My name is Matt. I'm new to StackOverflow and am fairly new to C++. Currently working my way through C++ Primer by Lippman.
I'm doing an exercise in the book, and task is to read integers in to the vector, and then multiply those integers by doing first and last, second and second to last, third and third last etc.
I did it myself without looking anything up, or else I'd barely learn if I just copied... my program compiles and acts as expected. My question is: did I do this correctly? is there a more efficient way of doing it?
I not only want to learn how to make working code, but I want to do it correctly. Thank you in advance!
#include <iostream>
#include <string>
#include <vector>
#include <cctype>
using std::cout; using std::cin; using std::vector; using std::endl;
using std::string;
int main()
{
vector<int> numbers;
int usernum = 0;
cout << "Enter some numbers: ";
while (cin >> usernum)
{
numbers.push_back(usernum);
}
unsigned maxElement = numbers.size() - 1;
unsigned minElement = 0;
for (auto i : numbers)
{
cout << numbers[minElement] << " * " << numbers[maxElement] << " = " << numbers[minElement] * numbers[maxElement] << "\n";
++minElement;
--maxElement;
}
return 0;
}
In a comment, you said:
I've noticed it doesn't actually work as fully expected. It goes through vector and multiplies everything twice as maxElement goes right through to beginning, and minElement goes through to end. Not sure how to stop it once it's only done the operation on each one time.
If you don't want to repeat the multiplications, you need to change the for loop a bit.
for ( ; minElement <= maxElement; ++minElement, --maxElement)
{
cout << numbers[minElement] << " * " << numbers[maxElement] << " = " << numbers[minElement] * numbers[maxElement] << "\n";
}
PS
When you use this logic, you'll need to make sure that minElement and maxElement are of a signed type. Otherwise, you will run into problems if numbers has only one element.
The first thing that feels strange for me is the namespaces you are using.
Instead of doing: using namespace std::vector; you can fairly just do using namespace std; because you call std::vector anyways: vector<int> numbers;***. This applies to any "used" namespace you work with. Just do using namespace std; once and for all.
***I am unsure that std::vector/std::cout/... is even a namespace. std - is a namespace. std::vector should be a class under std namespace:
namespace std
{
template<typename T> class vector<T> {...};
}
How come it "acts as expected". I do not get an idea of this loop: while(cin >> usernum). How do you know when the user input is finished from that point? For a first glimpse (did not compile/run it myself) I expect it either:
not compile
crash at runtime or having undefined behaviour
run the while-loop infinitely
Use this instead:
for (int i = 0, end_of_vector = numbers.size(); i < end_of_vector/2; i++)
{
cout << numbers[i] << " * " << numbers[end_of_vector - 1 - i] << " = " << numbers[i] * numbers[end_of_vector - 1 - i] << "\n";
}
Reasons:
In this case you do not need any special variables to store first and last vector indexes.
You iterate only through a half of an array.
Using for (auto i : numbers), is is expected to use i as an element of numbers vector. But you do not do this, instead, you use numbers as is. Thus, this for-loop is ambiguous
Related
So I got this code, I input three numbers. Now I want to display them in a for loop. From smallest to biggest, and in another loop from biggest to smallest. How can I do this?
int main()
{
int num1, num2, num3;
cout << "Enter first num" << endl;
cin >> num1;
cout << "Enter second num" << endl;
cin >> num2;
cout << "Enter third num" << endl;
cin >> num3;
}
I have done this like this but I think it's not a proper way to do it.
for(int i = 0; i <= 0; i++) {
cout << num1;
cout << num2;
cout << num3;
}
for(int i = 0; i <= 0; i++) {
cout << num3;
cout << num2;
cout << num1;
}
Edit:
is this better?
int main()
{
int numbers[3];
cout << "Enter first num" << endl;
cin >> numbers[0];
cout << "Enter second num" << endl;
cin >> numbers[1];
cout << "Enter third num" << endl;
cin >> numbers[2];
for (int i = 0; i <= 2; i++)
{
cout << numbers[i];
}
for (int i = 2; i >= 0; i--)
{
cout << numbers[i];
}
return 0;
}
What you are trying to do is very common in programming languages. I.e. enter data, process it and output it in one form or another.
