I am confused with this block of code:
ipPtr = ipPtr + 3; // 5
cout << *ipPtr << endl;
Why the cout is not 5 but some random large number? can anyone explain to me please. As my understanding I thought the cout << *ipPtr << endl; is pointed to the *ipPtr above it. Am I right ?
#include <iostream>
void main(){
using namespace std;
int iaArray[] = {1,2,3,4,5};
int* ipPtr = 0;
ipPtr = &(iaArray[1]);
cout << *ipPtr << endl;//2
++ipPtr;
cout << *ipPtr << endl;//3
ipPtr = ipPtr + 3; //not 5 but random number.
cout << *ipPtr << endl;
}
Because you have incremented the pointer past the end of the array. You seem to have forgotten that you wrote ++ipPtr before adding 3 to it.
&(iaArray[1])
|
iaArray = { 1, 2, 3, 4, 5 } ?
| |
++ipPtr ipPtr + 3
Because when you add 3 to the pointer it's already on the third position of the array, so it ends up after the last element.
ipPtr = &(iaArray[1]);
//Pointing to the second position (first one is 0)
++ipPtr;
//Pointing to the third position
//3 + 3 = 6
ipPtr = ipPtr + 3;
The array only has 5 positions so it prints whatever is in that memory location not 5 which is in the fifth position.
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Here is my program, it is using c++ and I done it on MacOS VSCode.
Library:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <string>
#include <iomanip>
using namespace std;
Function:
float* read_data(int& size){
int a_size = 0;
static float a[10];
float* temp = a;
cout << "Please enter values, and press 'Q' when finished: ";
for (int i = 0; i < size; i++){
cin >> a[i];
if (cin.fail()){
break;
}else{
a_size++;
}
}
cin.clear();
size = a_size;
return temp;
}
Main function:
int main(){
int size1 = 10;
int size2 = 10;
int size3 = 0;
float array[10];
float addarray[10];
float longarray[20];
float* ptr = array;
float* addptr = addarray;
float* longptr = longarray;
cout << fixed << setprecision(2);
ptr = read_data(size1);
cout << setw(15) << "Original array" << setw(5) << "=" << setw(5) << "{ " << *ptr;
for (int i = 0; i < size1; i++){
cout << ", " << *(ptr + i);
}
cout << " }" << endl;
for (int i = 0; i < size1; i++){
*(longptr + size3) = *(ptr + i);
size3++;
}
addptr = read_data(size2);
for (int i = 0; i < size2; i++){
*(longptr + size3) = *(addptr + i);
size3++;
}
cout << setw(15) << "New array" << setw(5) << "=" << setw(5) << "{ " << *longptr;
for (int i = 0; i < size3; i++){
cout << ", " << *(longptr + i);
}
cout << " }" << endl;
return 0;
}
The main objective of the program was to prompt the user for an array, maximum of 10 elements. Prompting the user was done using the float* read_data(int& size) function.
The program would echo or print out the inputted array.
After that, the user was prompt for a second time using the same function to get another array of elements, in this case is a list of float values.
Then, the program would use a dynamic data allocation (DMA) technique to combine the two array into one long array. The new array will be printed out and the program is terminated.
Problem
As I said before, the program are supposed to prompt the user for inputs every time the read_data() were called.
I called the function twice. It did run twice, but failed to prompt the user for input the second time around.
I thought the problem was due to the cin.fail() arguments. That is why I tried to mess with the cin.ignore() and cin.clear() either by removing them or only using one of them.
I am honestly at lost on what is the root of the problem.
The heart of the issue is that you want users to enter 'Q', a char, into your float variable.
Here's a small example:
#include <iostream>
int main() {
int a;
std::cin >> a;
if (std::cin.fail()) {
std::cout << "Fail.\n";
} else {
std::cout << "All clear.\n";
}
std::cin >> a;
std::cout << (2 * a) << '\n';
}
Output:
~/tmp
❯ ./a.out
Q
Fail.
0
~/tmp
❯ ./a.out
3
All clear.
3
6
So, you can see that you were on the right track. The issue is that if std::cin did fail, you are now responsible to clean up your mess.
std::cin.clear() is a good start. It resets the fail flag bits. But the stream is still in a bad state. What you haven't done is clean it up completely.
You're calling std::cin.ignore(), but leaving the parameter list empty.
