C++: Converting Hexadecimal to Decimal - c++

Question

I'm looking for a way to convert hex(hexadecimal) to dec(decimal) easily. I found an easy way to do this like :
int k = 0x265;
cout << k << endl;
But with that I can't input 265. Is there anyway for it to work like that:
Input: 265
Output: 613
Is there anyway to do that ?
Note: I've tried:
int k = 0x, b;
cin >> b;
cout << k + b << endl;
and it doesn't work.

Answer 1

#include <iostream>
#include <iomanip>
#include <sstream>
int main()
{
int x, y;
std::stringstream stream;
std::cin >> x;
stream << x;
stream >> std::hex >> y;
std::cout << y;
return 0;
}

Answer 2

Use std::hex manipulator:
#include <iostream>
#include <iomanip>
int main()
{
int x;
std::cin >> std::hex >> x;
std::cout << x << std::endl;
return 0;
}

Answer 3

Well, the C way might be something like ...
#include <stdlib.h>
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
printf("%X", n);
exit(0);
}

Answer 4

Here is a solution using strings and converting it to decimal with ASCII tables:
#include <iostream>
#include <string>
#include "math.h"
using namespace std;
unsigned long hex2dec(string hex)
{
unsigned long result = 0;
for (int i=0; i<hex.length(); i++) {
if (hex[i]>=48 && hex[i]<=57)
{
result += (hex[i]-48)*pow(16,hex.length()-i-1);
} else if (hex[i]>=65 && hex[i]<=70) {
result += (hex[i]-55)*pow(16,hex.length( )-i-1);
} else if (hex[i]>=97 && hex[i]<=102) {
result += (hex[i]-87)*pow(16,hex.length()-i-1);
}
}
return result;
}
int main(int argc, const char * argv[]) {
string hex_str;
cin >> hex_str;
cout << hex2dec(hex_str) << endl;
return 0;
}

Answer 5

I use this:
template <typename T>
bool fromHex(const std::string& hexValue, T& result)
{
std::stringstream ss;
ss << std::hex << hexValue;
ss >> result;
return !ss.fail();
}

Answer 6

std::cout << "Enter decimal number: " ;
std::cin >> input ;
std::cout << "0x" << std::hex << input << '\n' ;
if your adding a input that can be a boolean or float or int it will be passed back in the int main function call...
With function templates, based on argument types, C generates separate functions to handle each type of call appropriately. All function template definitions begin with the keyword template followed by arguments enclosed in angle brackets < and >. A single formal parameter T is used for the type of data to be tested.
Consider the following program where the user is asked to enter an integer and then a float, each uses the square function to determine the square.
With function templates, based on argument types, C generates separate functions to handle each type of call appropriately. All function template definitions begin with the keyword template followed by arguments enclosed in angle brackets < and >. A single formal parameter T is used for the type of data to be tested.
Consider the following program where the user is asked to enter an integer and then a float, each uses the square function to determine the square.
#include <iostream>
using namespace std;
template <class T> // function template
T square(T); /* returns a value of type T and accepts type T (int or float or whatever) */
void main()
{
int x, y;
float w, z;
cout << "Enter a integer: ";
cin >> x;
y = square(x);
cout << "The square of that number is: " << y << endl;
cout << "Enter a float: ";
cin >> w;
z = square(w);
cout << "The square of that number is: " << z << endl;
}
template <class T> // function template
T square(T u) //accepts a parameter u of type T (int or float)
{
return u * u;
}
Here is the output:
Enter a integer: 5
The square of that number is: 25
Enter a float: 5.3
The square of that number is: 28.09

Answer 7

This should work as well.
#include <ctype.h>
#include <string.h>
template<typename T = unsigned int>
T Hex2Int(const char* const Hexstr, bool* Overflow)
{
if (!Hexstr)
return false;
if (Overflow)
*Overflow = false;
auto between = [](char val, char c1, char c2) { return val >= c1 && val <= c2; };
size_t len = strlen(Hexstr);
T result = 0;
for (size_t i = 0, offset = sizeof(T) << 3; i < len && (int)offset > 0; i++)
{
if (between(Hexstr[i], '0', '9'))
result = result << 4 ^ Hexstr[i] - '0';
else if (between(tolower(Hexstr[i]), 'a', 'f'))
result = result << 4 ^ tolower(Hexstr[i]) - ('a' - 10); // Remove the decimal part;
offset -= 4;
}
if (((len + ((len % 2) != 0)) << 2) > (sizeof(T) << 3) && Overflow)
*Overflow = true;
return result;
}
The 'Overflow' parameter is optional, so you can leave it NULL.
Example:
auto result = Hex2Int("C0ffee", NULL);
auto result2 = Hex2Int<long>("DeadC0ffe", NULL);

Answer 8

only use:
cout << dec << 0x;

Answer 9

If you have a hexadecimal string, you can also use the following to convert to decimal
int base = 16;
std::string numberString = "0xa";
char *end;
long long int number;
number = strtoll(numberString.c_str(), &end, base);

Answer 10

I think this is much cleaner and it also works with your exception.
#include <iostream>
using namespace std;
using ll = long long;
int main ()
{
ll int x;
cin >> hex >> x;
cout << x;
}

Answer 11

std::stoi, stol, stoul, stoull can convert to different number systems
long long hex2dec(std::string hex)
{
std::string::size_type sz = 0;
try
{
hex = "0x"s + hex;
return std::stoll(hex, &sz, 16);
}
catch (...)
{
return 0;
}
}
and similar if you need return string
std::string hex2decstr(std::string hex)
{
std::string::size_type sz = 0;
try
{
hex = "0x"s + hex;
return std::to_string(std::stoull(hex, &sz, 16));
}
catch (...)
{
return "";
}
}
Usage:
std::string converted = hex2decstr("16B564");