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Alla [95]
3 years ago
9

Explain: The decode part of fetch decode execute cycle

Computers and Technology
2 answers:
zhenek [66]3 years ago
8 0

Answer: Picture below

Explanation:

egoroff_w [7]3 years ago
5 0

Answer:

Image below

Explanation:

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2 years ago
Write a function input_poly(ptr1) that reads in two polynomials at a time with coefficient and exponent from a file cp7_in.txt.
Tanya [424]

Answer:

try this

Explanation:

#include<stdio.h>

#include<malloc.h>

typedef struct poly{ double coff;

int pow;

struct poly *link; }SL;

void create(SL **head)

{

*head=NULL;

}

//Insertion  

void ins_beg(SL ** head, double c, int pw)

{

SL *ptr;

ptr=(SL *)malloc(sizeof(SL));

ptr->coff=c;

ptr->pow=pw;

ptr->link=*head;

*head=ptr;

}

void trav(SL **head)

{

SL *ptr;

ptr=*head;

printf("\n\t Cofficient Exponent\n");

while(ptr!=NULL)

{ printf("\t\t%3.2f \t%d\n",ptr->coff, ptr->pow);

ptr=ptr->link;

}

}

//addition of polynomial............

void mult(SL *pl1, SL *pl2, SL **res)

{

SL trav1, trav2, ptr, new1,*loc;

int x,y;

trav1=*pl1;

while(trav1!=NULL)

{

trav2=*pl2;

while(trav2!=NULL)

{

x=(trav1->coff)*(trav2->coff);

y=(trav1->pow)+(trav2->pow);

//printf("\t%d\t%d\n",x,y);

if(*res==NULL)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=x;

new1->pow=y;

new1->link=NULL;

*res=new1;

}

else

{

ptr=*res;

while((y<ptr->pow)&&(ptr->link!=NULL))

{

loc=ptr;

ptr=ptr->link;

}

if(y==ptr->pow)

ptr->coff=ptr->coff+x;

else

{

if(ptr->link!=NULL)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=x;

new1->pow=y;

loc->link=new1;

new1->link=ptr;

}

else

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=x;

new1->pow=y;

ptr->link=new1;

new1->link=NULL;

}

}

}

trav2=trav2->link;

}

trav1=trav1->link;

}

}

//...addition..............

void add(SL *pl1, SL *pl2)

{

SL trav1, ptr, *new1,*loc; trav1=*pl1;

while(trav1!=NULL)

{

loc=ptr=*pl2;

while((trav1->pow<ptr->pow)&&(ptr->link!=NULL))

{

loc=ptr; ptr=ptr->link;

}

if(loc==ptr)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=trav1->coff;

new1->pow=trav1->pow;

new1->link=*pl2;

*pl2=new1;

}

else

if(trav1->pow==ptr->pow)

{

ptr->coff=ptr->coff+trav1->coff;

//ptr->pow=ptr->pow+trav1->pow;

}

else

if(ptr->link!=NULL)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=trav1->coff;

new1->pow=trav1->pow;

new1->link=ptr;

loc->link=new1;

}

else

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=trav1->coff;

new1->pow=trav1->pow;

new1->link=NULL;

ptr->link=new1;

}

trav1=trav1->link;

}

}

int main()

{

SL first, sec, *res;

create(&first);

ins_beg(&first,10.25,0);

ins_beg(&first,4,1);

ins_beg(&first,5,2);

ins_beg(&first,4,4);

printf("\nFirst Polinomial:\n");

trav(&first);

create(&sec);

ins_beg(&sec,11,0);

ins_beg(&sec,6,1);

ins_beg(&sec,4,2);

ins_beg(&sec,3,3);

printf("\nSecond Polinomial:\n");

trav(&sec);

create(&res);

printf("\nMultiplication of two Polinomial:\n");

mult(&first,&sec,&res);

trav(&res);

printf("\nAddition of two Polinomial:\n");

add(&first,&sec);

trav(&sec);

//trav(&first);

return 0;

}

6 0
3 years ago
Discuss the different categories of computer hardware​
madam [21]

Answer:

Input devices: For raw data input.

Processing devices: To process raw data instructions into information.

Output devices: To disseminate data and information.

Storage devices: For data and information retention.

7 0
2 years ago
Read 2 more answers
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