Printing memory from variable, is something wrong?

**Subsonics** · 08-27-2009

Hey, I'm trying to print out the memory of a variable as binary. However I'm missing the signed bit for some reason. Can some one spot the problem, or is it something I'm missing about the signed bit lol.

Code:

#include <stdio.h>

int main()
{
	float test;
	int *memdump, i;

	printf("Enter value: ");
	scanf("%f", &test);
	printf("Entered:     %.0f\n", test);

	memdump = (int*)&test;
	
	printf("S EXP      MANTISSA\n");
	int binarray[32];
	for(i = 0; i < 32; i++){
		if(*memdump%2 == 0)
			binarray[31-i] = 0;
		else
			binarray[31-i] = 1;
		
		*memdump -= (*memdump%2);
		*memdump /= 2;
	}
	for(i = 0; i < 32; i++){
		printf("%d", binarray[i]);
		if(i == 0)
			printf(" ");
		if(i == 8)
			printf(" (1)");
	}
	printf("\n–––––––––––––––––––––––––––––––––––––\n");

	return 0;
}

Just to clarify, the signed bit is always zero.

**King Mir** · 08-27-2009

Use unsigned int instead of signed int. You get to 1 or -1 divided by 2, and both answers yield 0.

**Subsonics** · 08-27-2009

Aah, thank you. I will try that. :-)

**Subsonics** · 08-27-2009

Confirmed, it works! Thanks again.

**King Mir** · 08-27-2009

A few other points:
And you don't need to subtract (*memdump%2). that will get truncated when rounding anyway.
Consider using binarray[31-i] =*memdump%2.

**Subsonics** · 08-27-2009

Yes, that right, I did not think of that. It will make it more compact too. :-)

**Sebastiani** · 08-27-2009

Also, keep in mind that floats and ints aren't guaranteed to be the same size. Here's a fairly standard approach that works with any type of data:

Code:

#include <stdio.h>
#include <limits.h>

char* memory_to_binary_text_lsb_to_msb_order( void* dat, size_t siz, char* buf, size_t len )
{
    unsigned char
        msk,
        * seq = dat,
        * fin = seq + siz,
        * ptr = buf;
    if( siz == 0 || ( siz * CHAR_BIT + siz / CHAR_BIT ) >= len )
        return NULL;
    for( ;; )
    {
        for( msk = 0x1; msk; msk <<= 1 )
            *ptr++ = ( *seq & msk ) ? '1' : '0';
        if( ++seq == fin )
            break;
        *ptr++ = ' ';
    }
    *ptr = 0;
    return buf;        
}

int main( void )
{
    double
        val = exp( 1.0 );
    char
        buf[ 128 ];
    puts
    ( 
        memory_to_binary_text_lsb_to_msb_order
        ( 
            &val, 
            sizeof( val ), 
            buf, 
            sizeof( buf ) 
        ) 
    );
    return 0;
}

**Subsonics** · 08-28-2009

Hey, Sebastiani, thanks for the tip. I will digest, and learn. Had to leave the terminal for some sleep, lol.

**roaan** · 08-28-2009

I was having a doubt in the program above. When we do

memdump = (int*)&test;

it is putting the address of test into memdump. But when i was trying to execute the program and using it in debug mode memdump has the address of test say 0x0012ff60 and when i tried to figure out the value of *memdump i thought it should be 4 but the value is something very large 1082130432. I mean i input 4 into test so when i do *memdump if i am not wrong it implies value at address contained in memdump so shouldnt it be 4.

**King Mir** · 08-28-2009

If you input 4, the floating point value should be 4. But since your reading the float as an int, the number is different. Still the program in the first post (preferably fixed) will generate the number S=0 E=10 M=0.

**Subsonics** · 08-29-2009

Originally Posted by roaan

I was having a doubt in the program above. When we do

memdump = (int*)&test;

it is putting the address of test into memdump. But when i was trying to execute the program and using it in debug mode memdump has the address of test say 0x0012ff60 and when i tried to figure out the value of *memdump i thought it should be 4 but the value is something very large 1082130432. I mean i input 4 into test so when i do *memdump if i am not wrong it implies value at address contained in memdump so shouldnt it be 4.

That's true for an int, but the purpose for me was to be able to inspect how a floating point variable is represented internally.

**roaan** · 08-29-2009

i thought it should be 4 but the value is something very large 1082130432.

Do you mean to say the float 4 is stored as this large number when stored in an integer variable.

**Subsonics** · 08-29-2009

Originally Posted by roaan

i thought it should be 4 but the value is something very large 1082130432.

Do you mean to say the float 4 is stored as this large number when stored in an integer variable.

Yes, in a float you get the number represented as , signed bit, then a eight bit exponent, then a 23bit mantissa. If you print this as an int then it will be interpreted as a large number since an int have no idea about floating point representation.

**roaan** · 08-29-2009

Originally Posted by Subsonics

Yes, in a float you get the number represented as , signed bit, then a eight bit exponent, then a 23bit mantissa. If you print this as an int then it will be interpreted as a large number since an int have no idea about floating point representation.

Yes i did read about the representation of floats. But then just will there be no overflow problem with the integers if we are representing such a large value.

**Subsonics** · 08-29-2009

Originally Posted by roaan

Yes i did read about the representation of floats. But then just will there be no overflow problem with the integers if we are representing such a large value.

Both the int and the float are 32bits on my machine, so there should be no problem.

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