Что самый легкий путь состоит в том, чтобы получить интервал в консольном приложении?

scanf () - это ответ, но вы обязательно должны проверить возвращаемое значение, так как многие, многие вещи могут пойти не так, анализируя числа с внешнего ввода ...

int num, nitems;

nitems = scanf("%d", &num);
if (nitems == EOF) {
    /* Handle EOF/Failure */
} else if (nitems == 0) {
    /* Handle no match */
} else {
    printf("Got %d\n", num);
}

The standard library function scanf is used for formatted input: %d int (the d is short for decimal)

#include <stdio.h>
int main(void)
{
  int number;
  printf("Enter a number from 1 to 1000: ");

  scanf("%d",&number); 
  printf("Your number is %d\n",number);
  return 0;
} 

#include <stdio.h>

main() {

    int i = 0;
    int k,j=10;

    i=scanf("%d%d%d",&j,&k,&i);
    printf("total values inputted %d\n",i);
    printf("The input values %d %d\n",j,k);

}

из здесь

Aside from (f)scanf, which has been sufficiently discussed by the other answers, there is also atoi and strtol, for cases when you already have read input into a string but want to convert it into an int or long.

char *line;
scanf("%s", line);

int i = atoi(line);  /* Array of chars TO Integer */

long l = strtol(line, NULL, 10);  /* STRing (base 10) TO Long */
                                  /* base can be between 2 and 36 inclusive */

strtol is recommended because it allows you to determine whether a number was successfully read or not (as opposed to atoi, which has no way to report any error, and will simply return 0 if it given garbage).

char *strs[] = {"not a number", "10 and stuff", "42"};
int i;
for (i = 0; i < sizeof(strs) / sizeof(*strs); i++) {
    char *end;
    long l = strtol(strs[i], &end, 10);
    if (end == line)
        printf("wasn't a number\n");
    else if (end[0] != '\0')
        printf("trailing characters after number %l: %s\n", l, end);
    else
        printf("happy, exact parse of %l\n", l);
}