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C/C++ check if one bit is set in, i.e. int variable

Asked 2009-02-07T13:10:46.953
135
int temp = 0x5E; // in binary 0b1011110.

Is there such a way to check if bit 3 in temp is 1 or 0 without bit shifting and masking.

Just want to know if there is some built in function for this, or am I forced to write one myself.

c++ c bit-manipulation
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2 Answers

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According to this description of bit-fields, there is a method for defining and accessing fields directly. The example in this entry goes:

struct preferences {
    unsigned int likes_ice_cream : 1;
    unsigned int plays_golf : 1;
    unsigned int watches_tv : 1;
    unsigned int reads_books : 1;
}; 

struct preferences fred;

fred.likes_ice_cream = 1;
fred.plays_golf = 1;
fred.watches_tv = 1;
fred.reads_books = 0;

if (fred.likes_ice_cream == 1)
    /* ... */

Also, there is a warning there:

However, bit members in structs have practical drawbacks. First, the ordering of bits in memory is architecture dependent and memory padding rules varies from compiler to compiler. In addition, many popular compilers generate inefficient code for reading and writing bit members, and there are potentially severe thread safety issues relating to bit fields (especially on multiprocessor systems) due to the fact that most machines cannot manipulate arbitrary sets of bits in memory, but must instead load and store whole words.

answered 2009-02-07T13:20:45.997
0

if you just want a real hard coded way:

 #define IS_BIT3_SET(var) ( ((var) & 0x04) == 0x04 )

note this hw dependent and assumes this bit order 7654 3210 and var is 8 bit.

#include "stdafx.h"
#define IS_BIT3_SET(var) ( ((var) & 0x04) == 0x04 )
int _tmain(int argc, _TCHAR* argv[])
{
    int temp =0x5E;
    printf(" %d \n", IS_BIT3_SET(temp));
    temp = 0x00;
    printf(" %d \n", IS_BIT3_SET(temp));
    temp = 0x04;
    printf(" %d \n", IS_BIT3_SET(temp));
    temp = 0xfb;
    printf(" %d \n", IS_BIT3_SET(temp));
    scanf("waitng %d",&temp);

    return 0;
}

Results in:

1 0 1 0

answered 2009-02-07T14:30:59.380

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