C: Ampersand In Front Of A Number - Stack Overflow

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Learn more about Labs C: Ampersand in front of a number Ask Question Asked 12 years, 8 months ago Modified 12 years, 8 months ago Viewed 2k times 0

What does this line (x = n & 5;) mean in the code below? As far as I know ampersand is used for pointers. I was not expecting this code to compiled, but it compiled and ran fine. The results I got is

0,1,0,1,4,5,4,5,0,1,

#include <stdio.h> int main(void){ int x, n; for (n = 0; n < 10; n++){ x = n & 5; printf("%d,", x); } printf("\n"); return 0; }

• c

Share Improve this question Follow edited Apr 5, 2012 at 0:08 Jesse Good 52.3k1414 gold badges129129 silver badges173173 bronze badges asked Apr 5, 2012 at 0:06 lambalamba 1,64155 gold badges1919 silver badges2929 bronze badges 3

• I compiled using gcc -W -Wall -pedantic -std=c89 ctest.c – lamba Commented Apr 5, 2012 at 0:07
• 3 It's an operator. Open your introductory C++ textbook and read it until you're familiar with the language basics. – Kerrek SB Commented Apr 5, 2012 at 0:08
• Sorry, I had a long day. Can't believe I forgot about it. I even used it before. – lamba Commented Apr 5, 2012 at 0:10

Add a comment |

4 Answers 4

Sorted by: Reset to default Highest score (default) Trending (recent votes count more) Date modified (newest first) Date created (oldest first) 6

In this case it's bitwise AND.

x = n & 5;

will AND 5 (which is 0b101) with whatever is in n.

AND works on the bits that represent the values. The result will have a 1 if both values have a 1 in that position, 0 otherwise.

Since we're ANDing with 5, and there are only 4 values you can make with the two bits in 0b101, the only possible values for x are 1, 4, 5 and 0. Here's an illustration:

n & 5 = x 1 (0b0001) & 0b0101 = 1 (0b0001) 2 (0b0010) & 0b0101 = 0 (0b0000) 3 (0b0011) & 0b0101 = 1 (0b0001) 4 (0b0100) & 0b0101 = 4 (0b0100) 5 (0b0101) & 0b0101 = 5 (0b0101) 6 (0b0110) & 0b0101 = 4 (0b0100) 7 (0b0111) & 0b0101 = 4 (0b0100) 8 (0b1000) & 0b0101 = 0 (0b0000) 9 (0b1001) & 0b0101 = 1 (0b0001) Share Improve this answer Follow edited Apr 5, 2012 at 0:15 answered Apr 5, 2012 at 0:07 Timothy JonesTimothy Jones 22.1k66 gold badges6464 silver badges9494 bronze badges Add a comment | 5

That's the "bitwise and" operator. Normally, it takes two integers, and returns an integer that has only the bits set that are in both of it's parameters.

base10 base2 6 0110 3 0011 6&3 0010 (=2)

There's also "bitwise or" | which sets bits that either one has set.

base10 base2 6 0110 3 0011 6|3 0111 (=7)

and there's "bitwise xor" ^ which sets bits that are different.

base10 base2 6 0110 3 0011 6^3 0101 (=5) Share Improve this answer Follow answered Apr 5, 2012 at 0:12 Mooing DuckMooing Duck 66.6k1919 gold badges103103 silver badges165165 bronze badges 3

• I think you mean 6&3, 6|3 and 6^3 – torrential coding Commented Apr 5, 2012 at 0:14
• @torrentialcoding: Thanks. I started with 5 and 2, but then switched to more interesting numbers – Mooing Duck Commented Apr 5, 2012 at 0:16
• Right. 6 and 3 better illustrate the bit operations. – torrential coding Commented Apr 5, 2012 at 0:19

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It's a bitwise AND operator. The value of n is being ANDed with 5.

Share Improve this answer Follow answered Apr 5, 2012 at 0:07 torrential codingtorrential coding 1,76522 gold badges2424 silver badges3434 bronze badges 1

• 2 It's bitwise, not logical operator. – Cat Plus Plus Commented Apr 5, 2012 at 0:07

Add a comment | 2

You were already told that in this case it's a binary and. But this requires a bit more explanation. In C and other C-like languages there are two & operators. One is unary the other binary.

A unary operator acts on a single value alone, while a binary operator takes in two values. In C binary operators take precedence over unary ones. If you write

int b; int *a = &b;

Then b is the only value the operator can work on. If you write however

int c, d; int d = c & d;

then the operator has two values to work with and the binary interpretation takes precedence over the unary interpretation. Note that this does not only apply to the & operator, but also its counterpart *

int *f; int h = *f;

But also

int i,j; int k = i * j;

Like with all operators, precedence can be overridden with parentheses:

int l, *m; int n = l * (*m); Share Improve this answer Follow answered Apr 5, 2012 at 0:15 datenwolfdatenwolf 162k1313 gold badges190190 silver badges310310 bronze badges Add a comment |

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