I'm trying to implement a void function that takes a c string as its only parameter and reverses it and prints it. Below is my attempt at a solution however I'm not sure how to go about this problem.
void printBackwards(char forward[]) {
int i = 0;
char backwards[];
while (forward[i] != '\0') {
backwards[i] = forward[-i - 1];
i++;
}
cout << backwards;
}
Under such a condition, I guess you are expected to use recursion.
void printBackwards(char forward[]) {
if (!forward[0])
return;
printBackwards(forward + 1);
cout << forward[0];
}
Not being able to use strlen, we'll calculate it ourselves using a simple for loop. Then dynamically allocate a suitable buffer (add one character for the null terminating char, and I "cheated" by using calloc to zero the memory so I don't have to remember to set the null terminator. Then anoher simple loop to copy the original into the result in reverse.
#include <stdlib.h>
#include <stdio.h>
char *rev(char *s) {
size_t i;
char *s2 = s; // A pointer to the beginning as our first loop modifies s
for (i = 0; *s; s++, i++);
char *result = calloc(0, i + 1);
if (!result) return NULL; // In case calloc didn't allocate the requested memory.
for (size_t j = 0; j < i; j++)
result[j] = s2[i - j - 1];
return result;
}
Assuming you want to reverse the string rather than just printing it in reverse order, you first need to find the last character location (actually the position of the null terminator). Pseudo-code below (since this is an educational assignment):
define null_addr(pointer):
while character at pointer is not null terminator:
increment pointer
return pointer
Then you can use that inside a loop where you swap the two characters and move the pointers toward the center of the string. As soon as the pointers become equal or pass each other the string is reversed:
define reverse(left_pointer):
set right_pointer to null_addr(left_pointer)
while right_pointer > left_pointer plus one:
decrement right_pointer
swap character at left_pointer with character at right_pointer
increment left_pointer
Alternatively (and this appears to be the case since your attempt doesn't actually reverse the original string), if you need to print the string in reverse order without modifying it, you still find the last character. Then you run backwards through the string printing each character until you reach the first. That can be done with something like:
define print_reverse(pointer):
set right_pointer to null_addr(pointer)
while right_pointer > pointer:
decrement right_pointer
print character at right_pointer
That's probably better than creating a new string to hold the reverse of the original, and then printing that reverse.
One thing you should keep in mind. This very much appears to be a C-centric question, not a C++ one (it's using C strings rather than C++ strings, and uses C header files). If that's the case, you should probably avoid things like cout.
By using abstractions, like , your code will be much better at communication WHAT it is doing instead of HOW it is doing it.
#include <iostream>
#include <string>
#include <ranges>
int main()
{
std::string hello{ "!dlrow olleH" };
for (const char c : hello | std::views::reverse)
{
std::cout << c;
}
return 0;
}
Use a template
#include <iostream>
template<int N, int I=2>
void printBackwards(char (&forward)[N]) {
std::cout << forward[N-I];
if constexpr (I<N) printBackwards<N, I+1>(forward);
}
int main() {
char test[] = "elephant";
printBackwards(test);
}
While there seems to be several working answers, I thought I'd throw my hat in the stack (pun intended) since none of them take advantage of a FILO data structure (except #273K's answer, which uses a stack implicitly instead of explicitly).
What I would do is simply push everything onto a stack and then print the stack:
#include <stack>
#include <iostream>
void printBackwards(char forward[]) {
// Create a stack to hold our reversed string
std::stack<char> stk;
// Iterate through the string until we hit the null terminator
int i = 0;
while (forward[i] != '\0'){
stk.push(forward[i]);
++i;
}
// Iterate through the stack and print each character as we pop() it
while (stk.size() > 0){
std::cout << stk.top();
stk.pop();
}
// Don't forget the newline (assuming output lines should be separated)
std::cout << '\n';
}
int main(int argc, char* argv[]){
char s[] = "This is a string";
printBackwards(s);
return 0;
}
Hi guys as promised I have come back to add my own answer. This is my own way using array subscripts and using what I currently know.
