
char b='b';
char *front=&a;
char *rear=&b;
for (int i=0;i< (strlen(word)/2);i++)
{
*front=word[i];
*rear=word[strlen(word)-1-i];
word[i]=*rear;
word[strlen(word)-1-i]=*front;
}
return word;
}
it successfully works on my system ,i.e. on emacs+gcc on windows 7
Taken from C How To Program Deitel & Deitel 8th edition:
void reverse(const char * const sPtr)
{
if (sPtr[0] == '\0')
return;
else
reverse(&sPtr[1]);
putchar(sPtr[0]);
}
how do I pass a char vector to a char*? I know this problem could easily be solved with a predefined char[] array with a SIZE const, but I want the flexibility of a vector because there will be no predefined size.
using namespace std;
//prototype
void getnumberofwords(char*);
int main() {
//declare the input vector
vector<char> input;
/*here I collect the input from user into the vector, but I am omitting the code here for sake of brevity...*/
getnumberofwords(input);
//here is where an ERROR shows up: there is no suitable conversion from std::vector to char*
return 0;
}
void getnumberofwords(char *str){
int numwords=0;
int lengthofstring = (int)str.size();
//this ERROR says the expression must have a case
//step through characters until null
for (int index=0; index < lengthofstring; index++){
if ( *(str+index) == '\0') {
numwords++;
}
}
}
You can use data() member to get the pointer to the underlying array:
getnumberofwords(input.data());
The most obvious is to pass &your_vector[0]. Be sure to add a NUL to the end of your vector first though.
Alternatively, use std::string instead of std::vector<char>, in which case you can get a NUL-terminated string with the c_str member function.
Edit: I have to wonder, however, why getnmberofwords would be written to accept a char * unless it's some old C code that you just can't get away from using.
Given a typical definition of "word" counting some words that start out in a string can be done something like this:
std::istringstream buffer(your_string);
size_t num_words = std::distance(std::istream_iterator<std::string>(buffer),
std::istream_iterator<std::string>());
You should pass the reference of the vector to the function getnumberofwords.
void getnumberofwords(vector<char>& str){
int numwords=0;
int lengthofstring = str.size();
for (int index=0; index < lengthofstring; index++){
if ( str[index] == '\0') {
numwords++;
}
}
}
There is no method for converting the type from vector to pointer.
here's what I ended up doing which worked:
#include <iostream>
#include <cstring>
#include <string>
#include <iomanip>
using namespace std;
//prototype
void getnumberofwords(char*);
void getavgnumofletters(char*, int);
int main() {
const int SIZE=50;
char str[SIZE];
cout<<"Enter a string:";
cin.getline(str, SIZE);
getnumberofwords(str);
return 0;
}
void getnumberofwords(char *str){
int numwords=0;
int lengthstring=strlen(str);
//step through characters until null
for (int index=0; index < lengthstring; index++){
if (str[index] ==' ') {
numwords++;
}else{
continue;
}
}
numwords+=1;
cout<<"There are "<<numwords<<" in that sentence "<<endl;
getavgnumofletters(str, numwords);
}
void getavgnumofletters(char *str, int numwords) {
int numofletters=0;
double avgnumofletters;
int lengthstring=strlen(str);
//step through characters until null
for (int index=0; index < lengthstring; index++){
if (str[index] != ' ') {
numofletters++;
}else{
continue;
}
}
avgnumofletters = (double)numofletters/numwords;
cout<<"The average number of letters per word is "<<setprecision(1)<<fixed<<avgnumofletters<<endl;
}
/*
Can someone help me with this: This is a program to find all the permutations of a string of any length. Need a non-recursive form of the same. ( a C language implementation is preferred)
using namespace std;
string swtch(string topermute, int x, int y)
{
string newstring = topermute;
newstring[x] = newstring[y];
newstring[y] = topermute[x]; //avoids temp variable
return newstring;
}
void permute(string topermute, int place)
{
if(place == topermute.length() - 1)
{
cout<<topermute<<endl;
}
for(int nextchar = place; nextchar < topermute.length(); nextchar++)
{
permute(swtch(topermute, place, nextchar),place+1);
}
}
int main(int argc, char* argv[])
{
if(argc!=2)
{
cout<<"Proper input is 'permute string'";
return 1;
}
permute(argv[1], 0);
return 0;
}
Another approach would be to allocate an array of n! char arrays and fill them in the same way that you would by hand.