Something like this is best practice:
#include <iostream>
#include <limits>
int main() {
int a;
std::cin >> a;
if (std::cin.fail()) {
std::cout << "Fail.\n";
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
std::cin.clear();
} else {
std::cout << "All clear.\n";
}
std::cin >> a;
std::cout << (2 * a) << '\n';
}
Output:
~/tmp
❯ ./a.out
Q
Fail.
3
6
While that should fix address the question, your code is still fundamentally broken.
EDIT
Here's a mini-code review:
#include <iostream>
#include <algorithm> // These includes
#include <cmath> // are
#include <cstdlib> // not
#include <string> // used
#include <iomanip>
using namespace std; // Bad practice
// You should not attempt to return a C-style array
float* read_data(int& size){ // Why is the size being passed by reference?
int a_size = 0;
static float a[10]; // Not doing what you think
float* temp = a;
cout << "Please enter values, and press 'Q' when finished: ";
for (int i = 0; i < size; i++){ // Formatting; should be ") {"
cin >> a[i];
if (cin.fail()){ // Addressed above; root cause of your question
break;
}else{
a_size++;
}
}
cin.clear();
size = a_size;
return temp;
}
// While I assume the goal of the assignment is to get familiar with pointers,
// This assignment is trivial with vectors.
int main(){
int size1 = 10; // If declared const/constexpr, you'd only need one **
int size2 = 10; // These are bad names
int size3 = 0; // Front loading declarations is bad practice.
float array[10]; // ** And could use that variable here to avoid the magic number
float addarray[10];
float longarray[20]; // Not dynamic per your requirements.
float* ptr = array; // Still not dynamic, also unnecessary given the prior declarations
float* addptr = addarray;
float* longptr = longarray;
cout << fixed << setprecision(2);
ptr = read_data(size1);
// The first cout is not good, what do you think *ptr will print?
cout << setw(15) << "Original array" << setw(5) << "=" << setw(5) << "{ " << *ptr;
for (int i = 0; i < size1; i++){
cout << ", " << *(ptr + i); // The pointers can still use array syntax
}
cout << " }" << endl;
for (int i = 0; i < size1; i++){
*(longptr + size3) = *(ptr + i);
size3++;
}
addptr = read_data(size2);
for (int i = 0; i < size2; i++){
*(longptr + size3) = *(addptr + i);
size3++;
}
// As soon as you start repeating yourself, consider writing a function.
// The logic will only live in one place, and only need to be changed in
// one place.
cout << setw(15) << "New array" << setw(5) << "=" << setw(5) << "{ " << *longptr;
for (int i = 0; i < size3; i++){
cout << ", " << *(longptr + i);
}
cout << " }" << endl;
return 0;
}
A lot going on that is less than great. The biggest flub appears to be the static array. static in this scenario extends the lifetime of the variable until the end of the program. So, when you call this function the second time, you overwrite the first array with the second because there's only ever one array for every call of that function. In the future, I would refrain from grabbing random code online and just dumping it in your program unless you know what it does.
Now, let's move on to a working solution.
The first thing I'm going to do is state the requirements as I understood them.
Have the user enter data to fill two arrays of floats.
Each array should hold a maximum of ten elements.
The user enters "Q" to indicate that they are finished entering data.
Print the first array.
Tack the second array on to the end of the first, in a dynamic fashion.
Print the combined array.
Anytime you get an assignment, your first task should be to restate the problem in your own words. This will clarify requirements and you will demonstrate to yourself that you understand the problem to be solved.
What makes this interesting is the "Q" to quit. Note that I'm using double quotes. We will read all of our data as strings, and we have to convert to float as needed.
std::stof() exists, but it's not as simple as just calling it. That function can throw exceptions, and we want to ensure that only actual float values get converted.
So we'll wrap that call in a function of our own. Here's an example:
#include <exception>
#include <iostream>
#include <string>
class bad_user_input : public std::exception {
const char* what() const noexcept override {
return "User input was not valid.";
}
};
float convert_string_to_float(const std::string& val) {
std::size_t marker = 0;
float f = 0.0f;
try {
f = std::stof(val, &marker);
} catch(...) {
throw bad_user_input();
}
// Was the entire value entered a float?
if (marker != val.length()) {
throw bad_user_input();
}
// Getting here means a valid float was entered.
return f;
}
int main() {
float f = 0.0f;
// Test the function
try {
f = convert_string_to_float("5.6");
} catch(...) {
std::cerr << "Issue.\n";
}
std::cout << f << "\n\n";
f = 0.0f;
try {
f = convert_string_to_float("5.6cat");
} catch(...) {
std::cerr << "Issue.\n";
}
std::cout << f << "\n\n";
f = 0.0f;
try {
f = convert_string_to_float("Q");
} catch(...) {
std::cerr << "Issue.\n";
}
std::cout << f << '\n';
}
Output:
❯ ./a.out
5.6
Issue.
