Answer:
words.hasNext()
Explanation:
Given the code snippet below:
- while (inputFile.hasNextLine()) {
- String word = "";
- String line = inputFile.nextLine();
- Scanner words = new Scanner(line);
- while (words.hasNext()) {
- word = words.next();
- }
- System.out.println(word); }
- }
We have a inputFile Scanner object that can read data from a text file and we presume the inputFile has read several rows of data from the text file. So long as there is another line of input data available, the outer while loop will keep running. In each outer loop, one line of data will be read and assign to line variable (Line 3). Next, there is another Scanner object, words, which will take the current line of data as input. To get the last word of that line, we can use hasNext() method. This method will always return true if there is another tokens in its input. So the inner while loop will keep running so long as there is a token in current line of data and assign the current token to word variable. The word will hold the last token of current line of data upon exit from the inner loop. Then we can print the output (Line 8) which is the last word of the current line of data.
Answer:
The student’s actual bed is 75 inches long
Explanation:
Here in this question, we want to find out the length of the student’s actual bed.
The ratio of the model to the real is 1:25
Let the actual length of the bed be x inches
Thus;
1/25 = 3/x
By cross multiplying
1 * x = 3 * 25
x = 75 inches
Answer:
Output:
123456
123456
123456
123456
123456
123456
Explanation:
C Code:
#include <stdio.h>
int main() {
int n,i,j;
printf("Gimme a decimal value to use as n:");
scanf("%d",&n);
for(i=0;i<n;i++){
for(j=1;j<=n;j++){
printf("%d",j);
}
printf("\n");
}
return 0;
}
Equivalent assembly program:
.LC0:
.string "Gimme a decimal value to use as n:"
.LC1:
.string "%d"
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
lea rax, [rbp-12]
mov rsi, rax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call scanf
mov DWORD PTR [rbp-4], 0
.L5:
mov eax, DWORD PTR [rbp-12]
cmp DWORD PTR [rbp-4], eax
jge .L2
mov DWORD PTR [rbp-8], 1
.L4:
mov eax, DWORD PTR [rbp-12]
cmp DWORD PTR [rbp-8], eax
jg .L3
mov eax, DWORD PTR [rbp-8]
mov esi, eax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call printf
add DWORD PTR [rbp-8], 1
jmp .L4
.L3:
mov edi, 10
call putchar
add DWORD PTR [rbp-4], 1
jmp .L5
.L2:
mov eax, 0
leave
ret
Input:
6
The program was first written with a c code, anf and subsequently translated to an assembly language.
The answer to the question above is the last option: ETHICAL. The IT code of conduct, which is a written guideline, helps in determining whether a specific action is acceptable or ETHICAL. Basically, the IT code of conduct's purpose here is to maintain that the actions being done to the accepted and proper practices.
Answer:
monitor fileSharer
{
enum {THINKING, WAITING, READING} state[N];
condition self[N];
int total;
void open(int i) {
state[i] = WAITING;
if (i + total >= N)
{ self[i].wait(); } // Leaves monitor
state[i] = READING;
total += i;
}
void close(int i) {
state[i] = THINKING;
total -= i;
// Can signal one waiting proc whose ID won't break bank.
for (int x = N - total - 1; x >= 0; x--) {
if (state[x] == WAITING) {
self[x].signal(); break;
}
}
}
initialization.code() {
for (int i = 0; i < N; i++) {
state[i] = THINKING;
}
total = 0;
}
}