If the string is "abcd", put all of the "a" chars in position 0 for the first n-1! arrays, in position 1 for the next n-1! arrays, etc. Then put all of the "b" chars in position 1 for the first n-2! arrays, etc, all of the "c" chars in position 2 for the first n-3! arrays, etc, and all of the "d" chars in position 3 for the first n-4! arrays, etc, using modulo n arithmetic in each case to move from position 3 to position 0 as you are filling out the arrays.
No swapping is necessary and you know early on if you have enough memory to store the results or not.
A stack based non-recursive equivalent of your code:
#include <iostream>
#include <string>
struct State
{
State (std::string topermute_, int place_, int nextchar_, State* next_ = 0)
: topermute (topermute_)
, place (place_)
, nextchar (nextchar_)
, next (next_)
{
}
std::string topermute;
int place;
int nextchar;
State* next;
};
std::string swtch (std::string topermute, int x, int y)
{
std::string newstring = topermute;
newstring[x] = newstring[y];
newstring[y] = topermute[x]; //avoids temp variable
return newstring;
}
void permute (std::string topermute, int place = 0)
{
// Linked list stack.
State* top = new State (topermute, place, place);
while (top != 0)
{
State* pop = top;
top = pop->next;
if (pop->place == pop->topermute.length () - 1)
{
std::cout << pop->topermute << std::endl;
}
for (int i = pop->place; i < pop->topermute.length (); ++i)
{
top = new State (swtch (pop->topermute, pop->place, i), pop->place + 1, i, top);
}
delete pop;
}
}
int main (int argc, char* argv[])
{
if (argc!=2)
{
std::cout<<"Proper input is 'permute string'";
return 1;
}
else
{
permute (argv[1]);
}
return 0;
}
I've tried to make it C-like and avoided c++ STL containers and member functions (used a constructor for simplicity though).
Note, the permutations are generated in reverse order to the original.
I should add that using a stack in this way is just simulating recursion.
First one advice - don't pass std:string arguments by value. Use const references
string swtch(const string& topermute, int x, int y)
void permute(const string & topermute, int place)
It will save you a lot of unnecessary copying.
As for C++ solution, you have functions std::next_permutation and std::prev_permutation in algorithm header. So you can write:
int main(int argc, char* argv[])
{
if(argc!=2)
{
cout<<"Proper input is 'permute string'" << endl;
return 1;
}
std::string copy = argv[1];
// program argument and lexically greater permutations
do
{
std::cout << copy << endl;
}
while (std::next_permutation(copy.begin(), copy.end());
// lexically smaller permutations of argument
std::string copy = argv[1];
while (std::prev_permutation(copy.begin(), copy.end())
{
std::cout << copy << endl;
}
return 0;
}
As for C solution, you have to change variables types from std::string to char * (ugh, and you have to manage memory properly). I think similar approach - writing functions
int next_permutation(char * begin, char * end);
int prev_permutation(char * begin, char * end);
with same semantics as STL functions - will do. You can find source code for std::next_permutation with explanation here. I hope you can manage to write a similar code that works on char * (BTW std::next_permutation can work with char * with no problems, but you wanted C solution) as I am to lazy to do it by myself :-)
Have you tried using the STL? There is an algorithm called next_permutation which given a range will return true on each subsequent call until all permutations have been encountered. Works not only on strings but on any "sequence" type.
http://www.sgi.com/tech/stl/next_permutation.html
This solves the problem without recursion. The only issue is that it will generate duplicate output in the case where a character is repeated in the string.
#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<string.h>
int factorial(int n)
{
int fact=1;
for(int i=2;i<=n;i++)
fact*=i;
return fact;
}
char *str;
void swap(int i,int j)
{
char temp=str[i];
str[i]=str[j];
str[j]=temp;
}
void main()
{
clrscr();
int len,fact,count=1;
cout<<"Enter the string:";
gets(str);
len=strlen(str);
fact=factorial(len);
for(int i=0;i<fact;i++)
{
int j=i%(len-1);
swap(j,j+1);
cout<<"\n"<<count++<<". ";
for(int k=0;k<len;k++)
cout<<str[k];
}
getch();
}
#include <iostream>
#include <string>
using namespace std;
void permuteString(string& str, int i)
{
for (int j = 0; j < i; j++) {
swap(str[j], str[j+1]);
cout << str << endl;
}
}
int factorial(int n)
{
if (n != 1) return n*factorial(n-1);
}
int main()
{
string str;
cout << "Enter string: ";
cin >> str;
cout << str.length() << endl;
int fact = factorial(str.length());
int a = fact/((str.length()-1));
for (int i = 0; i < a; i++) {
permuteString(str, (str.length()-1));
}
}