A loop is what you use if you want to execute a piece of code a number of times... a number more than 1 that is. In your first example you execute the loop once, making the extra lines of code a bit redundant.
You already updated your code, showing you quickly realized how to used arrays. These types of arrays (i.e. int numbers[3];) are often referred to as C-style arrays, as they were inherited from C. In C++ we can now use std::array<int, 3> numbers;, which more adheres to the C++ style of working. (The std:: part of the type name indicates the type is define in the standard library namespace. The type is part of the C++ language standard.)
The problem with these two types is that they have a static(=fixed) size. I.e. the size has to be know at compile time. This can be quite a limitation if you don't know how many items the user wants to enter. However, the C++ standard defines other container types which can hold a variable amount of items. Of these, std::vector is the dynamic (variable) size counterpart of the array: these both store their items sequentially in memory.
So you can for instance use a loop to add (to the back = push_back()) a number of elements selected by the user to the vector.
#include <vector>
#include <iostream>
[...]
std::vector<int> numbers;
std::cout << "How many numbers do you want to enter?\n";
int N;
std::cin >> N;
numbers.reserve(N); // good practice to reserve memory if you know the number of elements
for (int i = 0; i < N; ++i) {
std::cout << "Enter a number: ";
int number;
std::cin >> number;
numbers.push_back(number);
}
[...]
Note that there is no check on the input: e.g. if the user would enter "-1" after the first question, things would break. I will not consider handling user error in my answer.
You can already see some code duplication in here: cout, type definition, cin. You can extract this in a separate function.
#include <string>
[...]
int ReadUserInput(std::string const& message) {
std::cout << message;
int value;
std::cin >> value;
return value;
}
or even better, you make a function template. I.e. a template for a function: the compiler will generate implementations of this function for you, depending on the type T inferred. I also use the std::string_view now, which can view to different types of strings (std::string, char*)
#include <string_view>
[...]
template<typename T>
T ReadUserInput(std::string_view message = "") {
if (!message.empty()) std::cout << message; //only print is there's an actual message
T value;
std::cin >> value;
return value;
}
Next, the C++ library has more to offer, including a number of algorithms that are commonly used in programming. One of these is a generator that repeatedly calls a function, of which the result is used to assign successive elements in a container. The object that points to a specific element in the container is called an iterator. The C++ standard library offers a convenient iterator type that executes a push_back: the std::back_inserter. The previous code can now be reduced to:
int const N = ReadUserInput<int>("How many numbers do you want to enter?\n");
std::vector<int> numbers;
numbers.reserve(N); // good practice to reserve memory if you know the number of elements
std::generate_n(back_inserter(numbers), N, ReadUserInputGenerator<int>("Enter a number: "));
"But wait", you might ask, "what is this ReadUserInputGenerator?". Well, to make the generate_n work, you need to pass a pointer/handle to a generator function, which is then executed for each element. If we'd just call ReadUserInput<int>("Enter a number: "), then the function would already have been evaluated. So we need to add another intermediate function object that makes this generator function. In the passed we would make a class for this
template<typename T>
class ReadUserInputGenerator {
private:
std::string_view message;
public:
ReadUserInputGenerator(std::string_view message = "") : message(message) {}
T operator()() const { return ReadUserInput(message); }
};
... but now we can do using lambda expressions
template<typename T>
auto ReadUserInputGenerator = [](std::string_view message = "") {
return [message]() { return ReadUserInput<T>(message); };
};
(note to more experienced readers: I'm not really sure about passing string_view by value. I based it on this)
SOOOOO, now we finally have our input. Next you wanted to sort it. Again, this is a common operation. There are actually many ways to sort a collection a values and it is a good excersize to implement these yourself... However, like I mentioned previously, as these kind of operations are common in programming, the C++ standard library actually offers an algorithm for sorting: std::sort.
std::sort(begin(numbers), end(numbers));
^ here begin and end refer to iterators pointing to the begin and end (or actually one past the end) of your vector. You could sort only part of your vector this way. However, the most common case is just begin to end, so in C++20 they've introduced the ranges algorithms, and the previous statement can be reduced to
std::ranges::sort(numbers);
... AAAND now its sorted... printing is next. You can print using a loop... but even there you will have a number of choices. And indexed loop:
for (int i = 0; i < numbers.size(); ++i) {
std::cout << numbers[i] << ' ';
}
An iterator based for loop:
for (auto it = cbegin(numbers); it != cend(numbers); ++it) {
std::cout << *it << ' ';
}
Note: the 'c' before begin and end denote that it is a "const" qualified iterator, i.e. it may not modify the contents of the object it points to.
