Reverse a string in place in C++ [duplicate] - c++

This question already has answers here:
How do you reverse a string in place in C or C++?
(21 answers)
Closed 8 years ago.
"Write a function that takes a string (character pointer) as input and returns the string reversed. The function should reverse the string in place and return it as the return value of the function."
char *strrev(char *str) {
char* end = str;
char tmp = 0;
if(str) {
while(*end) {
end++;
}
--end;
while(end > str) {
tmp = *end;
*end-- = *str;
*str++ = tmp;
}
}
}
I am new to C++. I am facing difficulty with this. Can you please correct my code.

Here's one for fun:
You will NOT want to turn this in as your assignment :/
See it Live On Coliru
#include <string>
std::string rev(std::string x)
{
auto a=x.begin();
auto b=x.rbegin();
while (a<b.base())
std::swap(*a++, *b++);
return x;
}
#include <iostream>
int main()
{
std::cout << rev("") << "\n";
std::cout << rev("1") << "\n";
std::cout << rev("12") << "\n";
std::cout << rev("123") << "\n";
std::cout << rev("Hello world!") << "\n";
}
Output:
 
1
21
321
!dlrow olleH

So your function, strrev claims to return a char * but it returns... well... nothing. That means that if you use the return value of this function, you're reading some random memory. You're in the realm of undefined behavior at that point and anything goes.
Your compiler should have warned you about this and you should learn to read and understand those warnings - they're there for a reason. If, perchance, it did not warn you, you need to crank up the warnings your compiler generates.
Once you fix your function to return the correct value, it will work as expected. Here's a hint: if you want to reverse the string in place then where would the start of the string be located in memory?

There is a standard algorithm in the STL which reverses a range.
char *strrev(char *str) {
const size_t n = strlen(str);
std::reverse(str, str+n);
return str;
}
Or better using C++ strings
void reverse(std::string& s) {
std::reverse(s.begin(), s.end());
}

Related

Cannot directly convert number to hex null-terminated string, has to convert to std::string then use .c_str()

I've tried to convert an integer to a hex null-terminated (or "C-style") string but I cannot use it with printf or my custom log function. It only works if I convert it to an std::string then use .c_str() when passing it as a parameter, which produces ugly, hard-to-understand code.
It's important to know that using std::string and appending to it with "str +=" does work.
const char* IntToHexString(int nDecimalNumber) {
int nTemp = 0;
char szHex[128] = { 0 };
char hex[] = { '0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F' };
while (nDecimalNumber > 0) {
nTemp = nDecimalNumber % 16;
sprintf(szHex, "%s%s", hex[nTemp], szHex);
nDecimalNumber = nDecimalNumber / 16;
}
sprintf(szHex, "0x%s", szHex);
return szHex;
}
I've tried to use Visual Studio Debugger but it doesn't show any error messages, because crashes somewhere in a DLL that has no symbols loaded
Your main problem is that you define a variable on the stack, locally in the function, and then return it.
After the function returns, the char* will point to "somewhere", to an undefined position. That is a major bug. You have also other bugs that have been commented on already. Like sprintf(szHex, "0x%s", szHex);, which is undefined behaviour (UB) or sprintf(szHex, "%s%s", hex[nTemp], szHex); which has the same problem + additionally a wrong format string.
The more C++ solution would be, as already shown in many posts:
#include <iostream>
#include <string>
#include <iomanip>
#include <sstream>
std::string toHexString(unsigned int hexValue)
{
std::ostringstream oss;
oss << "0x" << std::hex << hexValue;
return std::string(oss.str());
}
int main()
{
std::cout << toHexString(15) << '\n';
// or directly
std::cout << "0x" << std::hex << 15 << '\n';
return 0;
}
Of course a C-Style solution is also possible.
But all the following I would not recommend:
If you want to stick to C like solution with char *, you could make the char szHex[128] = { 0 }; static. Or, even better, pass in the pointer to a buffer and return its address, like in
#include <stdio.h>
#include <iostream>
char* toHexCharP(unsigned int hexValue, char *outBuffer, const size_t maxSizeOutBuffer)
{
snprintf(outBuffer,maxSizeOutBuffer-1,"0x%X",hexValue);
return outBuffer;
}
constexpr size_t MaxBufSize = 100U;
int main()
{
char buf[MaxBufSize];
std::cout << toHexCharP(15, buf, MaxBufSize) << '\n';
return 0;
}
But as said, I would not recomend. Too dangerous.
Your solution should look as follows:
std::string IntToHexString(int nDecimalNumber) {
std::ostringstream str;
str << std::hex << nDecimalNumber;
return str.str();
}
// ...
std::string transformed = IntToHexString(123);
You can then use transformed.c_str() to get your string as const char*.
Unless you have reasons to do so, you should never work with const char* in modern C++. Use std::string::c_str() if you need to.

