Viewing variable memory...

[xabhi]

hi, i wrote the following program to print out the contents of the memory of a variable. to handle different types, i have used macros definition combined with a void pointer...

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

#define type_scan(type, type_specifier)	 a_number = (type *)malloc(sizeof(type));\
					 scanf(#type_specifier, (type *)a_number);\
					 segment = sizeof(type) * 8;

int main()
{
	void *a_number;
	int segment;

	int choice;
	unsigned char *byte;
	char memory[1024];
	int i, j;

	printf("Enter type (i - int, f - float, l - long, u - unsigned int, d - double) : ");
	choice = getchar();

	printf("Enter any number: ");
	switch(choice) {
	case 'i':
		type_scan(int, %d);
		break;
	case 'f':
		type_scan(float, %f);
		break;
	case 'l':
		type_scan(long, %D);
		break;
	case 'u':
		type_scan(unsigned int, %u);
		break;
	case 'd':
		type_scan(double, %g);
		break;
	}

	j = i = 0;
	byte = (unsigned char *)a_number;
	
	while(j < segment) {
		for(i = 0; i < 8; i++) {
			memory[j++] = ((*byte & (1 << i)) > 0) ? '1' : '0';
		}
		byte++;
	}
	
	while(j--) { 
		printf("%c", memory[j]);
		if((j % 8) == 0) printf(" ");
	}

	printf("\n");
	exit(0);
}

the output works for int type, for eg,

Enter type (i - int, f - float, l - long, u - unsigned int, d - double) : i
Enter any number: 120
00000000 00000000 00000000 01111000

but if a double type input is given, it shows all the high order four bytes (double being 8 bytes long) = 0 irrespective of input.

why?? and is this even right way to print out the memory??

[Sebastiani]

The 'scanf' format string for double is "%lf". Also, creating a macro that references variables in 'main' is a bad practice, IMO.

[xabhi]

thanks...

[Elysia]

You're leaking memory...