Explain Quiz II No4.pdf

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val avg_ptr int numel max min std_ptr avg_ptr avg_ptr min max int numel int standard diviation

Explain Quiz II  4 
Main Function 
No. 
Code 
1 
//Finding a min max mean sd 
2 
#include<stdio.h> 
3 
#include<math.h> 
4 
void Find_Min_Max_Avg_std(int vals[], int numels, int *min_ptr, int *max_ptr, double 
*avg_ptr, double *std_ptr); 
5 
main() 
6 
{ 
7 
 
int temp,i; 
8 
 
int data[100]; 
9 
 
int min, max; 
10 
 
double avg, std; 
11 
 
i=0; 
12 
 
//collect a data 
13 
 
for(;;) 
14 
 
{ 
15 
 
 
scanf("%d", &temp); 
16 
 
 
if(temp!=-1) 
17 
 
 
{ 
18 
 
 
 
data[i]=temp; 
19 
 
 
 
i++; 
20 
 
 
} 
21 
 
 
else 
22 
 
 
{ 
23 
 
 
 
break; 
24 
 
 
} 
25 
 
} 
26 
 
//send to function 
27 
 
Find_Min_Max_Avg_std(data , i, &min, &max, &avg, &std); 
28 
 
printf("\nYou entered %d numbers \n", i); 
29 
 
printf("The maximum value is %d \n", max); 
30 
 
printf("The minimum value is %d \n", min); 
31 
 
printf("The average is %lf \n", avg); 
32 
 
printf("The standard diviation is %lf \n ", std); 
33 
} 
 
 
 
 
Find_Min_Max_Avg_std Function 
No. 
Code 
34 
void Find_Min_Max_Avg_std(int vals[], int numels, int *min_ptr, int *max_ptr, double 
*avg_ptr, double *std_ptr) 
35 
{ 
36 
 
int i, j; 
37 
 
int temp; 
38 
 
double temp_sd; 
39 
 
*avg_ptr =0; 
40 
 
*std_ptr =0; 
41 
 
//sorted find min max 
42 
 
for(i=0;i<numels;i++) 
43 
 
{ 
44 
 
 
for(j=0;j<numels;j++) 
45 
 
 
{ 
46 
 
 
 
if(vals[i]<vals[j]) 
47 
 
 
 
{ 
48 
 
 
 
 
temp = vals[i]; 
49 
 
 
 
 
vals[i] = vals[j]; 
50 
 
 
 
 
vals[j] = temp; 
51 
 
 
 
} 
52 
 
 
} 
53 
 
} 
54 
 
//then min max 
55 
 
 
*min_ptr = vals[0]; 
56 
 
 
*max_ptr = vals[numels-1]; 
57 
 
//then avgerage and standard diviation 
58 
 
//find average 
59 
 
for(i=0;i<numels;i++) 
60 
 
{ 
61 
 
 
*avg_ptr = *avg_ptr+vals[i]; 
62 
 
} 
63 
 
*avg_ptr = *avg_ptr/numels; 
64 
 
//standard diviation 
65 
 
for(i=0;i<numels;i++) 
66 
 
{ 
67 
 
 
temp_sd = pow((vals[i]-*avg_ptr),2); 
68 
 
 
*std_ptr = *std_ptr + temp_sd; 
69 
 
} 
70 
 
*std_ptr = sqrt(*std_ptr/numels); 
71 
} 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 16-24  #	




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 array index 
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• ก	
 39  *avg_ptr =0; +
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 40  *std_ptr =0; +
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• ก	
 42-56 ก" Bubble Sort Algorithm ก/
ก

//
ก ก1
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o *min_ptr = vals[0];  
 min = val[0]; * *min_ptr "1 min 
o *max_ptr = vals[numel-1]; " numel – 1 * array 
 index  0 
 max = val[numel-1]; * *max_ptr "1 max 
 
 
• ก	
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ก vals[] 2/"

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o 	
 61 *avg_ptr = *avg_ptr+vals[i]; ก
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กก loop 
o 	
 63 *avg_ptr = *avg_ptr/numels; ก
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• ก	
 65-70 ก
 standard diviation 2/

/
ก vals[] 2/"

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 array vals #	 
o 	
 67 temp_sd = pow((vals[i]-*avg_ptr),2); ก)(ก	0#	

  
 
o 	
 68 กก
 67 # loop 
 
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