#include <iostream>
using namespace std;
void printBackwards(char[]);
int main()
{
char word[] = "apples";
printBackwards(word);
return 0;
}
void printBackwards(char word[]) {
char* temp = word;
int count = 0;
while (*temp++ != '\0') {
count++;
}
for (int i = count - 1; i >= 0; i--) {
cout << word[i];
}
}
You can make a fixed-size buffer and create new ones if needed. Fill it reverse by moving the string offset back with every inserted character. Chars exceeding the buffer are returned to be processed later, so you can make a list of such buffers:
template<int SIZE>
struct ReversedCStr
{
static_assert(SIZE > 10); // just some minimal size treshold
// constexpr
ReversedCStr(char const* c_str, char const** tail = nullptr) noexcept
{
for(buffer[offset] = '\0'; *c_str != '\0';)
{
buffer[--offset] = *c_str++;
if(offset == 0) break;
}
if(tail) *tail = c_str;
}
//constexpr
char const* c_str() const noexcept { return buffer.data()+offset;};
private:
size_t offset = SIZE -1;
std::array<char,SIZE> buffer;
};
The tag is 'C++' so I assume you use C++ not C. The following code is C++11 so it should fit in every modern project. I posted the working example on godbolt.org.
It doesn't allocate memory, and is completely exception-free. The maximum memory wasted is {buffer_size + sizeof(char*)*number_of_chunks}, and can be easily turned into a list of reversed chunks like this:
char const* tail;
std::vector<ReversedCStr<11>> vec;
for(vec.emplace_back(str,&tail); *tail != '\0';)
vec.emplace_back(tail,&tail);
I'd like to know the fastest way for reading the last line in a std::string object.
Technically, the string after the last occurrence of \n in the fastest possible way?
This can be done using just string::find_last_of and string::substr like so
std::string get_last_line(const std::string &str)
{
auto position = str.find_last_of('\n');
if (position == std::string::npos)
return str;
else
return str.substr(position + 1);
}
see: example
I would probably use std::string::rfind and std::string::substr combined with guaranteed std::string::npos wrap around to be succinct:
inline std::string last_line_of(std::string const& s)
{
return s.substr(s.rfind('\n') + 1);
}
If s.rfind('\n') doesn't find anything it returns std::string::npos. The C++ standard says std::string::npos + 1 == 0. And returning s.substr(0) is always safe.
If s.rfind('\n') does find something then you want the substring starting from the next character. Again returning s.substr(s.size()) is safe according to the standard.
NOTE: In C++17 this method will benefit from guaranteed return value optimization so it should be super efficient.
I thought of a way that reads the string inversely (backwards) while storing what it reads
std::string get_last_line(const std::string &str)
{
size_t l = str.length();
std::string last_line_reversed, last_line;
for (--l; l > 0; --l)
{
char c = str.at(l);
if (c == '\n')
break;
last_line_reversed += c;
}
l = last_line_reversed.length();
size_t i = 0, y = l;
for (; i < l; ++i)
last_line += last_line_reversed[--y];
return last_line;
}
until it counters a '\n' character then reverse the stored string back and return it. If the target string is big and has a lot of new lines, this function would be very efficient.
I am creating a program that scans user input for words that are listed in an array. The find() function seems like it'll work, but I can't find anything showing how to implement it for what I want to do. I'm pretty new to programming (obviously).
#include <iostream>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
string subj [5]={"I","He","She","We","They"};
string verb [5]={" like"," hate"," sacrifice"," smell"," eat"};
string obj [5]={" bacon","s cats","s bagels","s children","s cops"};
string greeting [5]={"How are you","How's it going","Sup dude","Hey","Hello"};
string negvibe [4]={"bad","terrible","lousy","meh"};
string userfeeling;
int main()
{
srand(time(0));
int rando = rand() %5;//generates a random number between 0 and 4
int rando1 = rand() %5;
int rando2 = rand() %5;
cout << greeting [rando1] << "." << endl;
getline(std::cin,userfeeling);
if .... // What has to be done here?
find(negvibe, negvibe + 4, userfeeling) != negvibe + 4);
// Something like that?
// then ...
{
cout << subj[rando] << verb[rando1] << obj[rando2] <<"." <<endl;
}
return 0;
}
To make find work properly you should user iterators like so
if(find(std::begin(negvibe), std::end(negvibe), userfeeling) != std::end(negvibe)){
//code you want to happen if your word is found
}
Also in your current code, the if statement doesnt actually do anything since you end it with a semicolon and not {} or leave it blank if its one line. You can see an example of the if statement as well
Below is a link to find and iterators
http://www.cplusplus.com/reference/algorithm/find/
That find function will find some element of the negvibe array that is equal to userfeeling. If you are checking whether any element of negvibe is a substring of userfeeling, you should loop through negvibe and use the std::string::find method.
