How to scan integers until newline?

general
, ,
1

Is there a simpler way than taking the whole string as input and then separating all of them? Have been searching for a while and can’t find a shorter way.
I have T lines and they can have any number of values. These values can be negative also.
Thanks.

2

One way is to read input character-by-character instead of storing the whole string.

You start storing the input into a number. When you encounter a space, you move to a next number. When you encounter a newline (’\n’) or you reach end of input, you stop.

Here is the implementation which reads lines of input into an array and then displays the array.

#include<stdio.h>

int main()
{
	int t,i,a[100],n,num;
	/*I'm assuming that there are max 100 integers per line
	variable i will store the number of integers read successfully so far*/
	char ch,sign;
	scanf("%d",&t);
	getchar();
	/*here getchar() will clears out the '\n' in the input buffer
	which was left out while reading t*/
	while(t--)
	{
		i=0;num=0;sign='+';
		while(scanf("%c",&ch)==1)
		{
			if(ch=='-')sign='-';
			else if(ch==' ' || ch=='\n')
			{
				if(sign=='-')
				{num=-num;sign='+';}
				a[i++] = num;
				num=0;
				if(ch=='\n')break;
			}
			else num = 10*num + (ch-'0');
		}
		n=i;
		/*now we have successfully stored all numbers in a
		and the number of numbers in n*/
		for(i=0;i < n;++i)
			printf("%d ",a[i]);
		putchar('\n');
	}
	return 0;
}

This is the best I could do. I don’t know a shorter way to do it. This code is also not resistant to buggy input (like multiple minus signs in a single number, alphabets instead of numeric digits, etc)

3

If you are scanning integers, you could do this way

int n;
while((scanf("%d",&n)) != EOF)
{
    printf("%d",n);
    //other operations with n..
}

Sample code

4

Your code is fine but I want to distinguish between the entries of all the lines. Suppose there are 2 lines. First line with variable number of inputs corresponds to list x and I want to do some operation on x and similarly a different operation on list y.

5

Yeah that’s one day of dealing with it. I guess I’ll stick to raw_input().split(’ '). Python is better to use in this question I suppose.

6

I have also been looking for the answer for a while now. Earlier, I did not think properly because people gave hints to use dynamic memory allocation for array. But Actually, it can be done just with a simple do-while loop.
code:

#include <stdio.h>
int main(void) {
	int i=0,size,arr[10000];
	char temp; 
	do{
	  	scanf("%d%c", &arr[i], &temp); 
	  	i++; 
	  	} while(temp!= '\n');
  	
  	size=i; 
  	for(i=0;i<size;i++){ 
  		printf("%d ",arr[i]); 
  	} 
return 0;
}
3 Likes
7

@divyag2 Can you explain it in c++ ?

8

This code is written on c language and can scan for any number of lines and can take max of 100 entries per line…

# include <stdio.h>
 int main(void)
 {
   int get_number[][100],ety_per_row;//first index denote number of lines and second index denotes number of entries per row
   char character_status;
   for(int line=0;(character_status!='e'||character_status!='E')&&(ety_per_row<100);ent_per_row++){
      scanf("%d",&get_number[line][ety_per_row)]);
      if(get_number[line][ety_per_row]=='\n') {line++;ety_per_row=0;}
      character_status=get_number[line][ety_per_row];
    }
}

Since I typed this on cell phone I put maximum effort to eliminate errors…

9

My C++ solution

#include<bits/stdc++.h>

using namespace std;

int main() {

string input;

vector<vector> v; //Vector containing all integers of all lines scanned

int j=1;

while(getline(cin,input))
{

 stringstream x(input);

 vector<int> vt;    //Parsing integers scanned in current line

 int n;

 cout<<"\nLine "<<j<<" scanned: ";

 while(x>>n)
 {

cout<<n<<" ";

vt.push_back(n);

 }

 cout<<endl;

   v.push_back(vt);

j++;
}
}

Hope it helps

10

for(i=0;i<10;i++)
{
scanf("%d%c",&j[i],&c);
if(c==’\n’)
break;
}

this is what you need;