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// C Program to Simulate Banker's Algorithm - CODE BY Nived Kannada #include<stdio.h> void main() { //Here we need 3 arrays namely Allocation, Max and Available // arrays alloc and max are 2D arrays. array 'available' is a 1D array. int n, m, i, j, k, alloc[20][20], max[20][20], available[20]; int f[20],ans[20], ind=0, need[20][20]; //We need the Need matrix. //Reading the number of Processes from the input. printf("Enter number of processes: "); scanf("%d",&n); //Reading the number of Resources from the input. printf("Enter the number of Resources: "); scanf("%d",&m); //Reading the Allocation Values to the Matrix 'alloc[][]' printf("Enter the Values of Allocation Matrix: \n"); for(i=0;i<n;i++) { for(j=0;j<m;j++) { printf("Enter value at position (%d%d) :",i+1,j+1); scanf("%d",&alloc[i][j]); } } //Reading the Max values to the matrix 'max[][]' printf("Enter the Values of Max Matrix: \n"); for(i=0;i<n;i++) { for(j=0;j<m;j++) { printf("Enter value at position (%d%d) :",i+1,j+1); scanf("%d",&max[i][j]); } } //Reading the values of array available[] printf("Enter the values in available: \n"); for(j=0;j<m;j++) { printf("Enter value at position (%d) :",j+1); scanf("%d",&available[j]); } //We are using an array f to represent the finished status of each process. //Initially setting all processes as not finished. ie. Setting f[i]=0 for each process i. for(k=0;k<n;k++) { f[k]=0; } //Calculating values of the NEED MATRIX using its equation, for all processes //Equation is need[i][j] = max[i][j] - allocation[i][j] for(i=0;i<n;i++) //For each process { for(j=0;j<m;j++) //For each resource { need[i][j] = max[i][j] - alloc[i][j]; } } //Finding safe sequence int y=0; // ans[] array will be used to store the SAFE SEQUENCE in the end. for(k=0;k<n;k++) { for(i=0;i<n;i++) //For each process { if(f[i]==0) { int flag = 0; //setting flag as 0 or false. for(j=0;j<m;j++) //For each Resource { if(need[i][j] > available[j]) //If Need greater than Available, then { flag=1; //Setting flag as true or 1. //flag=1 means the Need is greater than what is Available for that particular resource. break; //Breaking out of this loop if need > available } } if(flag==0) { ans[ind++] = i; for(y=0;y<m;y++) //For each Resource { available[y] = available[y] + alloc[i][y]; //Setting availability to current availability + allocation } f[i]=1; //Declaring the current process as FINISHED. } } } } //Displaying the SAFE SEQUENCE. printf("The SAFE SEQUENCE is: \n"); for(i=0;i<n-1;i++) //Here loop ends at n-1 because we don't want to printf the arrowmark(->) at the end.printf(" P%d", answer[n-1) { printf(" P%d ->", ans[i]); } printf(" P%d", ans[n-1]); //Printing the final state in safe sequence without printing the arrowmark. } /* CODE ENDS HERE */ /* Code written and explicitly commented by Nived Kannada */

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