Plzz someone explain this

Question

#include <iostream>
using namespace std;

int main() {
int num = 1;
if(*(char*)&num==1)
cout<<"Little-endian"<<endl;
else
cout<<"Big-endian"<<endl;
return 0;
}

Answer 1

The code takes first byte of integer value of 1. If it is also 1 this means that left bytes are lower ones and it is little endian. If first byte of an int with value 1 is 0, this would mean left bytes are higher value (big endian)

Answer 2

&num - address of num
(char*) - cast type of to char pointer
(char*)&num - takes num address and interprets it as a char pointer(all pointers inside are 32bit values on x86)
*(char*)&num - returns the actual char value at address of num

For instance:
1 as int in memory looks like 1 bit and another 31 0 bits. If u take first byte of integer(which is 4 bytes BTW) u get 1 bit and 7 0bits, which is 1 char value on little endian.

On big endian 1 value integer is 31 0bits and 1 1bit. So if u take first byte of this int, u will get 8 0bits (0 value char)

Hope this clears everything

Answer 3

Max Charlington Can you explain the *(char*)&num==1 part step by step?

Answer 4

Max Charlington plzz explain the "if" part

Answer 5

Thank you Max Charlington