bool found_negvibe = false;
for (int i = 0; i < sizeof(negvibe) / sizeof(*negvibe); i++) {
found_negvibe = found_negvibe || userfeeling.find(negvibe[i]) != string::npos;
}
Also, you don't need to specify the size of the negvibe array, you can write this:
string negvibe[] = {"bad","terrible","lousy","meh"};
One more thing, you might prefer to use a std::vector over an array, if only because c++'s faculties for getting the size of a vector are slightly more succinct than those for getting the size of an array.
vector negvibe = {"bad","terrible","lousy","meh"};
bool found_negvibe = false;
for (int i = 0; i < negvibe.size(); i++) {
found_negvibe = found_negvibe || userfeeling.find(negvibe[i]) != string::npos;
}
As part of a homework assignment I need to be able to take an input string and manipulate it several ways using a list of string functions. The first function takes a string and reverses it using a for loop. This is what I have:
#include <iostream>
#include <string>
namespace hw06
{
typedef std::string::size_type size_type;
//reverse function
std::string reverse( const std::string str );
}
// Program execution begins here.
int main()
{
std::string inputStr;
std::cout << "Enter a string: ";
std::getline( std::cin, inputStr );
std::cout << "Reversed: " << hw06::reverse( inputStr )
<< std::endl;
return 0;
}
//reverse function definition
std::string hw06::reverse( const std::string str )
{
std::string reverseStr = "";
//i starts as the last digit in the input. It outputs its current
//character to the return value "tempStr", then goes down the line
//adding whatever character it finds until it reaches position 0
for( size_type i = (str.size() - 1); (i >= 0); --i ){
reverseStr += str.at( i );
}
return reverseStr;
}
The program asks for input, then returns this error:
terminate called after throwing an instance of 'std::out_of_range'
what(): basic_string::tat
I'm really at a loss as to what I'm doing wrong here. The loop seems correct to me, so am I misunderstanding how to reference the function?
Unless you really want to write a loop, it's probably easier to just do something like:
std::string reverse(std::string const &input) {
return std::string(input.rbegin(), input.rend());
}
The problem is that your loop never terminates. You have as your condition i >= 0, but size_type is unsigned, so 0 - 1 == 2^(sizeof(size_t) * 8) - 1, which is certainly out of the range of your string. Therefore, you need to pick something else as your termination condition. One option is you can use i != std::string::npos, but that feels wrong. You're probably better off with something like:
for (size_type i = str.size(); i != 0; ) {
reverseStr += str.at(--i);
}
EDIT: I did some checking on i != std::string::npos. It should be well-defined and OK. However, it still seems like the Wrong Way To Do It.
As Andreas Grapentin said, the problem is that std::string::size() returns a size_t which is required by the standard to be an unsigned type. So it will always be >= 0 and when you hit 0 and decrement it, you will go to some really large, positive number.
Consider something like this:
std::string hw06::reverse(const std::string &str)
{
std::string reverseStr;
for(size_t i = str.size(); i != 0; i--)
reverseStr += str.at(i - 1);
return reverseStr;
}
I'm not keen on answering homework questions, but seeing some of the answers, I couldn't resist this:
std::string hw06::reverse(const std::string &str)
{ return std::string(str.rbegin(), str.rend()); }
Simple, clean and least wasteful if you can't do it in-place.
As other answers say, the problem is in the loop. I'll suggest using the following "goes to" operator :)
for(size_t i = str.size(); i --> 0;)
{
}
use i-- and not --i. Or u will decrease i value before getting the char and get loop problems.
What is the effective way to replace all occurrences of a character with another character in std::string?
std::string doesn't contain such function but you could use stand-alone replace function from algorithm header.
#include <algorithm>
#include <string>
void some_func() {
std::string s = "example string";
std::replace( s.begin(), s.end(), 'x', 'y'); // replace all 'x' to 'y'
}
The question is centered on character replacement, but, as I found this page very useful (especially Konrad's remark), I'd like to share this more generalized implementation, which allows to deal with substrings as well:
std::string ReplaceAll(std::string str, const std::string& from, const std::string& to) {
size_t start_pos = 0;
while((start_pos = str.find(from, start_pos)) != std::string::npos) {
str.replace(start_pos, from.length(), to);
start_pos += to.length(); // Handles case where 'to' is a substring of 'from'
}
return str;
}
Usage:
std::cout << ReplaceAll(string("Number Of Beans"), std::string(" "), std::string("_")) << std::endl;
std::cout << ReplaceAll(string("ghghjghugtghty"), std::string("gh"), std::string("X")) << std::endl;
std::cout << ReplaceAll(string("ghghjghugtghty"), std::string("gh"), std::string("h")) << std::endl;
Outputs:
Number_Of_Beans
XXjXugtXty
hhjhugthty
EDIT:
The above can be implemented in a more suitable way, in case performance is of your concern, by returning nothing (void) and performing the changes "in-place"; that is, by directly modifying the string argument str, passed by reference instead of by value. This would avoid an extra costly copy of the original string by overwriting it.
