Day Of The Programmer, C (gcc)

Day Of The Programmer

/* This code accepts an year as input from stdin and processes the very date of the Day of the Programmer in that year, aware of the fact that there was a shift
   from the Julian to Gregorian Calendar in the year 1918, in Russia (Day of The Programmer was founded here)*/

#include <assert.h>
#include <ctype.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* dayOfProgrammer(int year) {
    char dateLeap[11] = "12.09.";
    char dateNonLeap[11] = "13.09.";
    char* chyear = (char*)malloc(10 * sizeof(char));
    char* date = (char*)malloc(15 * sizeof(char));
    sprintf(chyear, "%d", year);
    if(year < 1918) {  //julian calendar
        if(year%4 == 0) {
            date = strcat(dateLeap, chyear);
        }
        else {
            date = strcat(dateNonLeap, chyear);
        }
    }
    else if(year > 1918) {  //gregorian calendar
        if(((year%4 == 0) && (year%100 != 0)) || (year%400 == 0)) {
            date = strcat(dateLeap, chyear);
        }
        else {
            date = strcat(dateNonLeap, chyear);
        }
    }
    else {  //transitional year 1918
        date = "26.09.1918";
    }
    return date;
}

int main()
{
    int* year;
    year = (int*)malloc(sizeof(int));
    scanf("%d", year);
    printf("%s", dayOfProgrammer(*year));
    return 0;
}


run \| edit \| history \| help	1

Good code, but when you input a year more than four digits, it outputs this:

*** buffer overflow detected ***: 385308392/a.out terminated Abort signal from abort(3) (SIGABRT)

by Dami, 1 years ago

Please log in to post a comment.