字符混乱中的转义序列

Question

I came across this example here .

#include <iostream>

int main()
{
    int i = 7;

    char* p = reinterpret_cast<char*>(&i);
    if(p[0] == '\x7') //POINT OF INTEREST
        std::cout << "This system is little-endian\n";
    else
        std::cout << "This system is big-endian\n";
}

What I'm confused about is the if statement. How do the escape sequences behave here? I get the same result with p[0] == '\\07' ( \\x being hexadecimal escape sequence). How would checking if p[0] == '\\x7' tell me if the system is little or big endian?

Answer 1

Hex 7 and octal 7 happens to be the same value, as is decimal 7.

The check is intended to try to determine if the value ends up in the first or last byte of the int .

A little endian system will store the bytes of the value in "reverse" order, with the lower part first

07 00 00 00

A big endian system would store the "big" end first

00 00 00 07

By reading the first byte, the code will see if the 7 ends up there, or not.

Answer 2

The layout of a (32-bit) integer in memory is;

Big endian:

+-----+-----+-----+-----+
|  0  |  0  |  0  |  7  |
+-----+-----+-----+-----+
   ^ pointer to int points here

Little endian:

+-----+-----+-----+-----+
|  7  |  0  |  0  |  0  |
+-----+-----+-----+-----+
   ^ pointer to int points here

What the code basically does is read the first char that the integer pointer points to, which in case of little endian is \\x0 , and big endian is \\x7 .

Answer 3

7 in decimal is the same as 7 in hexadecimal and 7 in octal, so it doesn't matter if you use '\\x7' , '\\07' , or even just 7 (numeric literal, not a character one).

As for the endianness test: the value of i is 7, meaning it will have the number 7 in its least significant byte, and 0 in all other bytes. The cast char* p = reinterpret_cast<char*>(&i); makes p point to the first byte in the representation of i . The test then checks whether that byte's value is 7. If so, it's the least significant byte, implying a little-endian system. If the value is not 7, it's not a little-endian system. The code assumes that it's a big-endian, although that's not strictly established (I believe there were exotic systems with some sort of mixed endianness as well, although the code will probably not run on such in practice).