Code :
static inline void ReplaceAll2(std::string &str, const std::string& from, const std::string& to)
{
// Same inner code...
// No return statement
}
Hope this will be helpful for some others...
I thought I'd toss in the boost solution as well:
#include <boost/algorithm/string/replace.hpp>
// in place
std::string in_place = "blah#blah";
boost::replace_all(in_place, "#", "#");
// copy
const std::string input = "blah#blah";
std::string output = boost::replace_all_copy(input, "#", "#");
Imagine a large binary blob where all 0x00 bytes shall be replaced by "\1\x30" and all 0x01 bytes by "\1\x31" because the transport protocol allows no \0-bytes.
In cases where:
the replacing and the to-replaced string have different lengths,
there are many occurences of the to-replaced string within the source string and
the source string is large,
the provided solutions cannot be applied (because they replace only single characters) or have a performance problem, because they would call string::replace several times which generates copies of the size of the blob over and over.
(I do not know the boost solution, maybe it is OK from that perspective)
This one walks along all occurrences in the source string and builds the new string piece by piece once:
void replaceAll(std::string& source, const std::string& from, const std::string& to)
{
std::string newString;
newString.reserve(source.length()); // avoids a few memory allocations
std::string::size_type lastPos = 0;
std::string::size_type findPos;
while(std::string::npos != (findPos = source.find(from, lastPos)))
{
newString.append(source, lastPos, findPos - lastPos);
newString += to;
lastPos = findPos + from.length();
}
// Care for the rest after last occurrence
newString += source.substr(lastPos);
source.swap(newString);
}
A simple find and replace for a single character would go something like:
s.replace(s.find("x"), 1, "y")
To do this for the whole string, the easy thing to do would be to loop until your s.find starts returning npos. I suppose you could also catch range_error to exit the loop, but that's kinda ugly.
For completeness, here's how to do it with std::regex.
#include <regex>
#include <string>
int main()
{
const std::string s = "example string";
const std::string r = std::regex_replace(s, std::regex("x"), "y");
}
If you're looking to replace more than a single character, and are dealing only with std::string, then this snippet would work, replacing sNeedle in sHaystack with sReplace, and sNeedle and sReplace do not need to be the same size. This routine uses the while loop to replace all occurrences, rather than just the first one found from left to right.
while(sHaystack.find(sNeedle) != std::string::npos) {
sHaystack.replace(sHaystack.find(sNeedle),sNeedle.size(),sReplace);
}
As Kirill suggested, either use the replace method or iterate along the string replacing each char independently.
Alternatively you can use the find method or find_first_of depending on what you need to do. None of these solutions will do the job in one go, but with a few extra lines of code you ought to make them work for you. :-)
What about Abseil StrReplaceAll? From the header file:
// This file defines `absl::StrReplaceAll()`, a general-purpose string
// replacement function designed for large, arbitrary text substitutions,
// especially on strings which you are receiving from some other system for
// further processing (e.g. processing regular expressions, escaping HTML
// entities, etc.). `StrReplaceAll` is designed to be efficient even when only
// one substitution is being performed, or when substitution is rare.
//
// If the string being modified is known at compile-time, and the substitutions
// vary, `absl::Substitute()` may be a better choice.
//
// Example:
//
// std::string html_escaped = absl::StrReplaceAll(user_input, {
// {"&", "&"},
// {"<", "<"},
// {">", ">"},
// {"\"", """},
// {"'", "'"}});
#include <iostream>
#include <string>
using namespace std;
// Replace function..
string replace(string word, string target, string replacement){
int len, loop=0;
string nword="", let;
len=word.length();
len--;
while(loop<=len){
let=word.substr(loop, 1);
if(let==target){
nword=nword+replacement;
}else{
nword=nword+let;
}
loop++;
}
return nword;
}
//Main..
int main() {
string word;
cout<<"Enter Word: ";
cin>>word;
cout<<replace(word, "x", "y")<<endl;
return 0;
}
Old School :-)
std::string str = "H:/recursos/audio/youtube/libre/falta/";
for (int i = 0; i < str.size(); i++) {
if (str[i] == '/') {
str[i] = '\\';
}
}
std::cout << str;
Result:
H:\recursos\audio\youtube\libre\falta\
For simple situations this works pretty well without using any other library then std::string (which is already in use).
