Bitwise Operations

Given a number in hex, 0x1234
to extract an individual digit, you need to do bitwise manipulations.

WLOG, extract the third lowest digit (third from the right):

int x = 0x1234;
int third = (x >> (2*4)) & 0xf;
assert (third == 0x2);

The ">>" shift operator moves the number to the right by the specified number of bits. So:

   01001000110100b          0x1234
>>              8        >> 0x0008
------------------       ---------
           010010b          0x0012

The "&" integer bitwise operator masks the number. So:

  00010010b           0x0012
& 00001111b         & 0x000f
-----------         --------
  00000010b           0x0002

Pointer Example

C	Gold Assembly	Gold Machine	Notes
int a = -1;		1000: ffffffff	&a = 1000
int b = -1;		1004: ffffffff	&b = 1004
int* p = &a;		2000: 00001000	&p = 2000
void foo() {
*p = 5;	ld $0x5, r0 ld $0x2000, r1 ld 0x0(r1), r2 st r0, 0x0(r2)	0000 0000 0005 0100 0000 2000 1012 3002	r0 ~ *p r1 = &p r2 = p *p = *p
p = &c;	ld $0x1004, r2 st r2, 0x0(r1)	0200 0000 1004 3201	r2 = &c p = &c
*p = 8;	ld $0x8, r0 st r0, 0x0(r2)	0000 0000 0008 3002	r0 ~ *p *p = *p
}

TODO: insert slide show

Tracing Machine Execution

Short Example

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Store Example

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Mapping C to Assembly to Machine Language

Short Example

C	Gold Assembly	Gold Machine	Notes
void foo() {
int a = 0;	ld $0x0, r1	0100 0000 0000	r1 ~ a
a++;	inc r1	6301
a *= 2;	add r1, r1	6111
}

Store Example

C	Gold Assembly	Gold Machine	Notes
int a = 2;		1000: 0000 0002	&a = 1000
void foo() {
	ld $0x1000, r0 ld 0x0(r0), r1	0: 0000 0000 1000 1001	r0 = &a r1 = a
a++;	inc r1	6301
a *= 2;	add r1, r1	6111
	st r1, 0x0(r0)	3001	a = r1
}

Array Example

C	Gold Assembly	Gold Machine	Notes
int[] a = {0,1,2,3};		1000: 0000 0000 0000 0001 0000 0002 0000 0003	&a[] = 1000
int b = -1;		2000: ffff ffff	&b = 2000
void sum() {
b = 0;	ld $0x0, r2	0: 0200 0000 0000	r2 ~ b
	ld $0x1000, r0	0000 0000 1000	r0 = &a
b += a[0];	ld 0x0(r0), r1 add r1, r2	1001 6112
b += a[1];	ld 0x4(r0), r1 add r1, r2	1101 6112
b += a[2];	ld 0x8(r0), r1 add r1, r2	1201 6112
b += a[3];	ld 0xc(r0), r1 add r1, r2	1301 6112
	ld $0x2000, r0 st r2, 0x0(r0)	0000 0000 2000 3002	r0 = &b b = r2
}