convert word to binary

This is a discussion on convert word to binary within the C Programming forums, part of the General Programming Boards category; How can I convert a word such as 'A' to the number represented in ASCII ? Another question is how ...

  1. #1
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    convert word to binary

    How can I convert a word such as 'A' to the number represented in ASCII ? Another question is how can i convert a decimal number to binary number? Is any function in c allows us to do that?

    Thanx for any help

  2. #2
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    int num;
    num = (int) 'A';

    To convert a int to a binary string. One way is function itoa(), if your compiler has it.
    Code:
    #include <stdlib.h>
    int main(void)
    {
       int num = 255;
       char str[33];
       itoa(num,str,2);
    }
    And here's a clever function to print in binary that someone posted:
    Code:
    void printbits( int x )
    {
       int y;
       for( y = 31; y >= 0; y-- ) printf("%d",(x&(1<<y))?1:0);
          printf("\n");
    }

  3. #3
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    will this also work?
    int i;
    i='A';
    printf("%d",i)//prints ASCII value of 'A'

  4. #4
    Registered User Nutshell's Avatar
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    yes

  5. #5
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    >will this also work?
    Yes

  6. #6
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    big thanx for you guys!!
    now another question is how can I mask those binary #
    for ex:

    111000101
    if i want shift right to 3, it will become
    101111000,
    just like what shr does in assembly language, but how can i do that in c?
    is any function allows us to do that?
    Last edited by Supra; 03-29-2002 at 05:57 PM.

  7. #7
    Confused Magos's Avatar
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    X>>Y (Shifts X's bits Y steps to the right)
    X<<Y (Shifts X's bits Y steps to the left)
    MagosX.com

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    Teach a man to fish and you feed him for a lifetime.

  8. #8
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    asm
    {
    DW BINARY NUMBER = "=x"
    }

    As long as you can find a way of replacing x with the number, that works...

  9. #9
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    really big thanks to you all!!
    my last two questions are

    1. how can i "and" or "or" those binary in c#?

    ex:
    Code:
                1100110
       (and)    0011100
    ----------------------
                0000100
    2. how can i convert binary # back to decimal #?
    or where can i find the tutorial web site about them?

    Thanx again

  10. #10
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    >1. how can i "and" or "or" those binary in c#?

    Well in C:

    & - binary AND
    | - binary OR

    I don't know about C#, but I guess it's the same as in C.

    >2. how can i convert binary # back to decimal #?
    >or where can i find the tutorial web site about them?

    Binary number

    bn ... b3 b2 b1 b0

    Decimal

    2^0 b0 + 2^1 b1 + 2^2 b2 + ... + 2^n bn

  11. #11
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    Not again!!!

    Hereīs my code; itīs my first real C-program:

    Code:
    #include <stdio.h>
    #include <string.h>
    
    #define PUFFER 256
    
    
    void asctobin(void); /* converts ASCII-characters into binary digits */
    void bintoasc(void); /* converts binary digits into ASCII characters */
    
    unsigned char bin[]={1,2,4,8,16,32,64,128};
    
    int main(void)
    {
    	char chose;
    
    	for(chose=0;chose<40;chose++,printf("\n")) ; /* clear screen */
    
    again:
    	printf("Enter \'A\' to convert ASCII to binary or \'B\' to convert binary into ASCII: ");
    	scanf("%c",&chose);
    	if(chose=='A' || chose=='a') asctobin();
    	else
    	{
    		if(chose=='B' || chose=='b') bintoasc();
    		else{
    			rewind(stdin); /* back to the begin of stdin; if more then one character was entered */
    			goto again;
    		} /* only accepts 'A','a','B' and 'b' as inputs */
    	}
    
    	return 0;
    }
    
    
    void asctobin(void)
    {
            char string[MAX],c;
    	int count,len;
    
    	printf("Enter string (max length: 255 characters): ");
    	rewind(stdin); /* back to the begin of stdin */
    	fgets(string,PUFFER-1,stdin);
    
    	len=strlen(string);
    
    	for(;len>0;len--) /* as often as the value for the length of the inputed string */
    	{
    		c=*string; /* the character string points to at the moment */
    		for(count=7;count>=0;count--)
    		{        
    			if(c>=bin[count])
    			{
    				printf("1");
    				c-=bin[count];
    			}else
    			printf("0");
    		}
    		*string++;
    		printf(" ");
    	}
    }
    
    
    void bintoasc(void)
    {
    	char string[MAX],c;
    	int count=7,len;
    
    
    	printf("Enter digits (max 255):");
    	rewind(stdin); /* back to the begin of stdin */
    	fgets(string,PUFFER-1,stdin);
    
    	len=strlen(string);
    
    	for(;len>0;len--) /* as often as the value for the length of the inputed string */
    	{
    		if(*string=='1')
    		{
    			c+=bin[count], --count;
    		}else 
    		if(*string=='0') --count;
    
    		if(count==0)
    		{
    			printf("%c",c);
    			count=7;
    			c=0;
    		}
    
    		*string++;
    	}
    }
    klausi
    Last edited by klausi; 04-01-2002 at 10:50 AM.
    When I close my eyes nobody can see me...

  12. #12
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    thanx you all very much, but i still have questions about masking
    number
    I wrote a small program:
    Code:
    #include <stdlib.h>
    int main(void)
    {
       unsigned char i='a';
       unsigned char b='c';
       char str[33];
       i='a'+'c';
       printf("%d\n",i);
       itoa(i,str,2);
       printf("%s\n", str);
       i << 3;
       printf("%d\n",i);  //still can't shift
       i & 3;
       printf("%d\n",i); // still can't and
       return 0;
    }
    I can't mask the number
    cna anyone tell me what's wrong with it?
    really appreciate

  13. #13
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    Guest
    Originally posted by Supra
    i << 3;
    i & 3;
    I can't mask the number
    cna anyone tell me what's wrong with it?
    really appreciate
    That is the same as typing this somewhere in your code:

    1+6;

    As you see, it does nothing. You have to store your result somewhere:

    i=(i << 3); and i=(i & 3);

    Or in an even simplier way:

    i <<= 3; and i &= 3;

    Good luck!

  14. #14
    Confused Magos's Avatar
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    That was me replying above, for some reason you're not logged in if you enter this site from Hotmail.

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