将char字节转换为整数值

Question

For example , 130ABF (Hexadecimal) is equals to 1247935 (Decimal), So my byte array is

 char buf[3] = {0x13 , 0x0A , 0xBF};

and I need to retrieve the decimal value from the byte array.

Below are my sample code:

#include<iostream>
using namespace std;
int main()
{
    char buf[3] = {0x13 , 0x0A , 0xBF};
    int number = buf[0]*0x10000 + buf[1]*0x100 + buf[2];
    cout<<number<<endl;
    return 0;
}

and the result is : (Wrong)

1247679

Unless I change the

char buf[3] = {0x13 , 0x0A , 0xBF};

to

int buf[3] = {0x13 , 0x0A , 0xBF};

then It will get correct result.

Unfortunately, I must set my array as char type, anyone know how to solve this ?

Answer 1

Define the array as:

unsigned char buf[3];

Remember that char could be signed.

UPDATE: In order to complete the answer, it is interesting to add that "char" is a type that could be equivalent to "signed char" or "unsigned char", but it is not determined by the standard.

Answer 2

Array elements will be promouted to int before evaluating. So if your compiler treats char as signed you get next (assuming int is 32-bit):

int number = 19*0x10000 + 10*0x100 + (-65);

To avoid such effect you can declare your array as unsigned char arr[] , or use masking plus shifts:

int number = ((buf[0] << 16) & 0xff0000)
           | ((buf[1] <<  8) & 0x00ff00)
           | ((buf[2] <<  0) & 0x0000ff;

Answer 3

Since your char array is signed, when you want to initialize the last element ( 0xBF ), you are trying to assign 191 to it while the max it can store is 127: a narrowing conversion occurs... A workaround would be the following:

unsigned char[3] = { 0x13, 0x0A, 0xBF };

This will prevent the narrowing conversion. Your compiler should have given you a warning about it.