Reversing the Character Case of a string

I'm stuck on a particular problem. I'm trying to take a string, and reverse the character cases in the string.
For Example: "HaVinG FuN" should flip to "hAvINg fUn."
I think it has something to do with my loop or my If/Else statements. What am I missing? All capitalized characters come out capitalized still. All lower case characters also come out capitalized as well... My other two functions are behaving correctly, but not my reverseFunct function... FYI I've omitted the other functions to try to cut-down on clutter and focus on my problem.
#include "stdafx.h"
#include <string>
#include <iostream>
#include <cctype>
#include <cstring>
using namespace std;
// Function Prototypes
void upperFunct(char *);
void lowerFunct(char *);
void reverseFunct(char *);
int main()
{
cout << "Enter a string: " << endl;
char ltrs [300];
cin.getline(ltrs, 300);
char *ptr = nullptr;
ptr = ltrs;
upperFunct(ptr);
lowerFunct(ptr);
reverseFunct(ptr);
return 0;
}
//----------------------------------//
void upperFunct(char *ltrptr)
{
int count = 0;
while (ltrptr[count] != '\0')
{
ltrptr[count] = toupper(ltrptr[count]);
count++;
}
{
cout << "---> toupper function: " << ltrptr << endl;
}
}
//------------------------------------//
void lowerFunct(char *ltrptr)
{
int count = 0;
while (ltrptr[count] != '\0')
{
ltrptr[count] = tolower(ltrptr[count]);
count++;
}
cout << "---> tolower function: " << ltrptr << endl;
}
//------------------------------------//
void reverseFunct(char *ltrptr) // <-----NOT REVERSING CHARACTERS
{
int count = 0;
while (ltrptr[count] != '\0')
{
if (isupper(ltrptr[count]))
{
ltrptr[count] = tolower(ltrptr[count]);
}
else
{
ltrptr[count] = toupper(ltrptr[count]);
}
count++;
}
cout << "---> reverse function: " << ltrptr << endl;
}
Your check for lowercase letters reads as
else if (islower(ltrptr[count]));
Notice the extra semicolon.
This semicolon terminates the if statement, and thus the succeeding conversion to uppercase is not a then-clause to this if statement but rather is executed unconditionally on every character.
Change like this
// Function Prototypes "HaVinG FuN" should flip to "hAvINg fUn."
void reverseFunct(char *);
int main()
{
//cout << "Enter a string: " << endl;
char ltrs[300] = "HaVinG FuN";
//cin.getline(ltrs, 300);
char *ptr = nullptr;
ptr = ltrs;
reverseFunct(ptr);
ptr = nullptr;
return 0;
}
void reverseFunct(char *ltrptr) // <-----NOT REVERSING CHARACTERS
{
int count = 0;
while (ltrptr[count] != '\0')
{
if (isupper(ltrptr[count]))
{
ltrptr[count] = tolower(ltrptr[count]);
}
else
{
ltrptr[count] = toupper(ltrptr[count]);
}
count++;
}
cout << "---> reverse function: " << ltrptr << endl;
}
You're writing C code. Here's a C++ way to do it:
#include <string>
#include <algorithm>
char reverse_case_char(char c) {
const auto uc = static_cast<unsigned char>(c); // Sic.
return ::isupper(uc)? ::tolower(uc): ::toupper(uc);
}
void reverse_case(std::string& str) {