Or a range based for loop:
for (int number : numbers) {
std::cout << number << ' ';
}
There is also a special convenience iterator type that can push to the cout: std::ostream_iterator. You can copy the vector to this iterator using the algorithm std::copy
std::copy(cbegin(numbers), cend(numbers), std::ostream_iterator<int>(std::cout, " "));
Note, the second argument of the ostream_iterator is the delimiter, i.e. the string appended after each element. Of course there's also a C++20 ranges version.
std::ranges::copy(numbers, std::ostream_iterator<int>(std::cout, " "));
... FINALLY reversing.
One option is just to reverse all elements in the vector and print them out again using one of the above mentioned methods. There's of course an algorithm to do so: std::reverse.
std::reverse(begin(numbers), end(numbers));
However, this operation modifies the contents of the container(vector), which might be costly. If you don't want to to this, you'll have to loop though your vector in reverse order
for (int i = numbers.size() - 1; i >= 0; --i) {
std::cout << numbers[i] << ' ';
}
This looks complex, and it's easy to make errors.
You could instead use the reverse iterators of vector, to traverse through the vector in reverse order: (you need to add an 'r' to the begin/end)
for (auto it = crbegin(numbers); it != crend(numbers); ++it) {
std::cout << *it << ' ';
}
or
std::copy(crbegin(numbers), crend(numbers), std::ostream_iterator<int>(std::cout, " "));
For C++20, there's no range operation to reverse the vector. However, they introduced "views" that are used to observe the values in the vector in a specific way. One such a way is "observe it in reverse order": std::ranges::view::reverse. So in C++20 you will be able to do:
for (int number : numbers | std::views::reverse) {
std::cout << number << ' ';
}
or
std::ranges::copy(numbers | std::views::reverse, std::ostream_iterator<int>(std::cout, " "));
which both don't modify numbers.
The end code could look a little bit something like this (pre C++20 version):
#include <vector>
#include <iostream>
#include <string_view>
#include <iterator>
#include <algorithm>
template<typename T>
T ReadUserInput(std::string_view message = "") {
if (!message.empty()) std::cout << message; //only print is there's an actual message
T value;
std::cin >> value;
return value;
}
template<typename T>
auto ReadUserInputGenerator = [](std::string_view message = "") {
return [message]() { return ReadUserInput<T>(message); };
};
int main() {
int const N = ReadUserInput<int>("How many numbers do you want to enter?\n");
std::vector<int> numbers;
numbers.reserve(N); // good practice to reserve memory if you know the number of elements
std::generate_n(back_inserter(numbers), N, ReadUserInputGenerator<int>("Enter a number: "));
std::sort(begin(numbers), end(numbers));
std::copy(cbegin(numbers), cend(numbers), std::ostream_iterator<int>(std::cout, " "));
std::cout << '\n';
std::copy(crbegin(numbers), crend(numbers), std::ostream_iterator<int>(std::cout, " "));
std::cout << '\n';
}
Given that the input doesn't seem to have any meaning other than "some numbers", you should use a container, and the obvious choice would be:
std::vector<int> nums;
From c++20, you don't needs loops at all for this problem, because you can use ranges:
#include<ranges>
namespace rs = std::ranges;
namespace rv = std::views;
and now you can read in numbers like this:
rs::copy_n(std::istream_iterator<int>(std::cin), 3,
std::back_inserter(nums));
I'm not sure if you want to use the order that the user inputs the numbers, but if you want the actual smallest to largest, you can do:
rs::sort(nums);
Now to print out the numbers:
rs::copy(nums,
std::ostream_iterator<int>(std::cout, " "));
and in reverse:
rs::copy(nums | rv::reverse,
std::ostream_iterator<int>(std::cout, " "));
Here's a demo.