Replace all occurences of character a with character b in some_string:
for (size_t i = 0; i < some_string.size(); ++i) {
if (some_string[i] == 'a') {
some_string.replace(i, 1, "b");
}
}
If the string is large or multiple calls to replace is an issue, you can apply the technique mentioned in this answer: https://stackoverflow.com/a/29752943/3622300
here's a solution i rolled, in a maximal DRI spirit.
it will search sNeedle in sHaystack and replace it by sReplace,
nTimes if non 0, else all the sNeedle occurences.
it will not search again in the replaced text.
std::string str_replace(
std::string sHaystack, std::string sNeedle, std::string sReplace,
size_t nTimes=0)
{
size_t found = 0, pos = 0, c = 0;
size_t len = sNeedle.size();
size_t replen = sReplace.size();
std::string input(sHaystack);
do {
found = input.find(sNeedle, pos);
if (found == std::string::npos) {
break;
}
input.replace(found, len, sReplace);
pos = found + replen;
++c;
} while(!nTimes || c < nTimes);
return input;
}
I think I'd use std::replace_if()
A simple character-replacer (requested by OP) can be written by using standard library functions.
For an in-place version:
#include <string>
#include <algorithm>
void replace_char(std::string& in,
std::string::value_type srch,
std::string::value_type repl)
{
std::replace_if(std::begin(in), std::end(in),
[&srch](std::string::value_type v) { return v==srch; },
repl);
return;
}
and an overload that returns a copy if the input is a const string:
std::string replace_char(std::string const& in,
std::string::value_type srch,
std::string::value_type repl)
{
std::string result{ in };
replace_char(result, srch, repl);
return result;
}
This works! I used something similar to this for a bookstore app, where the inventory was stored in a CSV (like a .dat file). But in the case of a single char, meaning the replacer is only a single char, e.g.'|', it must be in double quotes "|" in order not to throw an invalid conversion const char.
#include <iostream>
#include <string>
using namespace std;
int main()
{
int count = 0; // for the number of occurences.
// final hold variable of corrected word up to the npos=j
string holdWord = "";
// a temp var in order to replace 0 to new npos
string holdTemp = "";
// a csv for a an entry in a book store
string holdLetter = "Big Java 7th Ed,Horstman,978-1118431115,99.85";
// j = npos
for (int j = 0; j < holdLetter.length(); j++) {
if (holdLetter[j] == ',') {
if ( count == 0 )
{
holdWord = holdLetter.replace(j, 1, " | ");
}
else {
string holdTemp1 = holdLetter.replace(j, 1, " | ");
// since replacement is three positions in length,
// must replace new replacement's 0 to npos-3, with
// the 0 to npos - 3 of the old replacement
holdTemp = holdTemp1.replace(0, j-3, holdWord, 0, j-3);
holdWord = "";
holdWord = holdTemp;
}
holdTemp = "";
count++;
}
}
cout << holdWord << endl;
return 0;
}
// result:
Big Java 7th Ed | Horstman | 978-1118431115 | 99.85
Uncustomarily I am using CentOS currently, so my compiler version is below . The C++ version (g++), C++98 default:
g++ (GCC) 4.8.5 20150623 (Red Hat 4.8.5-4)
Copyright (C) 2015 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
This is not the only method missing from the standard library, it was intended be low level.
This use case and many other are covered by general libraries such as:
POCO
Abseil
Boost
QtCore
QtCore & QString has my preference: it supports UTF8 and uses less templates, which means understandable errors and faster compilation. It uses the "q" prefix which makes namespaces unnecessary and simplifies headers.
Boost often generates hideous error messages and slow compile time.
POCO seems to be a reasonable compromise.
How about replace any character string with any character string using only good-old C string functions?
char original[256]="First Line\nNext Line\n", dest[256]="";
char* replace_this = "\n"; // this is now a single character but could be any string
char* with_this = "\r\n"; // this is 2 characters but could be of any length
/* get the first token */
char* token = strtok(original, replace_this);
/* walk through other tokens */
while (token != NULL) {
strcat(dest, token);
strcat(dest, with_this);
token = strtok(NULL, replace_this);
}
dest should now have what we are looking for.