int array regarding binary digits

Question

I am wondering why the results turn out to be 100, 8, 1 in the output. 010 doesn't even make any sense that it is not 2 in binary digits. Am I missing something?

#include "stdafx.h"
#include <iostream>

using namespace std;

int main()
{
    int code[3];
    code[0] = 100;
    code[1] = 010;
    code[2] = 001;
    printf("%d\n", code[0]);
    printf("%d\n", code[1]);
    printf("%d\n", code[2]);
}

Preceding a number with a 0 makes it an octal constant: https://msdn.microsoft.com/en-us/library/00a1awxf.aspx (010 (base 8) = 8 (base 10)) — jonsca, Jun 27 '18 at 02:45
It is standard -- so it will work the same regardless of OS or compiler. — David C. Rankin, Jun 27 '18 at 04:12

score 1 · Answer 1 · answered Jun 27 '18 at 03:58

1

100, 010 and 001 are not binary literals. 100 is a decimal literal; 010 and 001 are octal (base 8) literals. If you are using gcc there is an extension to support binary literals by using the prefix 0b, as in

code[0] = 0b100; // evaluates to 4

answered Jun 27 '18 at 03:58

meat

609
3
8

is it an extension? or is it fully-supported as specified in the [standard](https://timsong-cpp.github.io/cppwp/lex.literal#lex.icon-1)? – Joseph D. Jun 27 '18 at 04:14
It is an extension, e.g. [6.63 Binary Constants using the ‘0b’ Prefix](https://gcc.gnu.org/onlinedocs/gcc/Binary-constants.html) (but I get your point `:)` And see: [syntax - Why doesn't C have binary literals?](https://stackoverflow.com/questions/18244726/why-doesnt-c-have-binary-literals) – David C. Rankin Jun 27 '18 at 04:15
@DavidC.Rankin, sorry I thought this is a C++ question? – Joseph D. Jun 27 '18 at 04:19
1

@codekaizer - crud -- I'm getting old and senile -- it is a C++ question, and you are correct, but note from C++14 forward according to [cppreference.com - integer literal](https://en.cppreference.com/w/cpp/language/integer_literal) – David C. Rankin Jun 27 '18 at 05:05

int array regarding binary digits

1 Answers1