Thread: bit (or) operation,left shift

output is 12480
(0) is the problem i expect 16
i trace this code but i have problem at i=5

Code:

#include<stdio.h>
int main()
{
 
char c=48;// char c='0';
 
    int i, mask=1;
    for(i=1; i<=5; i++)
    {
        printf("%c",c|mask);
        
        mask = mask<<1;
    }
   
   
   
   
        system("pause");
    return 0;
}

c=0,mask=1
i=1
print 1
c=0,mask=2

i=2
print 2
c=0,mask=4

i=3
print 4
c=0,mask=8

i=4
print 8
c=0,mask=16

i=5
print (0 bit or 16)

2pown	128	64	32	16	8	4	2	1
bit id	7	6	5	4	3	2	1	0
c	0	0	0	0	0	0	0	0
mask	0	0	0	1	0	0	0	0
0 bitor 16	0	0	0	1	0	0	0	0

print 16 not 0 ???

c=0,mask=32

return 0 // end of main

i need to print char has value of 16(ASCII)

Code:

char c=16;
printf("%c",c);

out of this code is
ASCII code of 16 (data link escape)


all i expect is to get the ASCII of (char) but (not zero)

how the prog print "zero" instead ASCII code
thanks for your time

Originally Posted by Hassan Ahmed

how the prog print "zero" instead ASCII code

You start with c = 48, which is hex 30 or binary 00110000. Showing the numbers in binary, after i == 4, then mask = 00010000 and c | mask = 00110000 | 00010000 = 00110000 (no change since the bit is already set). The same thing happens for the next step where mask = 00100000.

i get it the output
ASCII Table and ASCII Code - Programming - Fclose Tutorials
ascii table

Code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
  char c=48;// char c='0';
  
    int i, mask=1;
    for(i=0; i<=7; i++)
    {
        char result=c|mask;
        printf("%c\n",result);
        printf("ascii code of c=%d\n",result);
        mask = mask<<1;
    }
  
  system("PAUSE");    
  return 0;
}

output

1
ascii code of c=49
2
ascii code of c=50
4
ascii code of c=52
8
ascii code of c=56
0
ascii code of c=48
0
ascii code of c=48
p
ascii code of c=112
░
ascii code of c=-80
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