Generate test cases for the following code by using basic path testing. a) Draw CFG. c) d) int test_prime() { How many basic paths for the CFG? List the basic paths. Generate test cases from it. int x=1,n,i; } scanf("%d",&n); if(n==2) else printf("prime"); for(i=2;i
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- Analyze errorIf (x < 100) && (x> 10)System.out.println(“x” is between 10 and 100”);HelpExplain the code (Do not use flowchart). See attached photo for the problem. It doesn't have to be long as long as you explain the important parts of the code int removeHead() { if(!isEmpty()) { return remove_between(head->next); } return 0; } int removeTail() { if(!isEmpty()) { return remove_between(tail->prev); } return 0; } int add(int num) { addTail(num); return index; } int remove(int num) { node* currnode = head->next; int pos = 0; while(currnode != tail) { pos++; if(currnode->element == num) { remove_between(currnode); return pos; } else { currnode = currnode->next; } } return 0; }…
- Bum 0; for (int i = 1; iQuestion 8 Consider the following code where n and m can be any number of more than 20. Select the correct time complexity. int iter_count = 0; for (int i = 0; i < n; i *= 2) { for (int j = 0; j < m; j++){ iter_count += 1; } for (int i = }) 0; iExplain what the code does step by step.briefly comment the code#include <stdio.h>#include <stdlib.h>/* ADJACENCY MATRIX int source,V,E,time,visited[20],G[20][20];void DFS(int i){int j;visited[i]=1;printf(" %d->",i+1);for(j=0;j<V;j++){if(G[i][j]==1&&visited[j]==0)DFS(j);}}int main(){int i,j,v1,v2;printf("\t\t\tGraphs\n");printf("Enter the no of edges:");scanf("%d",&E);printf("Enter the no of vertices:");scanf("%d",&V);for(i=0;i<V;i++){for(j=0;j<V;j++)G[i][j]=0;}/* creating edges :P */for(i=0;i<E;i++){printf("Enter the edges (format: V1 V2) : ");scanf("%d%d",&v1,&v2);G[v1-1][v2-1]=1; } for(i=0;i<V;i++){for(j=0;j<V;j++)printf(" %d ",G[i][j]);printf("\n");}printf("Enter the source: ");scanf("%d",&source);DFS(source-1);return 0;}A piece of programm code is given:int A[1200];...int find_in_arr(int * A, int x){ int found = 0;for(i = 0; i<1200; i++) if(A[i] == x) { found++; break } if ( found == 0) return 0;return 1;}- please optimize the code to run faster by using a barrier#include int main(void) { int a[3][3] = {1,3,5,7,9}; int i,j,sum=0; for (i = 0; i < 3; i++) for (j = 0; j < 3; j++) sum += a[i-i][j-j]; %3D printf("%d\n",sum); return 0; }Any time you design code with an artificial limit someonewill exceed it. For example, consider the code:void err(const char* const fmt, int a = 0, int b = 0, int c = 0, int d = 0, int e = 0){ fprintf(stderr, "Fatal Error:\n"); fprintf(stderr, fmt, a, b, c, d, e); fprintf(stderr, "\n"); abort();}Now this works if you wish to write simple messages:err(“Size parameter (%d) out of range”, size);But what happens when we wish to a slightly more complex call?err(“Point (%d,%d) outside of box (%d,%d), (%d,%d)”, point.x, point.y, box.x1, box.y1, box.x2, box.y2);Our function can take a format and up to five parameters. We just gave itsix. It's not going to work.Now we could fix the err function to add another parameter, but thatwould only work until we needed seven parameter. Another change would beneeded at eight and so on.Q9. Deleting the last item in a doubly circular linked list takes ......... time compared to deleting the first item in it. More O Less O Same O None of themRecommended textbooks for youComputer Networking: A Top-Down Approach (7th Edi…Computer EngineeringISBN:9780133594140Author:James Kurose, Keith RossPublisher:PEARSON
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