You don't need a loop, the one you have it's not really a loop in the sense that it only cycles once, you can use a chained ostream:
cout << num1 << " " << num2 << " " << num3 << "\n";
And
cout << num3 << " " << num2 << " " << num1 << "\n";
But if you want print them sorted by value and you can't use some container where you can apply a sorting algorithm, you'll need some conditionals.
EDIT: Is this better?
Storing it in an array makes it easier to deal with the data, for instance, it will allow you to sort it by value using something as simple as <algorithm> library std::sort.
I'm trying to reverse a string in my C++ code line below revStr.at(j) = str.at(size);
But it doesn't change any of the elements in revStr.
Is there another way to do it without using any libraries.
#include <iostream>
#include<sstream>
#include <iterator>
using namespace std;
int main() {
ostringstream d;
long long c = 123456789;
d << c;
//cout << c << endl;
string str = d.str();
//cout << str.at(0) << endl;
int size = str.size() - 1;
//cout << size << endl;
ostringstream e;
e << str;
string revStr = e.str();
for (int i = size; size==0; size--) {
//cout << str.at(size);
int j = 0;
revStr.at(j) = str.at(size);
j++;
} // End For
cout << "Original String is :" << str << endl;
cout << "Reversed String is :" << revStr << endl;
}
Use std::reverse:
#include <string>
#include <algorithm>
#include <iostream>
int main()
{
std::string test{"Hello"};
std::cout << "Original string: " << test << std::endl;
std::reverse(test.begin(), test.end());
std::cout << "Reversed string: " << test << std::endl;
return 0;
}
Output:
Original string: Hello
Reversed string: olleH
If you just want to reverse a string, you should use std::reverse, as described by Tyler Lewis. It is the best option.
If you want to learn C++, then writing your own version is good practice.
The line
for (int i = size; size==0; size--)
means “Create a new int called i and set it to size initially. Then, while size is zero, do the following and then decrement size”.
There are three problems with this:
Size is not zero unless you entered a one-character string
Since you never use i, there’s no point in declaring it
Inside the loop you use j which is set to zero each time.
You can fix the first by changing the middle part of the for loop to size >= 0 (but be careful—if you later change it so that size is an unsigned type, because it doesn’t make sense for it to be negative, that code won’t work; it’s generally better to increment going up instead). You can fix the second by using i everywhere in the loop statement, and not changing size. You can fix the third by using i in the loop body, and not declaring a new variable inside the loop.
I noticed you used std::string so I used std function swap and string. Depending on if you consider this as a 'library'. There are several definitions of 'reverse'. You could reverse the word order in a string, or a pure char to char reversal like I wrote. Reversal could also mean changing character case, etc... but this is simply swap first and last. Then swap the 2nd and 2nd to last, then swap the 3rd and 3rd to last, etc...
So some points from your code. You only need to loop half the string length. The swap is from the ith and the ith to last. So the last is numCharacters - 1, thus the ith to last would be Last - i or numCharacters - 1 - i. I believe this is what you intended by using a farLeft(i) and a farRight(j) index.
#include <iostream>
void reverseStringInPlace(std::string &stringToReverse)
{
int numCharacters = stringToReverse.length();
for (int i=0; i<numCharacters/2; i++)
{ std::swap(stringToReverse[i], stringToReverse[numCharacters-i-1]); }
}
int main()
{
std::string stringToReverse = "reversing a string";
std::cout << stringToReverse << std::endl;
reverseStringInPlace(stringToReverse);
std::cout << stringToReverse << std::endl;
return 0;
}
Output:
reversing a string
gnirts a gnisrever
Changes made to the piece of code in question, it works.
for (unsigned int i = size; size >= 0; size--) {
revStr[j] = str[size];
j++;
}
I'm currently having a problem making a code for a Coordinate system.
In the exercise I'm doing, I want to create a coordinate system with an Ordinate/Abscissa and a defined letter (for example dot A)
I must put information for 25 dots and it must control all dots with the same letter. They should be in a circle with a (0;0) coordinate beginning. If the information given about the 25 dots do not meet the set condition the selected dots must have new reentered information to meet the condition without changing the given values of the previous dots(which meet the expectations). It also should have all the information for dots which have 2 positive coordinates
here's the code I made. I'd be really thankful if someone helped me out.
