b. Consider the extract from a C program below, explain what would happen to the other processes on the system if this code is run: i. On a pre-emptive multitasking operating system il. On a co-operative multitasking operating system double h,w; char *p - ...; // Points to a block of memory to store the data for(h - 0; h < 600; h++) double y - (h • (2.0 / 600.0)) - 1.0; for (w - 0; w < 800; w++) double x - (w• (3.0 / 800.0)) - 1.5; int it - 4096; double r- 0, i - 0, r2 - 0, 12 - 0; do tmp - r2 - 12 + x; i - 2*r i+ y: r- tmp: r2 -r* r} 12 - i• i; } while((r2 + 12) <- 4.0 66 (--it) 1- 0); if(it - 0) *p++- 0; else *p++- 255;
b. Consider the extract from a C program below, explain what would happen to the other processes on the system if this code is run: i. On a pre-emptive multitasking operating system il. On a co-operative multitasking operating system double h,w; char *p - ...; // Points to a block of memory to store the data for(h - 0; h < 600; h++) double y - (h • (2.0 / 600.0)) - 1.0; for (w - 0; w < 800; w++) double x - (w• (3.0 / 800.0)) - 1.5; int it - 4096; double r- 0, i - 0, r2 - 0, 12 - 0; do tmp - r2 - 12 + x; i - 2*r i+ y: r- tmp: r2 -r* r} 12 - i• i; } while((r2 + 12) <- 4.0 66 (--it) 1- 0); if(it - 0) *p++- 0; else *p++- 255;
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Transcribed Image Text:b. Consider the extract from a C program below, explain what would happen to the other
processes on the system if this code is run:
i. On a pre-emptive multitasking operating system
ii. On a co-operative multitasking operating system
double h,w;
char *p = ...; // Points to a block of memory to store the data
for (h - 0; h < 600; h++)
{
double y = (h * (2.0 / 600.0)) - 1.0;
for (w - 0; w < 800; wt+)
double x = (w * (3.0 / 800.0)) - 1.5;
int it - 4096;
double r = 0, i - 0, r2 - 0, 12 - 0;
do
tmp - r2 - i2 + x;
i - 2 *r * i+ y;
r - tmp;
r2 -r* r;
12 - i* i;
} while( (r2 + i2) <- 4.0 66 (--it) !- 0);
if(it 0)
*p++ - 0;
else
*p++ - 255:
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