Assignment 4, due Sep 19

Assignment: The backspace, tab, vertical tab, linefeed and carriage return characters are the most important format effectors among the ASCII control characters. What do they do when you output them to the terminal window under C? What do they do when you output them from a Hawk program? Give your answer as a short sentence for each of these telling what it does and if the behavior is different or the same in the Hawk and C environments. (1.0 points, 0.2 per character)

Suggestion: You can design a single string that replaces "Hello world" and allows you to see the effect of each of these with a single experiment. You do not need to show your work, but if you show us your test strings for C and the Hawk environments, we can diagnose your errors.

Using this test string for C "BS\bBS HT\tHT VT\vVT LF\nLF CR\rCR\n" I got this output:

BBS HT	HT VT
             VT LF
CR CR

Translating the string to SMAL Hawk format is a bit ugly because you have to change, for example, "BS\bBS" to "BS",BS,"BS", but in the end, you get exactly the same output, so the following conclusion can be drawn for both programming environments:

• BS goes back one space so BS\bBS becomes BBS
• HT outputs spaces until the current column number is divisible by 8.
• VT goes down one line without changing the column number.
• LF goes down one line and back to the left margin.
• CR goes back to the left margin to begin overwriting the current line.

000000: 000230F0
000004: 30F00200
000008: FA00FFFA

A Problem: Give equivalent Hawk code, with labels such as A, B and C used instead of numeric memory addresses. First, try it by hand, then use the emulator to check your work, and figure out what your errors were. (0.6 points, 0.1 point per instruction or operand error)

A:      JUMP    C
B:      BR      D
C:      JUMP    B
D:      BR      A

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
int main() {
    int i;
    for (i = 0; i < 100; i++) {
	void * p = malloc(i);
	printf( "%lX\n", (intptr_t)p );
    }
    return 0;
}

This program just prints out the addresses of 100 blocks of memory. The %lX in the format string asks it to print a long integer in hexadecimal, and the (intptr_t) cast on the pointer forces the pointer to be passed to printf as an integer, using an integer representation large enough to hold one pointer.

a) When you call malloc on the divisional Linux machines, what alignment guarantee does it give you about the pointer it returns? That is, does it return an arbitrary address, an address divisible by 2, divisible by 4, or what? (0.7 points)

b) When you call MALLOC on the Hawk, what alignment guarantee does it give you about the pointer it returns? (0.7 points)

Suggestion: Instead of printing out 100 addresses and then trying to find a pattern, consider oring all of the addresses into a single value, then look at how many zero bits are on the bottom of the result after 100 allocations of various sizes of blocks. In Python, Java, C and C++, the a|b operator will or two values of any integer type; on the Hawk, the OR machine instruction does this for any two 32-bit registers.

If you use this suggestion, you can show your work by showing the value you got and your interpretation of that value.

#include 
#include 
#include 
int main() {
    int i;
    intptr_t acc = 0;
    for (i = 0; i < 100; i++) {
        void * p = malloc(i);
        acc = acc | (intptr_t)p;
    }
    printf( "%lX\n", acc );
    return 0;
}

[dwjones@fastx07 ~]$ a.out
13D1FF0
[dwjones@fastx07 ~]$ a.out
1A73FF0
[dwjones@fastx07 ~]$ a.out
1B9DFF0
[dwjones@fastx07 ~]$ a.out
4D1FF0
[dwjones@fastx07 ~]$ a.out
1107FF0