#include <iostream>
#include <cmath>
#include <stdio.h>
using namespace std;
int main(){
int dotX[23];//tri masiva
int dotY[23];
char dotName[23];
for (int i = 0; i<23; i++){// Cikal za vavejdane na masivite
cout << "Abscisa \t" << i + 1 << endl;
cin >> dotX[i];
cout << "Ordinata \t" << i + 1 << endl;
cin >> dotY[i];
cout << "Ime na tochkata" << endl;
cin >> dotName[i];
if (i >= 1){//IF operatora i cikula za obhozhdane na masiva i presmqtane na distanciite
bool flag = true;
while (flag){
double distance = sqrt(pow(dotY[i] - dotY[i - 1], 2) + pow(dotX[i] - dotX[i - 1], 2));//Formula za presmqtane na razstoqniqta
if (distance <= 6) {
char broi;
broi = broi++;
cout << "abscisa \t" << i + 1 << endl;
cin >> dotX[i];
cout << "ordinata \t" << i + 1 << endl;
cin >> dotY[i];
}
else{
flag = false;
}
}
}
}
float i;
for (float i = 0; i > 10, i++;){
float(dotX < 10);
cout << dotName[i] << endl;
}
}
There are a few big problems with your code.
First of all, the syntax for (float i = 0; i > 10, i++;) is completely wrong. It compiles, but that's just a coincidence. The different command in the for loop control structure should be separated by semicolons (;), not commas (,). The correct code would then be for (float i = 0; i > 10; i++). By the way, you made a typo, I think you meant for (float i = 0; i < 10; i++) (otherwise the for loop never runs since i is initialized to 0 and 0 > 10 is false from the beginning).
Second of all, you're initializing the variable i twice: once with float i; and once in the for loop. That shouldn't compile, although with some compilers it does. There are two options on how to do. The first option is to declare the variable outside of the for loop and just assign it without initializing it in the for loop:
float i;
for(i = 0; i < 10; i++){
//some stuff
}
The second option is to simply declare it in the for loop as you did in the first loop:
for(float i = 0; i < 10; i++){
//some stuff
}
Another mistake that you made is to declare i as a float and then try to access dotName[i]. Whatever you put inside the brackets has to be of type int or something similar (unsigned int, long, etc). Putting a float variable inside those brackets won't compile just like that. If you want to index an array with a float, you need to tell the compiler that you want to convert it to an int like this: dotName[(int)i] or dotName[int(i)]. This is called a cast. However, in your case, I would recommend just declaring i as an int.
Also, float(dotX < 10); is completely wrong, I don't really understand what you're trying to do there. I think you meant to do float(dotX[i] < 10);, but that still doesn't make any sense. What you would be doing there would be converting a bool to a float and then doing nothing with the result. That compiles and isn't wrong, but is completely useless. As I said, I don't understand what you want to do there.
Also, broi = broi++; is correct but useless. broi++; is enough. The ++ operator increments broi by one by itself and then returns the result. What the ++ operator does internally is basically this:
int operator++(int &x){
x = x + 1;
return x;
}
So it already increments the variable automatically without you having to do anything. What you did is the same as doing this:
broi = broi + 1;
broi = broi;
Here, the first line represents the ++ operator and the second line represents the = operator. It's clear that the second line is useless, so you can just remove it. In the same way, in your code, you can remove broi =, leaving simply broi++;.
You also did a few things that aren't recommended, but work just fine since the C++ standard supports them.
First of all, using namespace std; is bad practice. It's recommended to omit it and add std:: in front of cin, cout and endl. If you want to know why using namespace std; is bad practice, it's well explained here. However, I must admit that I personally still use using namespace std; since I think it's simpler.
Second of all, the main function is supposed to return 0, so it's recommended to add return 0; at the end of the main function. The return value of the main function tells what made the program close. The value 0 means that the program closed when it was supposed to. Any other values mean that the program crashed. A complete list of what each return value means is available here. Note that C++ supports omitting return 0; and most compilers add it automatically if it is omitted, but it's still recommended to have it. Also, C doesn't support omitting return 0; and in C it will return whatever happens to be in the memory, making it looked like the program crashed when it ended normally.
