Question

I’m working on a C programming assignment, I need to simulate the operation of a 3 bit decoder. My compiler is complaining, this Wikipedia article gives a list of C operators, but my code still seems not to be working.

Wikipedia: http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Bitwise_operators

Compiler error:

logic.c: In function ‘three_bit_decoder’:
logic.c:33: warning: statement with no effect
logic.c:33: error: expected ‘;’ before ‘~’ token
logic.c:34: warning: statement with no effect
logic.c:34: error: expected ‘;’ before ‘b0’
logic.c:35: warning: statement with no effect
logic.c:35: error: expected ‘;’ before ‘~’ token
logic.c:36: warning: statement with no effect
logic.c:36: error: expected ‘;’ before ‘b0’
logic.c:38: warning: statement with no effect
logic.c:38: error: expected ‘;’ before ‘b0’
logic.c:39: warning: statement with no effect
logic.c:39: error: expected ‘;’ before ‘b0’
logic.c:40: warning: statement with no effect
logic.c:40: error: expected ‘;’ before ‘~’ token
logic.c:41: warning: statement with no effect
logic.c:41: error: expected ‘;’ before ‘b0’
logic.c:43: warning: return makes integer from pointer without a cast
logic.c:43: warning: function returns address of local variable
make: *** [logic.o] Error 1

Code:

int three_bit_decoder(BIT b0, BIT b1, BIT b2) {
    /* Returns an 8 bit number, by 2^3
     * 
     */
        int myIntDecoder[8]; 

        myIntDecoder[0] ~b0 + ~b1 + ~b2;
        myIntDecoder[1] b0 + ~b1 + ~b2;
        myIntDecoder[2] ~b0 + b1 + ~b2;
        myIntDecoder[3] b0 + b1 + ~b2;

        myIntDecoder[4] b0 + ~b1 + ~b2;
        myIntDecoder[5] b0 + ~b1 + b2;
        myIntDecoder[6] ~b0 + b1 + b2;
        myIntDecoder[7] b0 + b1 + b2;

        return myIntDecoder;
}

Answer 1

If those are assignment statements you’re missing equal signs.

myIntDecoder[0] = ~b0 + ~b1 + ~b2;
myIntDecoder[1] = b0 + ~b1 + ~b2;
myIntDecoder[2] = ~b0 + b1 + ~b2;
myIntDecoder[3] = b0 + b1 + ~b2;

myIntDecoder[4] = b0 + ~b1 + ~b2;
myIntDecoder[5] = b0 + ~b1 + b2;
myIntDecoder[6] = ~b0 + b1 + b2;
myIntDecoder[7] = b0 + b1 + b2;

The next problem you’ll encounter is that int myIntDecoder[8] declares an array of 8 ints, which is not the same as an 8-bit int. A char is 8 bits wide on (almost) all platforms; or you can be more explicit and use one of the standard typedefs:

#include <stdint.h>

uint8_t myIntDecoder;

So as not to leave you hanging, I should mention that assigning to individual bits in a variable is not as simple as byte[5] = 1. Doing it requires some clever use of shifting and other bitwise operations.