Can you tell me where did you write the print function ? I wrote the print function in the main after the sort function was called and I was able to print the array elements just fine.
Here is another idea... It will probably get you started but you have to do more work on it to get the finish product
void display(int n)
{
// Right now all the X will be left justified. You have to figure out how to include spaces.
int i=0,j=1,num=1;
printf("In display %d\n",n);
for(i=1;i<=n;i++)
{
while(j<=2*i-1)
{
// Fill this part
}
j=1;
printf("\n");
}
}
int main(int argc, char* argv[])
{
int i=0,num=1,num1=2,toggle=0;
//Assume that user input is 3
for(i=0;i<=3;i++)
{
if(toggle==0)
{
display(num);
num=num+2;
toggle=1;
}
else if(toggle==1)
{
// Similarly write this one
}
}
return 1;
}
Well suppose after the first sort, this is your input array
int arr[10]={1,2,3,7,8};
if the user enters 6, all you got to do is move the elements 8,7 one step behind. You see 8 is greater than 6 copy 8 to the 6th location. Same applies for 7. 3 is not greater than 6, so the loop ends there
Hello People, I just came across this code as I was browsing through some C++ problem sets. This code gives compile time error but I am unable to understand the reason. I am C guy and my knowledge about C++ is not that great. So pardon if there is an obvious solution to this problem. Any help would be appreciated
class Base
{
public:
Base();
virtual ~Base();
};
class derived: public Base
{
public:
virtual ~derived();
};
int main()
{
Base *pb = new derived;
getch();
}
There are 2-3 mistakes in your code 1. You are storing the id is a[0][0], yet when you start storing the marks of the students you start again from a[0][0], the index of the loop on line 23 should start from 1. This mistake is repeated again in the function printinfo. 2. In the function printinfo(), the code for adding all the grades is wrong. Your sum statement is basically adding a[1][6] 5 times. See how you printed the array and compare how you are adding them 3. You have made the same functions in a couple of place is the findinfo function as well
What exactly do you mean by different from all numbers ??? Does it mean that the array has contains many elements with many duplicates, but one element has only 1 copy ?
Make a 2-D array temp of size n. This array will store the element and the number of times this element has occurred Initialize the temp array to 0
Scan through the input array. If you get an element say 10, temp.element=10 temp.count++; i++
Scan the temp array. If temp.count is 0, then the element i occurs just once in the input array else the element i occurs more than once
Run time of this algo is O(n). Space complexity is O(2n)
The objective of BFS is to print all the nodes at a given level, then move to the next level
In DFS we go depth wise as far as we can go. Then when we cant go any further we backtrack and then try again and so on.
DFS gives you a forest because you have many connected components as opposed to BFS where you have only 1 connected component. (Read the definition of connectivity for better understanding of this)
while(check till end of file)
{
/*
fread will put the data in array1
After that use string copy to move the data to character pointer array...lets say array2
*/
}
1. When you wanted to remove the number 6, you took the ASCII value of 6 ie 54 but forgot to subtract the ASCII value of 0. Hence your i variable became too big. 2. When you read the number 6 you have only 5 characters to read. So the for loop should not be <= but just < 3. When you start the second iteration, the value of c is 4 but the value of m is already greater than 4 so it does not enter the loop 4. Ensure that after you have finished reading the frame , the value of has decremented upto the correct value 5. After you have made all these changes , remove scanf and use fgets instead
Open the file Store the contents of the file, one line at a time into an array- Possibly an array of character pointers Sort the character pointer array
You do not need the fscanf function. At the point when end of the file is reached, the code will not enter the while loop. Consequently you can do away with the if on line 21 Also you instead of incrementing the count by 1 and then using the mod operation, another way could be to toggle the count values between 0 and 1 using if conditions
int main(int argc, char* argv[])
{
int a=3,b=2,c;
a=a==b==0;
printf("%d\n",a);
return 1;
}
I cannot understand why this code gives 1 as the answer.
According to me , this should be the order of evaluation b==0, which gives false & the expression becomes a=a==0 this should again give false & the expression should become a=0 thereby the value of a should be 0
Ok.... I think I have finally understood the question.. You could try to do some thing of this sort..
For each row For each column When you see a non zero number say a, check its 4 sides and push the positions of all the non zero numbers in a stack. Then pop the first element say b and push the positions of all of its non zero neighbors onto the same stack. Not the stack contains the positions of non zero neighbors of the both a and b. If you go on this way, by the time, the stack is empty you would have finished the first square
while(line[k] != '\0')
{
argv[i] = (char *)malloc(sizeof(char) * 11);
/*
while(line[k] != ' ')
This should be
while(line[k]!= ' '&& line[k]!='\0')
If you give the command of the type "/bin/ls , your loop keeps on
searching till it find the space character
*/
{
argv[i][j] = line[k];
k ++;
j ++;
}
argv[i][j] = '\0';
j = 0;
k ++;
i ++;
}
execve(argv[0], argv, envp);
return 0;
}
Also why are you using the argv array to store the parameters. Why not make a separate array ?
Well from WHAT I KNOW, most modern operating systems do not use SJFS. SJFS is more common in the batch computing days when the user had to explicitly submit his jobs. In such a scenario when the user submitted his job he was required to estimate the run time also Most of the current operating systems use prioritized round robin scheme. In this all processes start at the same priority. Then if the process used the entire time slot and does not finish, it is assumed that this is a batch job and the priority of this process decreases. Similarly if the process uses a part of it allotted time and then tell the OS that it needs IO, the process is assumed to be interactive and its priority is increased.
The cin part enables the user to input values which can then be stored in the variables a, b, c. Think of a,b, c as 3 containers. When you declare them
int a,b,c;
You are telling the computer that at some point some one will store a value in this container. When you use cin statements, the users provides the values which are then stored in the containers.
Hi All, Today I came across a peculiar piece of code.
int main(int argc, char* argv[])
{
printf("%lf\n",(0.99-0.90-0.09));
}
According to me the output for this code should be 0, but to my surprise when I ran this code the answer that I got was -0.000. I cannot understand the reason for the minus sign
