I have following code that gets and prints a string.
#include<iostream>
#include<conio.h>
#include<string>
using namespace std;
int main()
{
string str;
cout << "Enter a string: ";
getline(cin, str);
cout << str;
getch();
return 0;
}
But how to count the number of characters in this string using strlen() function?
For C++ strings, there's no reason to use strlen. Just use string::length:
std::cout << str.length() << std::endl;
You should strongly prefer this to strlen(str.c_str()) for the following reasons:
Clarity: The length() (or size()) member functions unambiguously give back the length of the string. While it's possible to figure out what strlen(str.c_str()) does, it forces the reader to pause for a bit.
Efficiency: length() and size() run in time O(1), while strlen(str.c_str()) will take Θ(n) time to find the end of the string.
Style: It's good to prefer the C++ versions of functions to the C versions unless there's a specific reason to do so otherwise. This is why, for example, it's usually considered better to use std::sort over qsort or std::lower_bound over bsearch, unless some other factors come into play that would affect performance.
The only reason I could think of where strlen would be useful is if you had a C++-style string that had embedded null characters and you wanted to determine how many characters appeared before the first of them. (That's one way in which strlen differs from string::length; the former stops at a null terminator, and the latter counts all the characters in the string). But if that's the case, just use string::find:
size_t index = str.find(0);
if (index == str::npos) index = str.length();
std::cout << index << std::endl;
Function strlen shows the number of character before \0 and using it for std::string may report wrong length.
strlen(str.c_str()); // It may return wrong length.
In C++, a string can contain \0 within the characters but C-style-zero-terminated strings can not but at the end. If the std::string has a \0 before the last character then strlen reports a length less than the actual length.
Try to use .length() or .size(), I prefer second one since another standard containers have it.
str.size()
Use std::string::size or std::string::length (both are the same).
As you insist to use strlen, you can:
int size = strlen( str.c_str() );
note the usage of std::string::c_str, which returns const char*.
BUT strlen counts untill it hit \0 char and std::string can store such chars. In other words, strlen could sometimes lie for the size.
If you really, really want to use strlen(), then
cout << strlen(str.c_str()) << endl;
else the use of .length() is more in keeping with C++.
Manually:
int strlen(string s)
{
int len = 0;
while (s[len])
len++;
return len;
}
#include<iostream>
#include<conio.h>
#include<string.h>
using namespace std;
int main()
{
char str[80];
int i;
cout<<"\n enter string:";
cin.getline(str,80);
int n=strlen(str);
cout<<"\n lenght is:"<<n;
getch();
return 0;
}
This is the program if you want to use strlen .
Hope this helps!
Simply use
int len=str.length();
Related
I am a complete beginner so the code may seem to be easy, but I cannot find a solution why it returns such values as:
input: kkkk
output:
14
kkkkřřřř╩┬ëŢ
Suprisingly the code works fine with online compilators, but not with the Visual Studio.
#include<iostream>
#include<string.h>
int main()
{
char word[20];
std::cin >> word;
int length = strlen(word);
int p = length - 1, i = 0;
char *var=new char [length];
while (i < length&&p>=0)
{
var[i]= word[p];
p--;
i++;
}
std::cout <<strlen(var)<<endl<< var;
if (!strcmp(var, word)) std::cout << "\nThe word is a palindrome";
return 0;
}
I can not use strings because my University doesn't allow to do so. I also know there are many different ways to attend to this problem but I just really want know what I have done wrong in this one :/
Your "copy routine" copies each character, but it does not copy the string termination character. Note that C-style strings as used in functions like strlen or strcmp need to be 0-terminated, and even cout <<, when getting a parameter of type char*, treats this as a C-style string: It will read until finding the terminating '\0', and if you do not write one, it will read beyond the boundaries you think it should do.
If your write
...
}
var[length] = '\0';
std::cout <<strlen(var)<<endl;
...
it should work.
The question requires combining two strings(the longer string in the front and the shorter one after the longer one) without using <string> header file.Each string inputted can't exceed 20 characters.
