Question

make an array of 3 rows and 5 columns with integers. choose the integers by hand. Print out the original array with the integers. Next print out the array in 1) hexadecimal and then in 2) ASCII characters.

Answer 1

Please find the code below:

#include <iostream>

using namespace std;

int main()

{

int array[3][5] = {

{61,62,63,64,65},

{66,67,68,69,70},

{91,92,93,94,95}};

int row=3;

int column = 5;

cout<<"Original array content"<<endl;

for(int i=0;i<row;i++){

for(int j=0;j<column;j++){

cout<<array[i][j]<<" ";

}

cout<<endl;

}

cout<<endl;

cout<<"Array in hexadecimal content"<<endl;

for(int i=0;i<row;i++){

for(int j=0;j<column;j++){

cout<<hex<<array[i][j]<<" ";

}

cout<<endl;

}

cout<<endl;

cout<<"Array in ACSII character"<<endl;

for(int i=0;i<row;i++){

for(int j=0;j<column;j++){

char ascii = array[i][j];

cout<<ascii<<" ";

}

cout<<endl;

}

return 0;

}

output:

c code::::

#include <stdio.h>

int main()

{

int array[3][5] = {

{61,62,63,64,65},

{66,67,68,69,70},

{91,92,93,94,95}};

int row=3;

int column = 5;

printf("Original array content\n");

int i,j;
for(i=0;i<row;i++){

for(j=0;j<column;j++){

printf("%d ",array[i][j]);

}

printf("\n");
}

printf("\n");

printf("Array in hexadecimal content\n");

for( i=0;i<row;i++){

for(j=0;j<column;j++){

printf("%x ",array[i][j]);

}

printf("\n");

}

printf("\n");

printf("Array in ACSII character\n");

for(i=0;i<row;i++){

for(j=0;j<column;j++){

char ascii = array[i][j];

printf("%c ",ascii);

}

printf("\n");

}

return 0;

}

output: