Scanning all integers from the standard input. Assume that each input line is an integer and an empty line denotes end of input

fgets
c read integers until end of line
getline c++
read until newline c++
c++ read integers until end of line
scanf
getchar() in c
c++ read integers separated by spaces from file

As your assumptions is given lines are integer so we can use below code. atoi function converts a string to int fgets used to take string input from the user

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(){
  char str[1024]={};
  while( fgets(str, 1024, stdin) && str[0] != 0 && str[0] != '\n' ){
      int n = atoi(str);
      printf("%d\n", n);
  }
  return 0;
}

How to scan integers until newline? - general, 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. /*I'm assuming that there are max 100 integers per line variable i will store� LANGUAGE: JAVA CHALLENGE: Write a loop that reads positive integers from standard input and that terminates when it reads an integer that is not positive. After the loop terminates, it prints out, on a line by itself and separated by spaces, the sum of all the even integers read, the sum of all the odd […]


I see a number of problems with your code:

  1. The code doesn't read integers. It reads chars and print them. However, it only prints every second char as the return value of second getchar isn't used.

  2. There is no error checking.

  3. The return type of getchar is int - not char.

There are many ways to do what you want. It seems you want a solution using getchar so something like this could be a way:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    while (1)
    {
        int n = 0;
        int multiplier = 1;
        int ch=getchar();
        if(ch=='\n') break;  // Empty line - stop reading integers
        if(ch=='-')
        {
            // Handle negative values
            multiplier = -1;
            ch=getchar();
            if(ch=='\n') exit(1);  // Error - unexpected newline
        }
        do
        {
            if(ch==EOF) exit(1); // Input error
            if (ch < '0' || ch > '9') exit(1); // Error - input is not a digit

            // "Add" current digit to the integer
            n = n * 10;
            n = n + ch - '0';

            ch=getchar();
        } while (ch != '\n');
        n = multiplier * n;
        printf("%d\n", n);
    }
    printf ("done\n");

    return 0;
}

A few comments:

The code above is not "production quality" but I wanted to keep the example simple. Some of the short commings are described below.

  1. Just calling exit(1) on error is probably not what you want in a real application. At least you would print some error message first but in general you would add better error handling when dealing with user input.

  2. The calculation of the integer value does not check for integer overflow. In a real application that is something you want to add.

  3. The line if(ch==EOF) exit(1); isn't really needed because the next if-statement will catch that case and exit anyway. However, I added to show the difference between an input error and non-digit input.

Tutorial 5 Solutions, When the end of input is reached it should print a count of how many digits int main(void) { char line[MAX_LINE]; int i; while (fgets(line, MAX_LINE, stdin) != upper_case.c which reads characters from its input and convert all lower case You can assume there are only 3 white space characters, space, tab & new-line. Write a loop that reads positive integers from standard input and that terminates when it reads an integer that is not positive. After the loop terminates , it prints out, separated by a space and on a single line, the sum of all the even integers read and the sum of all the odd integers read. Declare any variables that are needed.


The simplest way is to just read integers and stop when there is no more integers to read. scanf works fine:

int n;
while (scanf("%d", &n) == 1) {
    // do something with integer n
}
// last scanf call failed to find an integer, so that's probably the end of input

This will not work if the input is in form "integers, empty line, more integers that should be ignored". In that case the code should be a bit more complex.

How to take the input of a line of integers separated by a space in C , specified any input size, I will assume that the input contains an arbitrary number of integer inputs separated by a space and ending with a '\n' newline character. When should I use unsigned integers (positive only) in C++ & C? press Enter after typing each input value; as any white-space will act as a delimiter in C. Since the OP has not specified any input size, I will assume that the input contains an arbitrary number of integer inputs separated by a space and ending with a ' &#039; newline character.


Here is what I think you are looking for. In your original code snippet, you were only going to be printing every second integer because of the two getchar() functions, instead of removing that will read all characters. Then including the if statement before the print statement means you won't print out a newline character when printing the results.

            while (1)
            {
                char ch=getchar();
                if(ch=='\n')
                    break;
                printf("%c\n",ch);
            }

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