std::transform(str.begin(), str.end(), str.begin(), reverse_case_char);
}
#include <cassert>
int main()
{
std::string fun = "HaVinG FuN";
reverse_case(fun);
assert(fun == "hAvINg fUn");
return 0;
}
Others have already pointed out the mistake in your code so no need to repeat that. Instead this answer will give some alternative ways of implementing the task.
Your code is more C-style than C++ style. C++ has a number of functions/features that will allow you to write this in much shorter forms.
char ltrs[300] = "HaVinG FuN";
for (auto& ch : ltrs) ch = islower(ch) ? toupper(ch) : tolower(ch);
std::cout << ltrs << std::endl;
or
char ltrs[300] = "HaVinG FuN";
std::for_each(ltrs, ltrs + strlen(ltrs), [](char& ch)
{ ch = islower(ch) ? toupper(ch) : tolower(ch); });
std::cout << ltrs << std::endl;
or using the std::string
std::string str("HaVinG FuN");
for (auto& ch : str) ch = islower(ch) ? toupper(ch) : tolower(ch);
std::cout << str << std::endl;
Using these C++ functions/features makes the program shorter, easier to understand and the risk of bugs is lower.
Thanks for the help!!! I ended up figuring out my answer, while being able to maintain my less-than elegant code that is fitting with my class. Bipll ended up giving me what I was after, something to think about in terms that my original array was being modified each time.
I realize that my solution is sloppy and not appropriate for a work environment, but it is in-line with my homework assignment, as our teacher is encouraging us to learn C++ from the ground-up, not getting too much direct answers from places like SO. So I'm glad I learned a bit from here, as well as an indirect way to help me see my issues.
I ended up making a copy of my original array, and just passing that copy to my last reversing function. I was able to use the original array for the first 2 functions because the 1st function capitalized each character in the array, while the 2nd made them all lowercase. The 3rd function, the reverse, therefore had to have access to the original array, but in the 3rd order. The easiest way for a noob like me, given where I am in the class, was to make a copy of the 1st array and use that for the 3rd function.
//Snippet of code I needed
int main()
{
int index = 0;
cout << "Enter a string: " << endl;
const int Size = 300;
char ltrs[Size];
cin.getline(ltrs, Size);
char arrayCopy[Size];
char *ptr = nullptr;
char *ptr2 = nullptr;
ptr = ltrs;
//Copy of ltrs Array
//----------------------------------//
while (ptr[index] != '\0') //
{ //
arrayCopy[index] = ptr[index]; //
index++; //
} //
arrayCopy[index] = '\0'; //
//
ptr2 = arrayCopy; //
//----------------------------------//
return 0;
}
// Function to Reverse
void reverseFunct(char *ltrptr)
{
int count = 0;
while (ltrptr[count] != '\0')
{
if (isupper(ltrptr[count]))
{
ltrptr[count] = tolower(ltrptr[count]);
}
else
{
ltrptr[count] = toupper(ltrptr[count]);
}
count++;
}
cout << "---> reverse function: " << ltrptr << endl;
}