0
In convert_string_to_float
stof: no conversion
Issue.
0
Now that we can convert strings to floats, with guarantees, we are ready to read values. To test that we're reading correctly, we also want to be able to print our arrays:
#include <exception>
#include <iostream>
#include <string>
class bad_user_input : public std::exception {
const char* what() const noexcept override {
return "User input was not valid.";
}
};
float convert_string_to_float(const std::string& val) {
std::size_t marker = 0;
float f = 0.0f;
try {
f = std::stof(val, &marker);
} catch(...) {
throw bad_user_input();
}
// Was the entire value entered a float?
if (marker != val.length()) {
throw bad_user_input();
}
// Getting here means a valid float was entered.
return f;
}
void read_array_from_user(float* arr, int& size, const int maxCapacity, std::istream& sin = std::cin) {
size = 0;
std::string input;
while (size < maxCapacity && std::getline(sin, input)) {
if (input == "Q" || input == "q" || input.length() == 0) {
return;
}
try {
arr[size] = convert_string_to_float(input);
} catch(const std::exception& e) {
std::cerr << "Bad input. Reason: " << e.what() << "\nTry again.\n";
continue;
}
++size;
}
}
void print_array(float* arr, int size, std::ostream& sout = std::cout) {
sout << "{ ";
for (int i = 0; i < size; ++i) {
sout << arr[i] << (i == size - 1 ? " " : ", ");
}
sout << "}\n";
}
int main() {
constexpr int maxArrayCapacity = 10;
float arrayOne[maxArrayCapacity];
int arrayOneSize = 0;
read_array_from_user(arrayOne, arrayOneSize, maxArrayCapacity);
print_array(arrayOne, arrayOneSize);
}
Output:
~/tmp
❯ ./a.out
1
2
3
4
q
{ 1, 2, 3, 4 }
~/tmp took 3s
❯ ./a.out
1
2
Q
{ 1, 2 }
~/tmp took 2s
❯ ./a.out
1
2
3
4
5
6
{ 1, 2, 3, 4, 5, 6 }
~/tmp took 4s
❯ ./a.out
1
2
3
4
5
6
7
8
9
0
{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 }
~/tmp took 5s
❯ compilecpp tmp_repair.cpp
~/tmp
❯ ./a.out
1
2
3
4
5
6
7
8
9
0
{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 }
~/tmp took 4s
❯ ./a.out
1
2
3
q
{ 1, 2, 3 }
~/tmp took 2s
❯ ./a.out
1
2b
Bad input. Reason: User input was not valid.
Try again.
2
3
Q
{ 1, 2, 3 }
~/tmp took 3s
❯ ./a.out
1
2
3
Q
{ 1, 2, 3 }
~/tmp took 2s
❯ ./a.out
1
2
3
{ 1, 2, 3 }
So, we can read an array with a maximum of 10 values. The reading is robust enough to handle typos or obviously bad input. But we don't have to read 10 values, it's the maximum. We keep track of the actual size ourselves.
Take note on how simple the main() function is because we put our sub-tasks into their own functions.
Reading the second array only requires a few lines in our main() now.
int main() {
constexpr int maxArrayCapacity = 10;
float arrayOne[maxArrayCapacity];
int arrayOneSize = 0;
read_array_from_user(arrayOne, arrayOneSize, maxArrayCapacity);
print_array(arrayOne, arrayOneSize);
float arrayTwo[maxArrayCapacity];
int arrayTwoSize = 0;
read_array_from_user(arrayTwo, arrayTwoSize, maxArrayCapacity);
print_array(arrayTwo, arrayOneSize);
}
Easy-peasy there. Now we need to combine the arrays "dynamically."
We know the sizes of our two arrays, so we know the size of the final array. Now we just need to allocate the memory on the heap. I'm not going to use new, which is what I assume you're supposed to do. My reasoning is that C++ has had better methods for managing dynamic allocations since 2011.
