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MrRissso [65]
3 years ago
6

Given positive integer numinsects, write a while loop that prints that number doubled without reaching 100. follow each number w

ith a space. after the loop, print a newline. ex: if numinsects

Computers and Technology
2 answers:
ELEN [110]3 years ago
8 0
Please provide the language you're using when you ask for programming help, otherwise you aren't going to get the answer that you are looking for.

Here it is in Java, and I'm assuming the number is given via user input? Otherwise, just remove the user input function and replace the integer with a value of your choice. Note, that this isn't the full code; only what is relevant to the question.


public static void main(String[] args) {
    int num = numInput(10);
    printDoubles(num, 100);  // You can create a user input function for
                                             // maxValue if you wanted to.
}

/**
 * Receives user input between 0 and the absolute value of maxInput.
 * @param maxInput The largest absolute value that can be input.
 */
private static int numInput(int maxInput) {
    Scanner sc = new Scanner(System.in);

    maxInput = Math.abs(maxInput);

    int num = 0;
    while (!(num > 0 && num <= maxInput)) {
        num = sc.nextInt();

        if (!(num > 0 && num <= maxInput)) {
            System.out.println("Input too small or too large");
        }
    }

    return num;
}

/**
 * Continues to print out num doubled until maxValue is reached.
 * @param num The number to be printed.
 * @param maxValue The maximum value (not including in which num can                                       be doubled to.
 */
private static void printDoubles(int num, int maxValue) {
    if (num >= maxValue) {
        System.out.println("No output.");
    }

    while (true) {
        if (num >= maxValue) {
            break;
        }

        if (num < maxValue) {
            System.out.print(num + " ");
        }

            num *= 2;
    }

    System.out.println();
}

fredd [130]3 years ago
3 0

Answer:

while (numInsects < 100) {

         System.out.print(numInsects + " ");

        numInsects = numInsects * 2;

     }

     System.out.println();

Explanation:

Changed it to 100 to fit with your question

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miskamm [114]

Answer:

Human capital

Explanation:

It means the economic value of workers experience and skills

7 0
2 years ago
What shortcut keys do i use to print something on my keyboard ?
romanna [79]
CTRL+P. It's usually pressed when you want to print something.
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Read 2 more answers
Jim, the IT director, is able to complete IT management task very well but is usually two weeks late in submitting results compa
oee [108]

Answer:Effective but not efficient

Explanation:

Jim is effective because he was able to complete the IT tasks well but he is not efficient because he didn't submit the result on time because being efficient includes management of time.

5 0
3 years ago
Write an efficient C++ function that takes any integer value i and returns 2^i ,as a long value. Your function should not multip
Ilya [14]

Answer:

long power(int i)

{

   return pow(2,i);

}

Explanation:

The above written function is in C++.It does not uses loop.It's return type is long.It uses the function pow that is present in the math library of the c++.It takes two arguments return the result as first argument raised to the power of second.

for ex:-

pow(3,2);

It means 3^2 and it will return 9.

4 0
3 years ago
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
3 years ago
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