#include <vector>
#include <iostream>
#include <cstring>
#include <string.h>
using namespace std;
int main()
{
vector<string> test;
test.push_back("yasir");
test.push_back("javed");
for(int i=0; i!=test.end();i++)
{
cout << test[i];
}
}
Why is this code giving up an error? I am unable to identify the cause of the error.
Error: No Match for operator !=....
First of all, you are trying to compare int with the iterator of vector.
for(int i=0; i!=test.end();i++)
{
cout << test[i];
}
Here, the test.end() returns the iterator. There is no overloaded operator!= which can compare integer (int i = 0) with that iterator (test.end()).
So your loop should look more like:
for (std::vector<string>::iterator i = test.begin(); i != test.end(); i++)
{
cout << *i;
}
You can replace std::vector<string>::iterator with auto, if using C++11 or newer.
The next thing, you included <string.h> which contains old functions such as: strlen, strcpy. Similarly, <cstring> contains C-style strings.
If you want to you use operator<<, so if you want to write:cout << then you have to do: #include <string>.
As already mentioned, the problem is, that you try to compare an integer with an iterator in the "middle" of your for statement. Try this instead, it's more intuitive from my point of view
#include <vector>
#include <iostream>
#include <cstring>
#include <string.h>
using namespace std;
int main()
{
vector<string> test;
test.push_back("yasir");
test.push_back("javed");
for(int i=0; i<test.size();++i)
{
cout << test[i];
}
}
Related
I would like to print the strings at the top of columns with a 1 x 3 array.
I have edited this simple function several times, and this produces the least errors. New to C++, reading Deital Chap 6 Recursive.
What am I missing? I started with half brackes around strings, and brackets seemed to produce less errors.
Here is the code:
#include <iostream>
#include <string>
#include <array>
using namespace std;
int main() {
array a[1][3] = ["Car" "Hours" "Charge"]
cout<< a << endl;
}
Terminal produces errors as such:
parking_charges_6_12.cpp: In function ‘int main()’:
parking_charges_6_12.cpp:8:7: error: missing template arguments before ‘a’
8 | array a[1][3] = ["Car" "Hours" "Charge"]
^
This should work:
#include <array>
#include <iostream>
#include <string>
int main(){
std::array<std::string, 3> headlines = {"Car", "Hours", "Charge"};
for( auto const& elem : headlines ){
std::cout << elem << "\t";
}
}
It should be curly braces {} in the initializer, not []. And you need a comma between each element.
On the other hand, in later C++ revisions array can detect the type and number of elements, so you don't have to give that.
#include <iostream>
#include <array>
using namespace std;
int main() {
array a = {"Car", "Hours", "Charge"};
for (auto& item : a)
cout<< item << endl;
}
How about something like this:
#include<iostream>
using namespace std;
int main() {
string data[3] = {"Car", "Hours", "Charge"};
for (int i = 0; i < 3; i++)
cout << data[i] << " ";
}
Obviously it is not using the array header, but it's a working example. If you do need to use the array header, you can try something like :
#include <array>
#include <iostream>
#include <iterator>
#include <string>
using namespace std;
int main() {
array<string, 3> ar3 = {"Car", "Hours", "Charge"};
cout << ar3.size() << endl;
for (auto i : ar3)
cout << i << ' ';
return 0;
}
You can see it working online here
this is my code
the error is Segmentation fault,and i can't understand why
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
using namespace std;
int main(int argc, char* argv[])
{
char* szword[100];
int i = 0;
do
{
cin >> szword[i];
cout << szword[i];
i++;
}while(strcmp(szword[i - 1], "done"));
cout << i + 1;
return 0;
}
For starters neither declaration from headers <cstdio> and <string> is used in your program. So you should remove these directives
#include <cstdio>
#include <string>
You declared an initialized array with the element type char *. Thus this statement
cin >> szword[i];
invokes undefined behavior because the pointer szword[i] has indeterminate value.
Moreover this call even if the argument of the operator will be correct
cin >> szword[i];
can fail. You should check whether it was successful. And I think there is no great sense to output the string "done".
Also in this statement
cout << i + 1;
you are outputting a value that is greater than the number of inputted strings.
