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kolbaska11 [484]
3 years ago
5

What is one pass of a coding sequence called?​

Computers and Technology
2 answers:
Oksi-84 [34.3K]3 years ago
8 0

Answer:

In computer programming, a one-pass compiler is a compiler that passes through the parts of each compilation unit only once, immediately translating each part into its final machine code. ... This refers to the logical functioning of the compiler, not to the actual reading of the source file once only.

Explanation:

Maru [420]3 years ago
6 0

Answer:

an iteration

Explanation:

don't be like the guy above me nobody wants a song and a dane they just want the answers my guy

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Write a program that: Takes the list lotsOfNumbers and uses a loop to find the sum of all of the odd numbers in the list (hint:
Alika [10]

Answer:

Following are the code to this question:

#include <iostream>//defining header file

using namespace std;

int main()//defining main method

{

int x[]={2,3,4,6,7,8,9,1,11,12};//defining 1-D array and assign value

int i,sum=0;//defining integer variable

for(i=0;i<10;i++)//defining loop for count value

{

   if(x[i]%2==1)//defining if block to check odd value

   {

       sum=sum+x[i];//add value in sum variable

   }

}

cout<<sum;//print sum

return 0;

}

Output:

31

Explanation:

In the above-given program, an integer array "x" is declared that holds some integer values, and in the next line two integer variable "i and sum" is defined which is used to calculate the value.

In the next line, a for loop is declared, that counts all array value, and it uses the if block to check the odd value and add all the value into the sum variable.

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3 years ago
2. Assuming that the bits in an integer are numbered from 0, from right (least significant) to left (most significant), write an
3241004551 [841]

Answer:

Explanation:

Two ways of interpreting this question

(I) Shifting all the bits along with the choosen four,

in that case it will be i1>>2;

(ii) Just shifting the choosen four bits and keeping the remaining as it is.

It is equivalent to replacing the first four bits of i1 with i1>>2. let us call N = i1, and M = i1>>2.

max = ~0 //All one's

//1's through position 4, then all zeros

left = max - ((1<<4) - 1);

So we just have zeros between bits one to four and other are one.

Now final step;

(N&left) | m

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