My logic behind this is:
first use strlen to get the length of the str1 and str2,
use str3 to store the longer string, and str4 to store the shorter.
add str3 and str4 to str5
Here is my code:
#include<iostream>
using namespace std;
int main()
{
// combine two strings , longer in the front, and shorter
// after it. do not use strcat
char str1[20],str2[20],str3[20],str4[20],str5[40];
// str1 and str2 stores original data, str3 stores longer
// and str4 stores shorter, str5 stores total
int j=0;
cin.getline(str1,20);
cin.getline(str2,20);
if(strlen(str1)<=strlen(str2))
// give longer string value to str3,shorter to str2
{
for (int i=0;i<20;i++)
{
str3[i]=str2[i];
str4[i]=str1[i];
}
}
else
{
for (int i=0;i<20;i++)
{
str3[i]=str1[i];
str4[i]=str2[i];
}
}
for(j=0;str3[j]!='\0';j++)
{
str5[j]=str3[j];
}
for(int i=j;i<40;i++)
for(int m=0;m<20;m++)
{
str5[i]=str4[m];
}
cout<<str5<<endl;
return 0;
}
Here is the ouput:
What's my problem here? What are those characters in between the two strings? Thank you!!
Especially since you explicitly mentioned being a beginner, the solution is to use std::string:
#include <iostream>
#include <string>
int main() {
std::string a;
getline(std::cin, a);
std::string b;
getline(std::cin, b);
// Ensure that the longer string goes to the front.
if (a.size() < b.size());
swap(a, b);
std::string result = a + b;
std::cout << result << '\n';
// Or, simply:
std::cout << a << b << '\n';
}
The message here is that C++, despite its quirks, is a very high level language if you rely on its library instead of implementing every low level operation from scratch.
Everything is fine (!) up to this point
for(int i=j;i<40;i++)
for(int m=0;m<20;m++) // This loop runs m=0 to 20 for each position of i
{
str5[i]=str4[m];
}
For each index i you are copying in all 20 elements from str4, leaving just the value at str4[19] which could be anything
Just increment i and m by one together
int m = 0;
for(int i=j;i<40;i++)
{
str5[i]=str4[m++];
}
You are copying the entire 20 characters, 40 characters in the loop into the variables. stop copying when you find a '\0' character.
But using the std::string will make life simpler :)
Using std::string is nice and all but here's a few tips for working with char*:
1) You shouldn't copy strings to separate shorter and longer string, just use pointers and then work with these pointers, something along these lines:
const char *longer_string = 0, *shorter_string = 0;
if(strlen(str1)<=strlen(str2))
{
shorter_string = str1;
longer_string = str2;
}
else
{
shorter_string = str2;
storter_string = str1;
}
2) Using strcpy and strcat to combine strings could make life a lot easier:
char *combined_string = new char [strlen (shorter_string) + strlen (longer_string) + 1];
strcpy (combined_string, longer_string);
strcat (combined_string, shorter_string);
Some compilers would say that these functions aren't safe and you have to stick to _s versions, but I guess it's entirely up to you.
Since this is obviously homework: I'll just point out the existence of the function strcat, and the fact that you can use char* to the arrays, and just swap them, without having to recopy anything between the initial read and the concatenation (which means that you only need two arrays: one for each of the inputs, and one for the final value).
And also, when calculating sizes, etc. do not forget that C style strings have an extra '\0' at the end, and make allowances for it.
As #David Sykes has pointed out, the problem is with your for loop. So when you read input from cin ,it is not necessary that your input string contains 20 character. But in you form loop you are looping through those string beyond their length which may contains garbage characters. Example
char str1[20]
cin.getline(str1,20);
cout << str1[19] << endl;
Suppose your input for above code is "ABCD" which contains only 4 characters but your array has capacity of 20. So the remaining space has junk characters and when you will try to print any thing beyond actual length you will get wild character as you are getting in your code.
In C++, using printf-s I want to write a char array (etc char asd[50]) to console with a specified 50 space
(similar like "%.2d" method at decimals, if the string shorter fill it with spaces....)
Tried %50s and %.50s methods, both of them wronged my charachters....
I can (hardly) accept answers, but then calculat with the fact, i use a charachter array, so its not wrok to cout<
Try this if you find it useful,
#include <iomanip>
void prints(const char * s)
{
cout << std::setfill('0') << std::setw(50) << s;
}
A used a for loop with printfs (using %c) to write it, and after the \0 charachter it write spaces , so problem solved
for(int j=0;j<50;j++)
{
printf("%c",asd[j]);
}
#include<iostream>
#include<cmath>
#include<iomanip>
#include<string>
using namespace std;
int main()
{
string word;
int j = 0;
cin >> word;
while(word[j]){
cout << "idk";
j++;
}
cout << "nope";
system("pause");
return 0;
}
This is just a little trial program to test this loop out. The program I am working on is about vowels and printing vowels out from a sequence determined by the user. The string isn't defined until the user types in. Thank you for your guys help in advance.
Try this for your loop:
while(j < word.size()){
cout << "idk";
j++;
}
The size of an std::string is not unknown - you can get it using the std::string::size() member function. Also note that unlike C-strings, the std::string class does not have to be null-terminated, so you can't rely on a null-character to terminate a loop.
In fact, it's much nicer to work with std::string because you always know the size. Like all C++ containers, std::string also comes with built-in iterators, which allow you to safely loop over each character in the string. The std::string::begin() member function gives you an iterator pointing to the beginning of the string, and the std::string::end() function gives you an iterator pointing to one past the last character.
I'd recommend becoming comfortable with C++ iterators. A typical loop using iterators to process the string might look like:
for (std::string::iterator it = word.begin(); it != word.end(); ++it)
{
// Do something with the current character by dereferencing the iterator
//
*it = std::toupper(*it); // change each character to uppercase, for example
}
#include<iostream>
using namespace std;
int main()
{
char arr[200];
while(1) {
cin >> arr;
int i = sizeof(arr);
cout << "The arr input is "<< arr
<< " and the size of the array is "<< i << endl;
}
return 0;
}
For the input of 34,
This code outputs :The arr input is 34 and the size of the array is 200
while I want it to get the size of the used space of the array . So for The last input i want it to output :The arr input is 34 and the size of the array is 2
Can someone tell me how?
Maybe you want strlen(arr) here. It must be null terminated, otherwise the cout << arr would not have worked.
You would need to #include <cstring>
There's no automatic way to do what you want in the general case - you'll need to keep track somehow, either with your own counter, or by seeding the array with an 'invalid' value (that you define) and search for to find the end of the used elements (that's what the '\0' terminator character in a C-style string is).
In the example code you posted, the array should receive a null terminated C-style string, you can use that knowledge to count the number of valid elements.
If you're using C++ or some other library that has some more advanced data structures, you may be able to use one that keeps track of this kind of thing for you (like std::vector<>).
the size of the used space of the array
There is no such thing. If you have an array of 200 chars, then you have 200 chars. Arrays have no concept of "used" and "unused" space. It only works with C-strings because of the convention that those are terminated by a 0 character. But then again, the array itself cannot know if it is holding a C-string.
in a less involved manner, you can just count through each character till you hit a null with just a while loop. It will do the exact same thing strlen() does. Also, in practice, you should do type checking with cin, but i'll assume this was just a test.
#include <iostream>
using namespace std;
int main()
{
char arr[200];
int i;
while(1) {
cin >> arr;
i=0;
while (arr[i] != '\0' && i<sizeof(arr))
i++;
cout << "The arr input is "<< arr
<< " and the size of the array is "<< i << endl;
}
return 0;
}
Just for completeness, here is a much more C++ like solution that is using std::string instead of a raw char array.
#include <iostream>
#include <string>
int
main()
{
while (std::cin.good()) {
std::string s;
if (std::cin >> s) {
std::cout
<< "The input is " << s
<< " and the size is " << s.length()
<< std::endl;
}
}
return 0;
}
It doesn't use an array, but it is the preferable solution for this kind of problem. In general, you should try to replace raw arrays with std::string and std::vector as appropriate, raw pointers with shared_ptr (scoped_ptr, or shared_array, whatever is most appropriate), and snprintf with std::stringstream. This is the first step to simply writing better C++. You will thank yourself in the future. I wish that I had followed this advice a few years ago.
Try it
template < typename T, unsigned N >
unsigned sizeOfArray( T const (&array)[ N ] )
{
return N;
}