Arrays 231
Explanation In the above program two fo r loops are used. The inner fo r loop generates four binary
bits for each hexadecimal number and assigns it to array b in [ ] [ ]. The 0 to F hexadecimal numbers
are taken from the outer fo r loop. If the bit is 1 counter % c ' is increased. The value of c is checked for
even or odd condition. If c is odd the parity b it% 0' is displayed otherwise ' 1' .The output is displayed.
7.36 Write a progarm to convert binary to gray codes.
# include <stdio.h>
# include <conio.h>
void mpinQ
{
int bin[16] [4] ,x,y,asO,ceO;
clrscrO ;
for (x*0;x<16/x++)
{
y*x;
for (a=ยป0;a<4;a++)
{
bln[x][a]=y%2;
y-y/2;
}
}
c lrs c rO ;
printf (" n Binary Bits Gray B its");
printf C n ========== ===========");
fo r (x~0;x<16;x++)
I
printf (" n ");
for(y=3,-y>=0,-y~)
printf ("% 3d",bin[x]ly]);
printf ("% 5 d %2d %2d %2d",bin[x][3], bin [x][3]*bin[x][2 ],bin[x][2 ]Ab in [x ][l],b in [x ][l]Abm [x][0 ]);
OUTPUT:
Binary Bits Gray Bits
0000 0000
0001 0001
0010 0011
0011 0010
0100 0110
0101 0111
0110 0101
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.