Also, #include <stdio.h> is C and although it works in C++, it's not recommended. In C++, it's better to use #include <cstdio>. All standard libraries that end with .h in C can be used in C++ by removing .h and adding a c at the beginning. That's also the case with cmath: in C, it would be #include <math.h> and in C++, it's #include <cmath>.
A good version of your code would therefore be:
#include <iostream>
#include <cmath>
#include <cstdio>
int main(){
int dotX[23]; //tri masiva
int dotY[23];
char dotName[23];
for (int i = 0; i < 23; i++){ // Cikal za vavejdane na masivite
std::cout << "Abscisa \t" << i + 1 << std::endl;
std::cin >> dotX[i];
std::cout << "Ordinata \t" << i + 1 << std::endl;
std::cin >> dotY[i];
std::cout << "Ime na tochkata" << std::endl;
std::cin >> dotName[i];
if (i >= 1){ //IF operatora i cikula za obhozhdane na masiva i presmqtane na distanciite
bool flag = true;
while (flag){
double distance = sqrt(pow(dotY[i] - dotY[i - 1], 2) + pow(dotX[i] - dotX[i - 1], 2)); //Formula za presmqtane na razstoqniqta
if (distance <= 6) {
char broi;
broi++;
std::cout << "abscisa \t" << i + 1 << std::endl;
std::cin >> dotX[i];
std::cout << "ordinata \t" << i + 1 << std::endl;
std::cin >> dotY[i];
}
else{
flag = false;
}
}
}
}
for (int i = 0; i < 10; i++){
float(dotX[i] < 10); //Note that I don't understand what you're trying to do here, so I just changed it to something that compiles
std::cout << dotName[i] << std::endl;
}
}
I've recently started learning programming using the C++ language. I wrote a simple program that is supposed to reverse a string which I compile in the Terminal using gcc/g++.
#include <iostream>
using namespace std;
string reverse_string(string str)
{
string newstring = "";
int index = -1;
while (str.length() != newstring.length())
{
newstring.append(1, str[index]);
index -= 1;
}
return newstring;
}
int main()
{
string x;
cout << "Type something: "; cin >> x;
string s = reverse_string(x);
cout << s << endl;
return 0;
}
I've rewritten it multiple times but I always get the same output:
Type something: banana
��
Has anyone had a problem like this or know how to fix it?
Your code initializes index to -1, and then uses str[index] but a negative index has no rational meaning in C++. Try instead initializing it like so:
index = str.length() - 1;
I can see several issues with your code. Firstly, you are initializing index to -1, and then decrementing it. Maybe you meant auto index = str.length()-1;?
I recommend you look at std::reverse, which will do the job you're after.
Your main function then becomes:
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
int main()
{
string x;
cout << "Type something: ";
cin >> x;
reverse(x.begin(), x.end());
cout << x << endl;
return 0;
}
If you really want to write your own reverse function, I recommend iterators over array indices. See std::reverse_iterator for another approach.
Note, the above will simply reverse the order of bytes within the string. Whilst this is fine for ASCII, it will not work for multi-byte encodings, such as UTF-8.
You should use a memory debugger like valgrind.
It's a good practice to scan your binary with it, and will make you save so much time.
I've done some self learning in the past with c++ online but gave up, till I bought a textbook on it and giving it another go. In my past research, I never read anything on vector arrays (or maybe I did and don't remember, who knows).
Anyway it says like regular arrays, vector arrays can be created for any data type and I'm trying to get a char vector array going and I'm running into some compile errors take a look.
I want an array of 26 that houses all the letters in the alphabet capitalized. So 65 to 91 I think. If there is and easier way to initialize the array with the letter I'm interested in learning that way.
