# 程序代写 Name: – cscodehelp代写

Name:
ID#: X.500:
Nearby is a small C pro- gram which makes use of arrays, pointers, and func- tion calls. Fill in the tables associated with the ap- proximate memory layout of the running program at each position indicated. Assume the stack grows to lower memory addresses and that the sizes of C variable types correspond to common 64-bit systems.
Problem 1 (15 pts):
POSITION A SOLUTION
|——–+——–+———+——-|
| Frame | Symbol | Address | Value |
|——–+——–+———+——-|
#include
void flub(double *ap, double *bp){
|
|——–+——–+———+——-|
POSITION B SOLUTION
|——–+——–+———+——-|
| Frame | Symbol | Address | Value |
|——–+——–+———+——-|
1
2
3
4
5 6}
7 // POSITION B
8 return;
int c = 7;
if(*ap < c){ *ap = bp[1]; 9 } | 10 int main(){ | 11 double x = 4.5; 12 double arr[2] = {3.5, 5.5}; 13 double *ptr = arr+1; 14 // POSITION A 15 flub(&x, arr); 16 printf("%.1f ",x); 17 for(int i=0; i<2; i++){ 18 printf("%.1f ",arr[i]); 19 } 20 return 0; 21 } | arr[0] | #3048 Problem 2 (10 pts): Fill in the following ta- ble of equivalent ways to write these 8 bit quan- tities. There are a to- tal of 9 blanks to fill in and the first column indi- cates which blanks occur in which lines. Assume two’s complement encod- ing for the signed decimal column. |----------+-----------+------+-------+----------+---------| | SOLUTION | | | | Unsigned | Signed | | Blank #s | Binary | Hex | Octal | Decimal | Decimal | |----------+-----------+------+-------+----------+---------| CS 2021: Practice Exam 1 SOLUTION Fall 2021 University of Minnesota Exam period: 20 minutes Points available: 50 | main() | | | | x | #3064 | arr[1] | #3056 | arr[0] | #3048 | ptr | #3040 |i |#3036 | 4.5 | | 5.5 | | 3.5 | | #3056 | | ?| | main() | x | #3064 | | arr[1] | #3056 | 5.5 | | 5.5 | | 3.5 | | #3056 | | ptr | #3040 | |i|#3036|?| |--------+--------+---------+-------| | flub | ap | #3028 | #3064 | | | bp | #3020 | #3048 | | |c|#3016|7| |--------+--------+---------+-------| NOTES - Both Pos A and B are before i is assigned 0 so i remains undefined |||| |#1#2#3|00011011|0x1B| |||| |#4#5#6|10100101|0xA5| |~x+1 |01011011| | |#7#8#9|11000111|0xC7| |~x+1 |00111001| | |----------+-----------+------+-------+----------+---------| NOTES - Octal shows leading 0 which is not strictly necessary - Typical twos’ complement conversion technique show below binary representation: invert bits and add 1 | 0033 | | 0245 | | 0307 | | | 27| | 165 | | 199 | | | 27 | | -91 | | -57 | | 1/ 2 WRITE ON EVERY PAGE – Name: SOLUTION Backgound: Write a short C code fragment (1-5 lines) using a C I/O function call to accomplish the stated task. Assume in each case there is a variable FILE *fh which has been opened appropriately for the I/O operation. Also assume that any variables mentioned have already been declared. Problem 3 (5 pts): Read three text floating point values from fh formatted as standard decimal point values into double variables r,u,t. Check if the read fails due to reaching the end of file; if so print the message None left. // SOLUTION int res = fscanf(fh, "%lf %lf %lf", &r, &u, &t); if(res == EOF){ printf("None left "); } Problem 4 (5 pts): Write 8 characters from the beginning of array str to fh in binary format. // SOLUTION fwrite(str, sizeof(char), 8, fh); #include
#include
// Struct to count positive/negative
// numbers in arrays.
typedef struct {
int poss, negs;
} pn_t;
pn_t *get_pn(int *arr, int len);
// Allocates a pn_t and initializes
// its field to zero. Then scans array
// arr increment poss for every 0 or
// positive value and negs for every
// negative value. Returns the pn_t
// with poss/negs fields set. If arr
// is NULL or len is less than 0,
// returns NULL.
int main(){
int arr1[5] = {3, 0, -1, 7, -4};
pn_t *pn1 = get_pn(arr1, 5);
// pn1: {.poss=3, .negs=2}
free(pn1);
int arr2[3] = {-1, -2, -4};
pn_t *pn2 = get_pn(arr2, 3);
// pn2: {.poss=0, .negs=3}
free(pn2);
int *arr3 = NULL;
pn_t *pn3 = get_pn(arr3, -1);
// pn3: NULL
return 0; }
Problem 5 (15 pts): Nearby is a main() function demon- strating the use of the function get pn(). Implement this func- tion according to the documentation given. My solution is about 12 lines plus some closing curly braces.
// SOLUTION
pn_t *get_pn(int *arr, int len){
1
2
3
4
5}
6 pn_t *pn = malloc(sizeof(pn_t)); 7 pn->negs = 0;
8 pn->poss = 0;
9 for(int i=0; inegs++;
12 }
13 else{
14 pn->poss++; 15 }
16 }
17 return pn;
18 }
if(arr==NULL || len < 0){ return NULL; 2/ 2