Either wiped off the drive, or right next to the new ones, I do not recommend keeping the old files.
Answer:
The code is not dereferencing the pointers. You have to place an asterisk in front of the pointer to read the value the pointer points to.
Explanation:
So "if (str1 != str2)" must be "if (*str1 != *str2)".
likewise:
while (*str1 != 0 && *str2 != 0)
and
result = (*str1 == *str2);
Answer:
Explanation:
1. Write a program that declares an array named alpha with 50 components of the type double. Initialize the array so that the first 25 components are equal to the square of the counter (or index) variable and the last 25 components are equal to three times the index variable.
double alpha[50];
for (int i=0;i<25;i++)
{
alpha[i]=i*i;
alpha[i+25]=(i+25)*3;
}
2. Output the array so that exactly ten elements per line are printed.
for (int i=0;i<50;i++)
{
cout<<i+1<<". "<<alpha[i]<<" ";
if (((i+1)%10)==0)
{
cout<<endl;
}
}
3. Run your program again, but this time change the code so that the array is filled with random numbers between 1 and 100.
double alpha[50];
for (int i=0;i<50;i++)
{
alpha[i]=rand()%101;
}
for (int i=0;i<50;i++)
{
cout<<i+1<<". "<<alpha[i]<<" ";
if (((i+1)%10)==0)
{
cout<<endl;
}
}
4. Write the code that computes and prints the average of elements of the array.
double alpha[50],temp=0;
for (int i=0;i<50;i++)
{
alpha[i]=rand()%101;
temp+=alpha[i];
}
cout<<"Average :"<<(temp/50);
5. Write the code that that prints out how many of the elements are EXACTLY equal to 100.
double alpha[50],temp=0;
for (int i=0;i<50;i++)
{
alpha[i]=rand()%101;
if(alpha[i]==100)
{
temp++;
}
}
cout<<"Elements Exacctly 100 :"<<temp;
Please note: If you put each of above code to the place below comment it will run perfectly after compiling
#include <iostream>
using namespace std;
int main()
{
// If you put each of above code here it will run perfectly after compiling
return 0;
}
Answer:
See the code below and the algorithm explanation on the figure.
Explanation:
The explanation in order to get the answer is given on the figure below.
Solving this problem with C. The program is given below:
#include <stdio.h>
int main(void) {
int n, Even=0, Odd=0, Zeros=0;
for (;;) {
printf("\nEnter the value the value that you want to check(remember just integers): ");
//IF we input a non-numeric character the code end;
if (scanf("%d", &n) != 1) break;
if (n == 0) {
Zeros++;
}
else {
if (n % 2) {
Even++;
}
else {
Odd++;
}
}
}
printf("for this case we have %d even, %d odd, and %d zero values.", Even, Odd, Zeros);
return 0;
}
Answer:
miles_gallon = float(input("Enter car's miles/gallon: "))
dollars_gallon = float(input("Enter gas dollars/gallon: "))
print("Gas cost for 20 miles is $", (20 / miles_gallon) * dollars_gallon)
print("Gas cost for 75 miles is $", (75 / miles_gallon) * dollars_gallon)
print("Gas cost for 500 miles is $", (500 / miles_gallon) * dollars_gallon)
Explanation:
*The code is in Python.
Ask the user to enter the car's miles/gallon and gas dollars/gallon
Calculate the gas cost for 20 miles, divide 20 by miles_gallon and multiply the result by dollars_gallon, then print it
Calculate the gas cost for 75 miles, divide 75 by miles_gallon and multiply the result by dollars_gallon, then print it
Calculate the gas cost for 500 miles, divide 500 by miles_gallon and multiply the result by dollars_gallon, then print it
