Example 01 1. Assume int x[5] = {0, 1, 2, 3, 4} What are the values of x after calling: swap(x+1, x+4)? void swap(int *pa, int *pb) { int t; t=*pa; *pa=*pb; *pb=t; }
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![Example 01
1.
Assume int x[5] = {0, 1, 2, 3, 4} What are the values of x after calling: swap(x+1, x+4)?
void swap(int *pa, int *pb)
{ int t;
t=*pa; *pa=*pb; *pb=t;
}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff15fa3fd-ab30-4521-bb90-f014debd02b6%2F4d75ef8b-2119-4162-bef4-027c6cd2d224%2Fhl7dtkk_processed.png&w=3840&q=75)
Step by step
Solved in 3 steps with 1 images
- 2- Let a = [123; 2 3 4; 12 5], b = [ 2133 points ; 50 -2; 2 3 -1] prod = a* b. What is the value of prod?Consider the following code: int main() { char *s = strdup("This is a test"); char *d = " "; while (s) { char *token = strsep(&s, d); puts(token); } free(s); return 0;} Which of the following causes an error or segmentation fault? A. free is called on a pointer that was not malloc'ed B. strsep must be called with s instead of &s C. The while loop is an infinite loop D. The string s points to is in read-only memory E. Something else*68. What is the return value of f (p, p), if the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value. int f(int &x, int c) { c = c 1; if X = X + 1; return f(x, c) } (a) 3024 (c==0) return 1; * X; (b) 6561 (c) 55440 (d) 161051
- // add.ll define void @add(i32* %ptr1, i32* %ptr2, i32* %val) {ret void} Fill add.ll function to do the following operation: void add(int *ptr1, int *ptr2, int *val) { *ptr1 += *val; *ptr2 += *val; } This is the full question. It is related to LLVM. If it's going to help there is one more code given which is: #include <stdio.h> void add(int *ptr1, int *ptr2, int *val); int main(int argc, char **argv) {FILE *f = fopen(argv[1], "r");int a, b, c;fscanf(f, "%d %d %d", &a, &b, &c);add(&a, &b, &c);printf("%d %d\n", a, b);fclose(f); return 0;}Topical Information Use C++. This lab will help you practice with dynamic memory (NOT mixed with classes). Program Information One statistic of interest to many researchers is the median of a set of data. The median is the value in the middle. (This is NOT the average, mind you.) Being in the middle, of course implies that the data are in order (sorted). If there are an odd number of data items, the middle is simple to find. If there are an even number of data items, you are supposed to take the average of the two middle values. You'll need to find the median of a set of data from a file. You don't know how many data are in the file, of course. You'll have to count the data, then read them in to sort, then find the median, and finally print it (the median) out. Do NOT use any more memory than is necessary!!Q1/ find code optimization of this code then find type to this code. for (i=1;iC++int func(int a, int b) { return (aint* p: int a[3]{1, 2, 3}; p = a; What is the value of *(p+2)?DO NOT COPY FROM OTHER WEBSITES Correct and Detailed answer will be Upvoted else downvoted. Thank you!//Assignment 06 */public static void main[](String[] args) { String pass= "ICS 111"; System.out.printIn(valPassword(pass));} /* public static boolean valPassword(String password){ if(password.length() > 6) { if(checkPass(password) { return true; } else { return false; } }else System.out.print("Too small"); return false;} public static boolean checkPass (String password){ boolean hasNum=false; boolean hasCap = false; boolean hasLow = false; char c; for(int i = 0; i < password.length(); i++) { c = password.charAt(1); if(Character.isDigit(c)); { hasNum = true; } else if(Character.isUpperCase(c)) { hasCap = true; } else if(Character.isLowerCase(c)) { hasLow = true; } } return true; { return false; } }#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill Education
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