Difficulties with string declaration/reference parameters (c++)

Last week I got an homework to write a function: the function gets a string and a char value and should divide the string in two parts, before and after the first occurrence of the existing char.
The code worked but my teacher told me to do it again, because it is not well written code. But I don't understand how to make it better. I understand so far that defining two strings with white spaces is not good, but i get out of bounds exceptions otherwise. Since the string input changes, the string size changes everytime.
#include <iostream>
#include <string>
using namespace std;
void divide(char search, string text, string& first_part, string& sec_part)
{
bool firstc = true;
int counter = 0;
for (int i = 0; i < text.size(); i++) {
if (text.at(i) != search && firstc) {
first_part.at(i) = text.at(i);
}
else if (text.at(i) == search&& firstc == true) {
firstc = false;
sec_part.at(counter) = text.at(i);
}
else {
sec_part.at(counter) = text.at(i);
counter++;
}
}
}
int main() {
string text;
string part1=" ";
string part2=" ";
char search_char;
cout << "Please enter text? ";
getline(cin, text);
cout << "Please enter a char: ? ";
cin >> search_char;
divide(search_char,text,aprt1,part2);
cout << "First string: " << part1 <<endl;
cout << "Second string: " << part2 << endl;
system("PAUSE");
return 0;
}
I would suggest you, learn to use c++ standard functions. there are plenty utility function that can help you in programming.
void divide(const std::string& text, char search, std::string& first_part, std::string& sec_part)
{
std::string::const_iterator pos = std::find(text.begin(), text.end(), search);
first_part.append(text, 0, pos - text.begin());
sec_part.append(text, pos - text.begin());
}
int main()
{
std::string text = "thisisfirst";
char search = 'f';
std::string first;
std::string second;
divide(text, search, first, second);
}
Here I used std::find that you can read about it from here and also Iterators.
You have some other mistakes. you are passing your text by value that will do a copy every time you call your function. pass it by reference but qualify it with const that will indicate it is an input parameter not an output.
Why is your teacher right ?
The fact that you need to initialize your destination strings with empty space is terrible:
If the input string is longer, you'll get out of bound errors.
If it's shorter, you got wrong answer, because in IT and programming, "It works " is not the same as "It works".
In addition, your code does not fit the specifications. It should work all the time, independently of the current value which is stored in your output strings.
Alternative 1: your code but working
Just clear the destination strings at the beginning. Then iterate as you did, but use += or push_back() to add chars at the end of the string.
void divide(char search, string text, string& first_part, string& sec_part)
{
bool firstc = true;
first_part.clear(); // make destinations strings empty
sec_part.clear();
for (int i = 0; i < text.size(); i++) {
char c = text.at(i);
if (firstc && c != search) {
first_part += c;
}
else if (firstc && c == search) {
firstc = false;
sec_part += c;
}
else {
sec_part += c;
}
}
}
I used a temporary c instead of text.at(i) or text\[i\], in order to avoid multiple indexing But this is not really required: nowadays, optimizing compilers should produce equivalent code, whatever variant you use here.
Alternative 2: use string member functions
This alternative uses the find() function, and then constructs a string from the start until that position, and another from that position. There is a special case when the character was not found.
void divide(char search, string text, string& first_part, string& sec_part)
{
auto pos = text.find(search);
first_part = string(text, 0, pos);
if (pos== string::npos)
sec_part.clear();
else sec_part = string(text, pos, string::npos);
}
As you understand yourself these declarations
string part1=" ";
string part2=" ";
do not make sense because the entered string in the object text can essentially exceed the both initialized strings. In this case using the string method at can result in throwing an exception or the strings will have trailing spaces.
From the description of the assignment it is not clear whether the searched character should be included in one of the strings. You suppose that the character should be included in the second string.
Take into account that the parameter text should be declared as a constant reference.
Also instead of using loops it is better to use methods of the class std::string such as for example find.
The function can look the following way
#include <iostream>
#include <string>
void divide(const std::string &text, char search, std::string &first_part, std::string &sec_part)
{
std::string::size_type pos = text.find(search);
first_part = text.substr(0, pos);
if (pos == std::string::npos)
{
sec_part.clear();
}
else
{
sec_part = text.substr(pos);
}
}
int main()
{
std::string text("Hello World");
std::string first_part;
std::string sec_part;
divide(text, ' ', first_part, sec_part);
std::cout << "\"" << text << "\"\n";
std::cout << "\"" << first_part << "\"\n";
std::cout << "\"" << sec_part << "\"\n";
}
The program output is
"Hello World"
"Hello"
" World"
As you can see the separating character is included in the second string though I think that maybe it would be better to exclude it from the both strings.
An alternative and in my opinion more clear approach can look the following way
#include <iostream>
#include <string>
#include <utility>
std::pair<std::string, std::string> divide(const std::string &s, char c)
{
std::string::size_type pos = s.find(c);
return { s.substr(0, pos), pos == std::string::npos ? "" : s.substr(pos) };
}
int main()
{
std::string text("Hello World");
auto p = divide(text, ' ');
std::cout << "\"" << text << "\"\n";
std::cout << "\"" << p.first << "\"\n";
std::cout << "\"" << p.second << "\"\n";
}
Your code will only work as long the character is found within part1.length(). You need something similar to this:
void string_split_once(const char s, const string & text, string & first, string & second) {
first.clear();
second.clear();
std::size_t pos = str.find(s);
if (pos != string::npos) {
first = text.substr(0, pos);
second = text.substr(pos);
}
}
The biggest problem I see is that you are using at where you should be using push_back. See std::basic_string::push_back. at is designed to access an existing character to read or modify it. push_back appends a new character to the string.
divide could look like this :
void divide(char search, string text, string& first_part,
string& sec_part)
{
bool firstc = true;
for (int i = 0; i < text.size(); i++) {
if (text.at(i) != search && firstc) {
first_part.push_back(text.at(i));
}
else if (text.at(i) == search&& firstc == true) {
firstc = false;
sec_part.push_back(text.at(i));
}
else {
sec_part.push_back(text.at(i));
}
}
}
Since you aren't handling exceptions, consider using text[i] rather than text.at(i).

