Homework & Coursework Questions

#include<stdio.h>
#include<conio.h>
main()
{
   int i,j,a=0,b=0,c=0,f[10],t[10][10],al[10][10],ta[10][10];
   int a1[10][10], max[10][10], n[10][10], n1,p,k=0;
   printf(“\n enter no.of resources”);
   scanf(“%d”,n1);
   printf(“\nenter the max no .of resources for each type”);

  for(i=0;i<n1;i++)
 {
    scanf(“%d”,&t[b][i]);
  }
  printf(“\nenter no .of process”);
  scanf(“%d”,&p);
  printf(“\nenter allocation resources”);

  for(i=0;i<p;i++)
  {
    f[i]=0;
    for(j=0;j<n1;j++)
    {
       scanf(“%d”,&a1[i][j]);
     }
   }

 for(i=0;i<p;i++)
 {
    for(j=0;j<n1;j++)
   {
      if(a1[i][j]<=t[b][j])
      {
         t[b][j]+=a1[i][j];
        continue;
       }
     else
    printf(“\n wrong resources allocation”);
 }
    printf(“\n chance of deadlock occurrence after allocation”);
  for(j=0;j<n1;j++)
  {
     printf(“%d”,a1[b][j]);
   }
  printf(“\n enter the max resources for every process”);
for(i=0;i<p;i++)
{
   for(j=0;j<n1;j++);
   {
     scanf(“%d”,&max[i][j]);
     n[i][j]=max[i][j]-a1[i][j];
    }
}
printf(“\n needed resources for every process to start execution”);
for(i=0;i<p;i++)
{ 
   printf(“\n%d %d%d”,n[i][j],n[i][j+1],n[i][j+2]);
}
printf(“\n safe sequence the sequence of process to compute their execution”);
for(a=0;a<(p-c);)
{
   for(i=0;i<p;i++)
   {
      j=0;
      b=0;
       if(f[i]==0)
       {
           if(n[i][j] <= a1[b][j] && n[i][j+1] <= a1[b][j+1] && n[i][j+2] <= a1[b][j+2])
             {
                  printf(“\n process %d execution started and completed”,i+1);
                  for(k=0;k<n-1;k++)
                  {
                       a1[b][k]+=a1[i][k];
                   }
                    f[i]=1;
                   c++;
              }
           else
          f[i]=0;
         }
  }
  getch();
}

  #!/bin/bash
  echo "enter no.of resources: "
    read n1 
    echo -n "enter the max no .of resources for each type:  " 
    
    for(( i=0; i <$n1; i++ ))
    do 
      read ${t} 
    done
    echo -n "enter no .of process: " ​
    read p 
    echo -n "enter allocation resources: " 
    for((i=0;i<p;i++)) 
    do 
     f=0 
     for((j=0;j<$n1;j++)) 
     do
      read ${a1} 
     done
    done

for (( i=1; i<=5 ; i++ ))
do
  for (( j=1; j<=5 ; j++ ))
  do
    eval a_${i}_[$j]=$(( i + j*10 ))
  done
done

echo "${a_3_[4]}"

k=3 l=4
eval echo "\"\${a_${k}_[l]}\""

eval "val=\${a_${k}_[l]}"
echo "$val"