#include <vector>
#include <iostream>
using namespace std;
int main()
{
vector <char> vchChar(26, 65);
for (int iii = 0; iii < vchChar.size(); iii++ )
{
for (int jjj = 65; jjj < 91; jjj++)
{
vchChar(iii) = jjj;
cout << "vchChar(" << iii+1 << ") is:\t" << vchChar(iii) << endl;
}
//cout << "vchChar(" << iii+1 << ") is:\t" << vchChar(iii) << endl;
}
return 0;
}
Originally I had square brackets instead of the parenthesise, fixed that and had hoped it would work but that got a whole set of new problems when I tried changing them in the cout statements. When I had them in square brackets it printed out in the terminal fine no compile errors. So now I have the cout statements like
cout << "vchChar(" << iii+1 << ") is:\t" << vchChar[iii] << endl;
I thought vector arrays where incremented by one from the element before it. But all I got when I printed vchChar into the terminal where all 'A's. So I tried playing around with another for loop to assign them by one from the element before it. I got some different outputs then I'd desired, and cant find the right algorithm to do it.
I'll keep at it, but an answer on this post is just as good for me. I have little idea what I'm doing so post everything you've got, but keep in mind that I probably wont have any idea what you're talking about :S. I've probably left something out because I've change the code a bit when troubleshooting, so if there are any question ask, and thank for your time.
I do not see any sense in your code. If I have understood correctly what you need is the following
#include <iostream>
#include <vector>
#include <numeric>
int main()
{
std::vector<char> vchChar(26);
std::iota( vchChar.begin(), vchChar.end() , 'A' );
for ( char c : vchChar ) std::cout << c << ' ';
std::cout << std::endl;
return 0;
}
Or you can write it even the following way
#include <iostream>
#include <vector>
#include <numeric>
int main()
{
std::vector<char> vchChar( 'Z' - 'A' + 1 );
std::iota( vchChar.begin(), vchChar.end() , 'A' );
for ( char c : vchChar ) std::cout << c << ' ';
std::cout << std::endl;
return 0;
}
If your compiler does not support standard algorithm std::iota and the range-based for statement then you can write
#include <iostream>
#include <vector>
#include <numeric>
int main()
{
std::vector<char> vchChar( 'Z' - 'A' + 1 );
for ( char c = 'A'; c <= 'Z'; ++c ) vchChar[c - 'A'] = c;
for ( std::vector<char>::size_type i = 0; i < vchChar.size(); ++i )
{
std::cout << "vchChar[" << i << "] is:\t" << vchChar[i] << std::endl;
}
return 0;
}
Take into account that it is a bad idea to use magic numbers as 65 or 91. For example if the program will run in an IBM mainframe then the result will be unexpected because there is another coding system, that is EBCDIC instead of ASCII.
As for statement
vchChar(iii) = jjj;
then it is invalid. Expression vchChar(iii) means a call of the operator function with one argument that is not defined in class std::vector.
You have at least the following issues in your code:
Setting a vector element is not done by myVector(index), but myVector[index], so basically the operator[]. In this special case, however, you can just push them in a row to the back.
You are trying to print one element of the array with vchChar(iii), but you should use the .at(index) method.
It is not crucial, but in this special case, you could use the iterator pattern to go through the vector rather than dealing with the indexing. Even if you do not do that, it is needless to use iii and jjj for variable names instead of the regular i and j.
I would prefer to use size_t or the vector<char>::size_type for the loop counters as you compare one of them against the vector size.
You are setting the elements more than once because you have a nested loop.
You are needlessly constructing the vector differently than the default.
You are using hard coded integers rather than actual characters.
Therefore, your correct code would look like this:
#include <vector>
#include <iostream>
using namespace std;
int main()
{
vector <char> vchChar;
for (char c = 'A'; c <= 'Z'; ++c)
vchChar.push_back(c);
for (vector<char>::size_t i = 0; i < vchChar.size(); ++i)
cout << "vchChar(" << i+1 << ") is:\t" << vchChar.at(i) << endl;
return 0;
}
Disclaimer: this is just compilation and runtime fix. I have not dealt with use case and design issues. There are better solutions as I partially mentioned them, but I decided to make your code with the least impact.
You do not need nested loops.
{
vector<char> vchChar;
for (char letter = 'A'; letter <= 'Z'; ++letter) {
vchChar.push_back(letter);
}
for (int i = 0; i < (int) vchChar.size(); ++i) {
cout << vchChar[i];
}
}