Solution for Exercise 05_03-sizes.c

05_03-sizes.c

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

int main()
{
	double *d;

	/* allocate storage for 100 double values */
	d = malloc( sizeof(double) * 100 );
	/* always test the result */
	if( d==NULL )
	{
		fprintf(stderr,"Unable to allocate memory\n");
		exit(1);
	}

	/* output data */
	printf("The buffer's address is %p\n",d);
	printf("The buffer's size is %zu bytes\n",sizeof(double)*100 );

	/* clean-up */
	free(d);
	return 0;
}

Output

The buffer's address is 0x55bbbce832a0
The buffer's size is 800 bytes

Notes

* Based on the output, a double on this system occupies 8 bytes of storage.

* Yes, the address you see when you run the program will doubtless be different from what's shown above.

* Pointer variable d is declared as a double. It's assigned an address from the malloc() function equal to the size of a double multiplied to 100. This quantity ensures enough storage for 100 double values.

* An if test confirms that the buffer is properly allocated. If not, the program exits.

* The two printf() statements output the results as requested in the book: The buffer's address and the buffer's size.

* The %p placeholder is used to output an address.

* The %zu placeholder is used to output a size_t value, the number of bytes in a buffer.

* The buffer's size is calculated by using the same expression found the malloc() statement: sizeof(double)*100. the result is the exact number of bytes required to store 100 double values in memory.

* As usual, your solution need not look exactly like mine.

Exercise 05_09-yourname.c →