#include <memory>
// ...
int main() {
constexpr int maxArrayCapacity = 10;
float arrayOne[maxArrayCapacity];
int arrayOneSize = 0;
read_array_from_user(arrayOne, arrayOneSize, maxArrayCapacity);
print_array(arrayOne, arrayOneSize);
float arrayTwo[maxArrayCapacity];
int arrayTwoSize = 0;
read_array_from_user(arrayTwo, arrayTwoSize, maxArrayCapacity);
print_array(arrayTwo, arrayTwoSize);
int combinedArraySize = arrayOneSize + arrayTwoSize;
// This is considered dynamic; it's allocated on the heap
auto combinedArray = std::make_unique<float[]>(combinedArraySize);
int idx = 0;
for (int i = 0; i < arrayOneSize; ++i) {
combinedArray[idx] = arrayOne[i];
++idx;
}
for (int i = 0; i < arrayTwoSize; ++i) {
combinedArray[idx] = arrayTwo[i];
++idx;
}
print_array(combinedArray.get(), combinedArraySize);
}
Output:
~/tmp
❯ ./a.out
1
2
q
{ 1, 2 }
3
4
5
q
{ 3, 4, 5 }
{ 1, 2, 3, 4, 5 }
Hopefully, the most gnarly stuff is over where you get the user input, mostly due to the requirement that users can enter a letter to signal they're done. And that makes sense since now you have to do type conversions. Other languages like python make it possible with a lot less LOC (Lines Of Code) on your part. The task itself is not difficult, but ensuring users behave is a whole other beast. If you're allowed to assume well-behaved input will always be provided, you can simplify the code quite a bit.
I am trying to increment the value of an element inside an array, using a pointer storing that element's address.
When I write it as " p++; " it does not work, whereas if I write it as *p=*p+1 it works.
*p=*p+1; // this increments the value
*p++; //while this does not
I am expecting both ways to work
operator ++ has higher precedence than dereference so *p++ increments the pointer and then performs dereference. If you want to increment the value it initially points to then you need to add parentheses: (*p)++;
#include <iostream>
using namespace std;
int main (int argc, char *argv[])
{
int numbers[] = { 100, 200, 300, 400, 500, 600 };
int *p1 = &numbers[0];
int *p2 = &numbers[2];
int *p3 = &numbers[4];
*p1 = *p1 + 1;
*p2 = *p2++;
++*p3;
cout << numbers[0] << " " << numbers[1] << " " << numbers[2] << " " << numbers[3] << " " << numbers[4] << " " << numbers[5];
}
Running this through Xcode gives me a clue as to what's going wrong, it says: "temp/test.cxx:11:12: warning: unsequenced modification and access to 'p2' [-Wunsequenced]".
The output is 101 200 300 300 501 600.
The first method works, as you said in your question.
The second does something completely different from what you were expecting: it takes the value pointed to by p2 (300) then increments the pointer and saves the value back to that new address.
The third example with p3 is closer to what you're trying to achieve I think.
I would like to represent a std::vector of a structure containing several integers as a "flatten" vector of integers, without copying the data.
I tried something with a reinterpret_cast as shown below:
#include <vector>
#include <iostream>
struct Tuple
{
int a, b, c;
};
int main()
{
// init
std::vector<Tuple> vec1(5);
for(size_t i=0; i<vec1.size(); ++i)
{
vec1[i].a = 3 * i + 0;
vec1[i].b = 3 * i + 1;
vec1[i].c = 3 * i + 2;
}
// flattening
std::vector<int>* vec2 = reinterpret_cast<std::vector<int>*>(&vec1);
// print
std::cout << "vec1 (" << vec1.size() << ") : ";
for(size_t i=0; i<vec1.size(); ++i)
{
std::cout << vec1.at(i).a << " " << vec1.at(i).b << " " << vec1.at(i).c << " ";
}
std::cout << std::endl;
std::cout << "vec2 (" << vec2->size() << ") : ";
for (size_t j = 0; j < vec2->size(); ++j)
{
std::cout << vec2->at(j) << " ";
}
std::cout << std::endl;
return 0;
}
which works well since the output is:
vec1 (5) : 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
vec2 (15) : 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
My questions are :
Is this behavior compiler dependent? (I am using g++ 6.3.0)
How vec2 knows that the size of the vector is 15 and not 5?