If to use character arrays then your program could look the following way
#include <iostream>
#include <cstring>
int main()
{
const size_t N = 100;
char szword[N][N];
size_t i = 0;
while ( std::cin.getline( szword[i], sizeof( szword[i] ) ) &&
std::strcmp( szword[i], "done" ) != 0 )
{
std::cout << szword[i++] << '\n';
}
std::cout << i << '\n';
return 0;
}
The program output might look like
Hello
World
2
This below code works fine, if you want to use char *, for C++ string you can use the C++ version
C Version:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
using namespace std;
int main(int argc, char* argv[])
{
char *tmp;
int i = 0;
do
{
cin >> tmp;
cout << tmp;
i++;
}while(strcmp(tmp, "done"));
cout << i + 1;
return 0;
}
C++ Version:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
using namespace std;
int main(int argc, char* argv[])
{
string tmp;
int i = 0;
do
{
cin >> tmp;
cout << tmp;
i++;
}while(tmp != "done"));
cout << i + 1;
return 0;
}
Please explain why is this giving an error but the other on is running fine
The following code gives the error:
#include <iostream>
#include <string>
#include <conio.h>
#include <math.h>
#include <iomanip>
#include <string.h>
using namespace std;
int main()
{
string s1,s2;
int i;
cout << "Enter the string to copy into another string : ";
getline(cin,s1);
for(i=0; s1[i]!='\0'; ++i)
{
s2[i]=s1[i];
}
s2[i]='\0';
cout<<"\n\nCopied String S2 is : "<<s2;
return 0;
}
Error looks like this
But this works perfectly fine
#include <iostream>
#include <string>
#include <conio.h>
#include <math.h>
#include <iomanip>
#include <string.h>
using namespace std;
int main()
{
char s1[100], s2[100], i;
cout << "Enter the string to copy into another string : ";
cin>>s1;
for(i=0; s1[i]!='\0'; ++i)
{
s2[i]=s1[i];
}
s2[i]='\0';
cout<<"\n\nCopied String S2 is : "<<s2;
return 0;
}
In your case, s2 is initialized to an empty string with a length of 0, so you can't write past the bounds. If you want to, you must first resize it:
s2.resize(s1.length());
for(i=0; s1[i]!='\0'; ++i)
{
s2[i]=s1[i];
}
Also, c++ std::string does not need a terminating nullbyte, unlike C strings.
This question already has answers here:
STL vector reserve() and copy()
(5 answers)
Closed 4 years ago.
I made a function for printing a vector using interators. The function is part of program meant to copy a vector of strings to another vector of strings. The original function was a simple for loop using the .at() member function and that worked. So, I am not sure what's wrong with this one.
The code:
#include <string>
#include <string>
#include <fstream>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
void print(vector<string> &v);
int main() {
ifstream inFS;
inFS.open("sentence.txt");
vector<string> wordList; vector<string> newVect;
string s;
while (inFS >> s) {
wordList.push_back(s);
}
copy(wordList.begin(), wordList.end(), newVect.begin());
print(wordList);
print(newVect);
}
void print(vector<string> &v) {
for (vector<string>::iterator i = v.begin(); i != v.end(); i++) {
cout << *i << " ";
}
cout << endl;
}
The output:
C:\Users\westt\Documents\PROJECT>a.exe
C:\Users\westt\Documents\PROJECT>
Try this simple change to initialize the newVect. As the comment noted, you could also resize() newVect
#include <string>
#include <string>
#include <fstream>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
void print(vector<string> &v);
int main() {
ifstream inFS;
inFS.open("sentence.txt");
vector<string> wordList;
string s;
while (inFS >> s) {
wordList.push_back(s);
}
vector<string> newVect(wordList);
print(wordList);
print(newVect);
}
void print(vector<string> &v) {
for (vector<string>::iterator i = v.begin(); i != v.end(); i++) {
cout << *i << " ";
}
cout << endl;
}
EDIT: Sorry, i didn't put any result. I expected to remove the (10th, 11th), then the (20th, 21st), then the (30th, 31st) and so forth but instead I get this http://imgur.com/TbVmLo4
I have a vector composed by several vectors of the same size and i want to get rid of part of their end using remove if and a lambda function but there seems to be a displacement in the removal. Here is the code that I am running:
#include <vector>
#include <numeric>
#include <iostream>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <typeinfo>
int main()
{
int sizeOfSpectrum=10;
int desiredSizeOfSpectrum=8;
int count=-1;
std::vector<int> v(100);
std::iota (std::begin(v), std::end(v), 0);
auto newend=std::remove_if(v.begin(),v.end(),[&](int i)->bool{
count++;
int temp=count%sizeOfSpectrum;
bool test=(temp) > desiredSizeOfSpectrum;
return ( test);
});
v.erase(newend, v.end());
std::cout << "\n";
std::cout << "v[i]: \n";
for(int i : v)std::cout <<v[i]<<": ";
std::cout << "\n";
return 0;
}
The problem is in this line:
for(int i : v)std::cout <<v[i]<<": ";
When this line runs, i is given the value of the contents of each element of the vector. You are treating this as an index into the vector, rather than the value.
Instead use:
for(int i : v)
std::cout <<i<<": ";
Note: you still won't get quite the output you are after, but this should give you enough to work that out for yourself.