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4 years ago

In the context of the box model, what is the difference between a margin and padding

Computers and Technology

1 answer:

Andrej [43]4 years ago

3 0

Answer:

It is good to know about the differences between margin and padding .Margin is the outer space of an element, while padding is the inner space of an element. In other words, margin is the space outside of an element's border, while padding is the space inside of its border. You can set auto value to margin.

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Hdhdhrhdh you are not the intended recipient. xmp-qite-xsy.

Aleks04 [339]

Answer:

ummmm... okay

Explanation:

3 0

3 years ago

Programs for embedded devices are often written in assembly language. Some embedded processors have limited instructions, like M

oksano4ka [1.4K]

Answer:

#include <iostream>

using namespace std;

int main()

{

char address[2];

int tag, firstBit, secondBit, setNumber;

int cache[4][2]={{1,5}, {2,4}, {3,2}, {6,0}};

cout << "Enter the address as hex(in small letters: "<

cin >> address;

for (int i < 0; i < 8; i++){

if (address[0] == '0'){

tag = 0;

firstBit = 0;

} else if (address[0] == '1'){

tag = 0;

firstBit = 1;

} else if (address[0] == '2'){

tag =1;

firstBit = 0;

} else if (address[0] == '3'){

tag = 1;

firstBit = 1;

} else if (address[0] == '4'){

tag = 2;

firstBit = 0;

} else if (address[0] == '5'){

tag = 2;

firstBit = 1;

} else if (address[0] == '6'){

tag = 3;

firstBit = 0;

} else if (address[0] == '7'){

tag = 3;

firstBit = 1;

} else if (address[0] == '8'){

tag = 4;

firstBit = 0;

} else if (address[0] == '9'){

tag = 4;

firstBit = 1;

} else if (address[0] == 'A'){

tag = 5;

firstBit = 0;

} else if (address[0] == 'B'){

tag = 5;

firstBit = 1;

} else if (address[0] == 'C'){

tag = 6;

firstBit = 0;

} else if (address[0] == 'D'){

tag = 6;

firstBit = 1;

} else if (address[0] == 'E'){

tag = 7;

firstBit = 0;

} else if (address[0] == 'F'){

tag = 7;

firstBit = 1;

} else{

cout<<"The Hex number is not valid"<< endl;

}

if(address[1]>='0' && address[1]<'8'){

secondBit = 0;

} else if(address[1]=='8'|| address[1]=='9'||(address[1]>='a' && address[1]<='f')){

secondBit = 1;

} else{

cout<<"The Hex number is not valid"<< endl;

return 0;

}

setNumber = firstBit * 2 + secondBit;

if(cache[setNumber][0]==tag || cache[setNumber][1]==tag){

cout<<"There is a hit";

} else{

cout<< "There is a miss";

}

return 0;

}

Explanation:

The C++ source code prompts the user for an input for the address variable, then the nested if statement is used to assign the value of the firstBit value given the value in the first index in the address character array. Another if statement is used to assign the value for the secondBit and then the setNumber is calculated.

If the setNumber is equal to the tag bit, Then the hit message is printed but a miss message is printed if not.

4 0

3 years ago

Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth

telo118 [61]

Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.

public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}

So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.

4 0

4 years ago

Write a program to calculate the

marishachu [46]

Answer:

blep blep blep belp belp belp belp belp belp BELPPPPPPPPPP

5 0

3 years ago

Petra has an interview with an IT company. What technique can help prepare her?

SVEN [57.7K]

I would go C because that way it gets her prepared to answer any techie questions they would have for her and would make her a better option.

3 0

3 years ago

Read 2 more answers