Why am I getting run-time error? The code is still giving me the correct output

Question

Why am I getting run-time error when running this code? Here's a Test Input Case:

#include <stdio.h>
long long int s[999999];
// Complete the arrayManipulation function below.
int main()
{
    int n,m,i;
    long int a,b,val,cval;
    scanf("%d %d",&n,&m);
    for(;m>0;m--)
    {
        scanf("%ld %ld %ld",&a,&b,&val);
        s[a-1]+=val;
        if(b!=n)
            s[b]-=val;
    }
    val=s[0];
    cval=s[0];
    for(i=1;i<n;i++)
    {
        cval+=s[i];
        if(val<cval)
            val=cval;
    }
    printf("%ld",val);
    return val;
}

Expected and actual results match but compiler giving run-time error.

Compilers don't give run-time errors. When you run the program, what is the error you get? — Jonathan Leffler, Aug 01 '19 at 06:23
Is the problem simply that your program is exiting with `return val;` and `val` is not zero (it's 200 for the test data), so the exit status is treated as an error status? I've run the code, and I don't see anything else to cause trouble. — Jonathan Leffler, Aug 01 '19 at 06:29
OP, Please use `gdb` to find where its crashing. Do share `backtrace` if you're able to get it. — Achal, Aug 01 '19 at 06:33

score 2 · Answer 1 · answered Aug 01 '19 at 06:55

The return statement of your application should usually return EXIT_SUCCESS (0) or EXIT_FAILURE (1).

You return a very positive value - which is means for the shell that this was an error. Just change the return value to "0" or EXIT_SUCCESS to remove that "run time error".

See also: What should main() return in C and C++?

Why am I getting run-time error? The code is still giving me the correct output

1 Answers1