How adding a hexadecimal value with integer work?

Question

In the below code ,variable 'b' holds value '133' which is quite straight forward. How about variable 'a' ? Why is it '131' ?

I see the only difference is '015' instead of '15'.

#include<stdio.h>
int main()
{
  int a,b,c;
  a=015 + 0x71 +5;
  printf("%d\n",a);  // prints '131'
  b=15 + 0x71 +5;
  printf("%d\n",b);  // prints '133'
}

Can someone let me know whats going on in here ?

**hexadecimal** = base 16 = `0x{9-9A-Fa-f_]+`, every "digit" represents 4 bits, **octal** = base 8 = `0[0-7_]+`, every digit represents 3 bits. 015 is octal. Hexadecimal has won in computer history. — Joop Eggen, Nov 27 '15 at 10:31

Chris Maes · Answer 1 · 2015-11-27T10:33:28.210

7

Leading zeros indicate that the number is expressed in octal. 015 in octal is 13 in decimal notation.

octal-literal is the digit zero (0) followed by zero or more octal digits (0, 1, 2, 3, 4, 5, 6, 7)

(from cppreference)

edited Nov 27 '15 at 10:33

answered Nov 27 '15 at 10:25

Chris Maes

35,025
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Mentioning `strtol` misguided in this context. Octal integer constants have nothing to do with said standard library function. – user694733 Nov 27 '15 at 10:31
@user694733 you're right, I adapted that. – Chris Maes Nov 27 '15 at 10:33

score 3 · Answer 2 · answered Nov 27 '15 at 10:27

3

015 is actually a indicator for octal number system representation, so

015 means 1x8^1 + 5*8^0 = 13

So

  a=015 + 0x71 +5;

Strictly speaking means (1*(8^1) + 5*(8^0) ) + (7*(16^1) + 1*(16^0)) + (5*(10^0)) = 133

answered Nov 27 '15 at 10:27

Magisch

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How adding a hexadecimal value with integer work?

2 Answers2