Is their any other solution avoiding the use of reinterpret_cast? (If I "accidentally" add a double member to Tuple, the resulting issue could be hard to track...)
If vec1 has a specific allocator: std::vector<Tuple,A<Tuple>>, what should be the type of vec2? std::vector<int> or std::vector<int,A<int>> or std::vector<int,A<Tuple>>?
You can't legally reinterpret_cast the entire vector to a different type of vector. But you can legally cast a pointer to struct to a pointer to the first element of that struct. So this works:
std::vector<Tuple> vec1(5);
int* vec2 = &vec1.front().a;
size_t vec2_size = vec1.size() * sizeof(vec1[0]) / sizeof(vec2[0]);
for (size_t j = 0; j < vec2_size; ++j)
{
std::cout << vec2[j] << " ";
}
You need to make sure there's no padding in Tuple, so:
static_assert(sizeof(Tuple) == 3 * sizeof(int), "Tuple must be 3 ints");
To answer your bulleted questions:
Is this behavior compiler dependent?
Your code was illegal.
How vec2 knows that the size of the vector is 15 and not 5?
You got lucky, your code was illegal.
Is their any other solution avoiding the use of reinterpret_cast?
See above.
If vec1 has a specific allocator: std::vector>, what should be the type of vec2?
Same as above, int*.
Imagine I have the following code:
int* firstArray[4];
fill_n(firstArray, 4, 0);
//firstArray values are: 0, 0, 0, 0
int* secondArray[4];
fill_n(secondArray, 4, 1);
//secondArray values are: 1, 1, 1, 1
Would there be a way to link the first value of firstArray to the first value of secondArray so that if I do:
secondArray[0] = 20;
//secondArray values would be: 20, 1, 1, 1 (because I just changed it)
//firstArray values would be: 20, 0, 0, 0 (because it's pointing to the value I just changed)
As far as I've tested and researched I can't do something along those lines.
P.S.: This is to create a Samurai Sudoku (in which some chunks of numbers are shared), so when I modify a value that 2 sudokus share it would be updated in both.
You are mixing up concepts. You don't even need to declare a second array just do this:
int myarray [20];
int * mypointer;
mypointer = myarray;
// then you can use both mypointer[] myarray[] the same way to access array elements
The concept of arrays is related to that of pointers. In fact, arrays work very much like pointers to their first elements, and, actually, an array can always be implicitly converted to the pointer of the proper type.
The [] operator in arrays acts the same way as a as a de-referencing operator, but with the added ability of automatically advancing the pointer according to the data type. This is why Array[1] references the same value as *(Array+1)
however you are declaring an arrays of pointers to integers, which means that you can't "store" integer values in this array, but rather store the value of the address where the integer is located.
Additionally when you declare an array you are essentially declaring a constant pointer, so you can not "steal" it and make it pointer to another location.
Study this Code and it's output (RUN THIS CODE)
#include <iostream>
using namespace std;
void printarray (int arg[], int length) {
for (int n=0; n<length; ++n)
cout << arg[n] << ' ';
cout << '\n';
}
int main ()
{
int first[] = {5, 10, 15, 14, 13};
printarray (first,3);
// int* third[] = {1,1,1}; Not accepted because int is not int*
// storing the addresses of first as pointers in 2 different arrays
int* third[] = {first,first+2,first+3};
int* forth[] = {first,first+2,first+3};
// the memory adress where the pointers TO first is stored
cout << third << endl;
cout << forth << endl;
cout << &third << endl;
cout << &forth << endl;
// the memory adress where the pointer TO the value of first[0] is stored
cout << *third << endl;
cout << *forth << endl;
cout << third[0] << endl;
cout << forth[0] << endl;
// you are defrencing twice
cout << *third[0] << endl;
cout << *forth[0] << endl;
cout << **third << endl;
cout << **forth << endl;
// assign once
first[0] = 77;
// applys to all values
cout << first[0] << endl;
cout << *third[0] << endl;
cout << *forth[0] << endl;
// better yet declare a int* and use it same way your array
int* second;
second = first;
cout << first[0] << endl;
cout << second[0] << endl;
// again change value and the change is reflected everywhere
second[0] = 99;
cout << first[0] << endl;
cout << second[0] << endl;
cout << *third[0] << endl;
cout << *forth[0] << endl;
}
OUTPUT
5 10 15
0x786378c0b860
0x786378c0b880
0x786378c0b860
0x786378c0b880
0x786378c0b840
0x786378c0b840
0x786378c0b840
0x786378c0b840
5
5
5
5
77
77
77
77
77
99
99
99
99
Store int**s instead of int*s this way you can have each entry in the array point to the same address.
