C++ convert hex string to signed integer

C++ Problem Overview

I want to convert a hex string to a 32 bit signed integer in C++.

So, for example, I have the hex string "fffefffe". The binary representation of this is 11111111111111101111111111111110. The signed integer representation of this is: -65538.

How do I do this conversion in C++? This also needs to work for non-negative numbers. For example, the hex string "0000000A", which is 00000000000000000000000000001010 in binary, and 10 in decimal.

C++ Solutions

Solution 1 - C++

use std::stringstream

unsigned int x;   
std::stringstream ss;
ss << std::hex << "fffefffe";
ss >> x;

the following example produces -65538 as its result:

#include <sstream>
#include <iostream>

int main() {
	unsigned int x;   
	std::stringstream ss;
	ss << std::hex << "fffefffe";
	ss >> x;
    // output it as a signed type
	std::cout << static_cast<int>(x) << std::endl;
}

In the new C++11 standard, there are a few new utility functions which you can make use of! specifically, there is a family of "string to number" functions (http://en.cppreference.com/w/cpp/string/basic_string/stol and http://en.cppreference.com/w/cpp/string/basic_string/stoul). These are essentially thin wrappers around C's string to number conversion functions, but know how to deal with a std::string

So, the simplest answer for newer code would probably look like this:

std::string s = "0xfffefffe";
unsigned int x = std::stoul(s, nullptr, 16);

NOTE: Below is my original answer, which as the edit says is not a complete answer. For a functional solution, stick the code above the line :-).

It appears that since lexical_cast<> is defined to have stream conversion semantics. Sadly, streams don't understand the "0x" notation. So both the boost::lexical_cast and my hand rolled one don't deal well with hex strings. The above solution which manually sets the input stream to hex will handle it just fine.

Boost has some stuff to do this as well, which has some nice error checking capabilities as well. You can use it like this:

try {
    unsigned int x = lexical_cast<int>("0x0badc0de");
} catch(bad_lexical_cast &) {
    // whatever you want to do...
}

If you don't feel like using boost, here's a light version of lexical cast which does no error checking:

template<typename T2, typename T1>
inline T2 lexical_cast(const T1 &in) {
    T2 out;
    std::stringstream ss;
    ss << in;
    ss >> out;
    return out;
}

which you can use like this:

// though this needs the 0x prefix so it knows it is hex
unsigned int x = lexical_cast<unsigned int>("0xdeadbeef");

Solution 2 - C++

For a method that works with both C and C++, you might want to consider using the standard library function strtol().

#include <cstdlib>
#include <iostream>
using namespace std;

int main() {
	string s = "abcd";
	char * p;
	long n = strtol( s.c_str(), & p, 16 );
	if ( * p != 0 ) { //my bad edit was here
		cout << "not a number" << endl;
	}
	else {
		cout << n << endl;
	}
}

Solution 3 - C++

Andy Buchanan, as far as sticking to C++ goes, I liked yours, but I have a few mods:

template <typename ElemT>
struct HexTo {
    ElemT value;
    operator ElemT() const {return value;}
    friend std::istream& operator>>(std::istream& in, HexTo& out) {
        in >> std::hex >> out.value;
        return in;
    }
};

Used like

uint32_t value = boost::lexical_cast<HexTo<uint32_t> >("0x2a");

That way you don't need one impl per int type.

Solution 4 - C++

Working example with strtoul will be:

#include <cstdlib>
#include <iostream>
using namespace std;




int main() {
string s = "fffefffe";
char * p;
long n = strtoul( s.c_str(), & p, 16 );
if ( * p != 0 ) {

cout << "not a number" << endl;
}    else {

cout << n << endl;
}
}
int main() {
string s = "fffefffe";
char * p;
long n = strtoul( s.c_str(), & p, 16 );
if ( * p != 0 ) {

cout << "not a number" << endl;
}    else {

cout << n << endl;
}
}

strtol converts string to long. On my computer numeric_limits<long>::max() gives 0x7fffffff. Obviously that 0xfffefffe is greater than 0x7fffffff. So strtol returns MAX_LONG instead of wanted value. strtoul converts string to unsigned long that's why no overflow in this case.

Ok, strtol is considering input string not as 32-bit signed integer before convertation. Funny sample with strtol:

#include <cstdlib>
#include <iostream>
using namespace std;




int main() {
string s = "-0x10002";
char * p;
long n = strtol( s.c_str(), & p, 16 );
if ( * p != 0 ) {

cout << "not a number" << endl;
}    else {

cout << n << endl;
}
}
int main() {
string s = "-0x10002";
char * p;
long n = strtol( s.c_str(), & p, 16 );
if ( * p != 0 ) {

cout << "not a number" << endl;
}    else {

cout << n << endl;
}
}

The code above prints -65538 in console.

Solution 5 - C++

Here's a simple and working method I found elsewhere:

string hexString = "7FF";
int hexNumber;
sscanf(hexString.c_str(), "%x", &hexNumber);

Please note that you might prefer using unsigned long integer/long integer, to receive the value. Another note, the c_str() function just converts the std::string to const char* .

