Can someone help me with this? I want to write a code that will print random even number "i" times. Here "i" is variable and input (i is always greater than 0). Example: Input (4) I want output to be (4,4,4,4), Input (2) I want output to be (2,2). I want to print the number many times mentioned in the input. Here is the question

CodePudding user response：

Here, the term "random" may not be appropriate.

In order to achieve this, you can:

• get the numerical representation of the user input. To do so you can use strtol which returns a number.
• use a for-loop print the N occurrences using the standard printf function along with the number formatter %d.

For instance:

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

int main(int argc, char **argv)
{
    int number;
    int i;

    /* `argc` contains the total number of arguments passed to the
     * application. There is always at least one argument (i.e.: the
     * application name). Here, we ensure the user has given a argument
     * which we expect to be the number (i).
     *
     * If there are less than two arguments given then the user has
     * not passed anything => the command line was "$ ./a.out". */
    if (argc < 2) {
        goto usage;
    }

    /* Get the user input number by parsing the user given argument.
     * As arguments are string of characters, this function transforms
     * the string "1234" into 1234. The result is a number we will be
     * able to manipulate (add, sub, mul, div, etc.). */
    char *p;
    number = strtol(argv[1], &p, 10);

    /* We want to ensure the input was really a number.
     * The `strtol` function provides a way to verify whether the
     * given string is correctly formatted by giving the last character
     * of the string. In case the number is not formatted correctly,
     * then the last char is not the NULL terminating char. */
    if (*p != '\0') {
        goto usage;
    }

    /* Ensure the number is positive. */
    if (number < 0) {
        goto usage;
    }

    /* This for loop executes "number" of times. We have a counter
     * `i` which value will be incremented from [ 0 to "number" [.
     * Each time we execute the loop, we print the number. */
    for (i = 0; i < number; i  ) {
        printf("%d ", number);
    }
    printf("\n");

    return 0;

usage:
    // TODO
    return 1;
}

CodePudding user response：

for me, the easiest way is to read input from the user and just display the number "n" times. Here's an example:

#include <stdio.h>

int main(int argc, char **argv) {
    int user_input, is_correct; // user_input is a number to be displayed

    // scanf reads input from the user, %d is an integer
    is_correct = scanf("%d", &user_input);

    // check if user entered a number
    if(!is_correct) {
        printf("Incorrect input, enter number!");;
        return -1;
    }

    // check if number is even
    if(user_input % 2 != 0) {
        printf("Number is not even, try again!");
        return -1;
    }

    // display number
    for(int i = 0; i < user_input;   i) {
        printf("%d ", user_input);
    }

    return 0;
}