You would initialize the first element lets say as:
int *first = new int;
*first = 1;
firstArray[0] = first;
secondArray[0] = first;
Then if you write *first = 20; both arrays will be updated.
There is a problem with your question that you initialize first to 0 and then to 1. If they were the same address then setting to 1 would overwrite the setting to 0.
You question is not self coherent :
you write int* secondArray[4]; : secondArray is an array of 4 pointers to int
you write secondArray[0] = 20 : here secondArray is an array of 4 int values
Hyp 1 : arrays of 4 int values
&secondArray[1] is &secondArray[0] + 1. Full stop. That the way arrays work in C++. In that case if secondArray[0] is the same as firstArray[0] then the 2 arrays are at same address and are in fact the same array and your example becomes :
int firstArray[4];
fill_n(firstArray, 4, 0);
//firstArray values are: 0, 0, 0, 0
/* int secondArray[4]; first elements could not be related and you should have : */
int *secondArray = firstArray;
fill_n(secondArray, 4, 1);
//secondArray values are: 1, 1, 1, 1
//firstArray values are also: 1, 1, 1, 1
Hyp 2 : arrays of 4 pointers
You can have first element of both array point to same value while the three others are independent
int array[5];
int *firstArray[4];
int *secondArray[4];
firstArray[0] = &(array[0]);
for(int i=1; i<4; i++)
firstArray[i] = secondArray[i] = &(array[1]);
}
secondArray[0] = &(array[4]);
Then :
int* firstArray[4];
fill_n(firstArray, 4, 0);
//firstArray values are: 0, 0, 0, 0
int* secondArray[4];
fill_n(secondArray, 4, 1);
//secondArray values are: 1, 1, 1, 1
//firstArray values are: 1, 0, 0, 0
and
*(secondArray[0]) = 20;
//secondArray values would be: 20, 1, 1, 1 (because I just changed it)
//firstArray values would be: 20, 0, 0, 0 (because it's pointing to the value I just changed)
That is what you asked for, but it is a very unusual requirement and I'm not sure it is really what you need.
I'm having trouble using recursion to add letters to a base 10 - base 12 conversion. How would I go about adding letters into my function? I was thinking about adding an if statement in, but i have no idea where and how to go about this. pointers are appreciated Thanks!
Given a count from 1 to 12:
Dec 1 2 3 4 5 6 7 8 9 10 11 12
Duo 1 2 3 4 5 6 7 8 9 X E 10
my function:
template<class myType>
myType convertDec(myType number){
if(number == 0)
return number;
//if statement somewhere in here? not sure considering i can't touch the return statement
return (number % 12) + 10*convertDec(number / 12);
}
example ideal output:
65280 = 31940 (works fine)
2147483626 = 4EE23088X (doesnt work!)
#include <iostream>
#include <string>
using namespace std;
string ConvertToDuodecimal(unsigned long long n)
{
if (n < 12)
return string() + "0123456789XE"[n];
return ConvertToDuodecimal(n / 12) + ConvertToDuodecimal(n % 12);
}
int main()
{
cout << ConvertToDuodecimal(0) << endl;
cout << ConvertToDuodecimal(1) << endl;
cout << ConvertToDuodecimal(10) << endl;
cout << ConvertToDuodecimal(11) << endl;
cout << ConvertToDuodecimal(12) << endl;
cout << ConvertToDuodecimal(13) << endl;
cout << ConvertToDuodecimal(65280) << endl;
cout << ConvertToDuodecimal(2147483626) << endl;
return 0;
}
Output (ideone):
0
1
X
E
10
11
31940
4EE23088X
The base is a property of the way a number is displayed and nothing to do with it's internal representation. You need to write a function that prints a normal int using base 12.
a = 12; // A = 12 (in base 10)
printf("%d",a); // Prints a in base 10 (still 12)
printf("%x",a); // Prints a in base 16 (now C)
You code is changing the actual value, which isn't the right thing to do.
(and yes, before the pedants strike, printf isn't good C++...)