C++ Palindrome Creator Recursion Program

#include <iostream>
using namespace std;
void palindrome(char *s){
if (s[1] == 0)
{
return;
}
cout << s[0];
palindrome(++s);
cout << s[0];
}
int main(){
char s[30]="foobar";
palindrome(s);
cout << endl;
}
I have to create a recursive function that turns any cstring into a palindrome. This is a homework question and the only code I can modify is inside the palindrome function. This code almost works. The problem is it returns "foobaraboo" and leaves off the last letter. Any tips to point me in the right direction? I've played around with the code for quite a while and can't seem to figure out how to change it so the last character of the palindrome shows up.
You have two bugs in your logic. The first one:
if (s[1] == 0)
{
return;
}
This should be:
if (*s == 0)
{
return;
}
if your entire input is the string "x", your expected result is the palindrom "xx", but this bug will result in an empty string getting printed.
You also have a more fundamental bug:
cout << s[0];
palindrome(++s);
cout << s[0];
Your intent here is to print the current character, recursively print the rest of the palindrome string, and then re-print the same character.
But because of the ++s, you're printing the wrong character, the second time around.
This should be:
cout << s[0];
palindrome(s+1);
cout << s[0];
EDIT: a question has been raised whether you want the last character printed twice. If not, then this should be:
cout << s[0];
if (s[1])
{
palindrome(s+1);
cout << s[0];
}
Here you are
#include <iostream>
void palindrome( const char *s )
{
if ( s[0] )
{
std::cout << s[0];
if ( s[1] )
{
palindrome( s + 1 );
std::cout << s[0];
}
}
}
int main()
{
const char *s = "foobar";
palindrome( s );
std::cout << std::endl;
}
The program output is
foobaraboof

C++ segmentation fault and recurrent function

In this program, I used a recurrent function which is Strlen.
However, prompt tells me that it is
zsh:
segmentation fault ./Strlen5
Does anyone know why it happens ?
My Code:
#include <iostream>
using namespace std;
size_t Strlen(char* str) {
size_t i = 0;
if (str[i] != '\0') {
i += 1;
return 1 + Strlen(str);
} else {
return 0;
}
}
void ShowLength(char* str) {
cout << "length:\"" << str << "\" : " << Strlen(str) << endl;
}
int main() {
ShowLength("Hello");
ShowLength("Titan");
}
You never increment your pointer. You need to start one index ahead each time you recurse.
return 1 + Strlen(str + 1);
Also you should use const char* instead of char*
#include <iostream>
using namespace std;
size_t Strlen(const char* str) {
if (str[0] != '\0') {
return 1 + Strlen(str + 1);
} else {
return 0;
}
}
void ShowLength(const char* str) {
cout << "length:\"" << str << "\" : " << Strlen(str) << endl;
}
int main() {
ShowLength("Hello");
ShowLength("Titan");
}
Working demo
By the way the reason you got a segmentation fault is due to how a function call is constructed.
Since you need to pass on the stack the arguments and the pointer to next the instruction after the call you "waste" a portion of your stack every time you call a function (until you return).
This usually is not a problem, but if you call an infinite number of function one after the other you face a stack overflow.
Further reading:
Why infinite recursion leads to seg fault