Converting 4 bytes in little endian order into an unsigned integer

Question

I have a string of 256*4 bytes of data. These 256* 4 bytes need to be converted into 256 unsigned integers. The order in which they come is little endian, i.e. the first four bytes in the string are the little endian representation of the first integer, the next 4 bytes are the little endian representation of the next integer, and so on.

What is the best way to parse through this data and merge these bytes into unsigned integers? I know I have to use bitshift operators but I don't know in what way.

Answer 1

Hope this helps you

unsigned int arr[256];
char ch[256*4] = "your string";
for(int i = 0,k=0;i<256*4;i+=4,k++)
{
arr[k] = ch[i]|ch[i+1]<<8|ch[i+2]<<16|ch[i+3]<<24;
}

Answer 2

Alternatively, we can use C/C++ casting to interpret a char buffer as an array of unsigned int. This can help get away with shifting and endianness dependency.

#include <stdio.h>
int main()
{
    char buf[256*4] = "abcd";
    unsigned int *p_int = ( unsigned int * )buf;
    unsigned short idx = 0;
    unsigned int val = 0;
    for( idx = 0; idx < 256; idx++ )
    {
        val = *p_int++;
        printf( "idx = %d, val = %d \n", idx, val );
    }
}

This would print out 256 values, the first one is idx = 0, val = 1684234849 (and all remaining numbers = 0).

As a side note, "abcd" converts to 1684234849 because it's run on X86 (Little Endian), in which "abcd" is 0x64636261 (with 'a' is 0x61, and 'd' is 0x64 - in Little Endian, the LSB is in the smallest address). So 0x64636261 = 1684234849.

Note also, if using C++, reinterpret_cast should be used in this case:

const char *p_buf = "abcd";
const unsigned int *p_int = reinterpret_cast< const unsigned int * >( p_buf );