So if you have a const char* ready, just go ahead with using that variable name directly, as shown below [I am also showing the usage of the unsigned long variable for a larger hex number. Do not confuse it with the case of having const char* instead of string]:

const char *hexString = "7FFEA5"; //Just to show the conversion of a bigger hex number
unsigned long hexNumber; //In case your hex number is going to be sufficiently big.
sscanf(hexString, "%x", &hexNumber);

This works just perfectly fine (provided you use appropriate data types per your need).

Solution 6 - C++

I had the same problem today, here's how I solved it so I could keep lexical_cast<>

typedef unsigned int    uint32;
typedef signed int      int32;

class uint32_from_hex   // For use with boost::lexical_cast
{
    uint32 value;
public:
    operator uint32() const { return value; }
    friend std::istream& operator>>( std::istream& in, uint32_from_hex& outValue )
    {
        in >> std::hex >> outValue.value;
    }
};

class int32_from_hex   // For use with boost::lexical_cast
{
    uint32 value;
public:
    operator int32() const { return static_cast<int32>( value ); }
    friend std::istream& operator>>( std::istream& in, int32_from_hex& outValue )
    {
        in >> std::hex >> outvalue.value;
    }
};

uint32 material0 = lexical_cast<uint32_from_hex>( "0x4ad" );
uint32 material1 = lexical_cast<uint32_from_hex>( "4ad" );
uint32 material2 = lexical_cast<uint32>( "1197" );

int32 materialX = lexical_cast<int32_from_hex>( "0xfffefffe" );
int32 materialY = lexical_cast<int32_from_hex>( "fffefffe" );
// etc...

(Found this page when I was looking for a less sucky way :-)

Cheers, A.

Solution 7 - C++

This worked for me:

string string_test = "80123456";
unsigned long x;
signed long val;

std::stringstream ss;
ss << std::hex << string_test;
ss >> x;
// ss >> val;  // if I try this val = 0
val = (signed long)x;  // However, if I cast the unsigned result I get val = 0x80123456

Solution 8 - C++

Try this. This solution is a bit risky. There are no checks. The string must only have hex values and the string length must match the return type size. But no need for extra headers.

char hextob(char ch)
{
    if (ch >= '0' && ch <= '9') return ch - '0';
    if (ch >= 'A' && ch <= 'F') return ch - 'A' + 10;
    if (ch >= 'a' && ch <= 'f') return ch - 'a' + 10;
    return 0;
}
template<typename T>
T hextot(char* hex)
{
    T value = 0;
    for (size_t i = 0; i < sizeof(T)*2; ++i)
        value |= hextob(hex[i]) << (8*sizeof(T)-4*(i+1));
    return value;
};

Usage:

int main()
{
    char str[4] = {'f','f','f','f'};
    std::cout << hextot<int16_t>(str)  << "\n";
}

Note: the length of the string must be divisible by 2

Solution 9 - C++

just use stoi/stol/stoll for example:

std::cout << std::stol("fffefffe", nullptr, 16) << std::endl;

output: 4294901758

Solution 10 - C++

For those looking to convert number base for unsigned numbers, it is pretty trivial to do yourself in both C/C++ with minimal dependency (only operator not provided by the language itself is pow() function).

In mathematical terms, a positive ordinal number d in base b with n number of digits can be converted to base 10 using:

Example: Converting base 16 number 00f looks like:

= 0*16^2 + 0*16^1 + 16*16^0 = 15

C/C++ Example:

#include <math.h>

unsigned int to_base10(char *d_str, int len, int base)
{
    if (len < 1) {
        return 0;
    }
    char d = d_str[0];
    // chars 0-9 = 48-57, chars a-f = 97-102
    int val = (d > 57) ? d - ('a' - 10) : d - '0';
    int result = val * pow(base, (len - 1));
    d_str++; // increment pointer
    return result + to_base10(d_str, len - 1, base);
}

int main(int argc, char const *argv[])
{
    char n[] = "00f"; // base 16 number of len = 3
    printf("%d\n", to_base10(n, 3, 16));
}

Content Type	Original Author	Original Content on Stackoverflow
Question	Clayton	View Question on Stackoverflow
Solution 1 - C++	Evan Teran	View Answer on Stackoverflow
Solution 2 - C++	anon	View Answer on Stackoverflow
Solution 3 - C++	Mike Lundy	View Answer on Stackoverflow
Solution 4 - C++	Kirill V. Lyadvinsky	View Answer on Stackoverflow
Solution 5 - C++	AamodG	View Answer on Stackoverflow
Solution 6 - C++	Andy J Buchanan	View Answer on Stackoverflow
Solution 7 - C++	mike	View Answer on Stackoverflow
Solution 8 - C++	Johannes Beutel	View Answer on Stackoverflow
Solution 9 - C++	Wang	View Answer on Stackoverflow
Solution 10 - C++	justinokamoto	View Answer on Stackoverflow