CODEWITHSUNDEEP
cargvargc
idjuradj
idjuradj

Reputation: 1465

Explain the output of this program?

What is the output of the following program, if we pass to it the following parameters through the command line:

bcd abcd ab abc

So, since we pass 4 arguments, argc is 4? We initialize i to 2 and then go and checked argv's from 1 to 3 - my guess would be we add i = 2, and later, in the next iteration i = 3, and that's 5, so the output would be 5?

void main(int argc, char* argv[])
{
    char *p, *q; 
    int i = 2, j = 0, k = 0; 

    for (; i < argc; i++)
    {
        p = argv[i-1];
        q = argv[i];

        for (j = 0; *q && *p; j++, p++, q++)
        {
            if (*p != *q)
            {
                break;
            } 
        }

        if (!*p || !*q)
        {
            k += i; 
        }
    } 

    printf("%d",k); 
}

Upvotes: 0

Views: 200

Answers (2)

Lidong Guo
Lidong Guo

Reputation: 2857

Can you explain why is argc 5, and not 4? and what would be argv[0]?

The argv[0] is you program's name. like a.out or something else you named. argv[1] ... is the params you passed to the program. so argc is 1+ paramNumberYouPassed.

Upvotes: 0

SheetJS
SheetJS

Reputation: 22905

argc is 5.

This program checks each pair of consecutive arguments and counts how many are substrings of each other (either the first is a substring of the second or vice versa):

bcd abcd // i = 2
abcd ab  // i = 3, good
ab abc   // i = 4, good

In this case, since i=3 and i=4 fit the criteria, k is 7.

Breaking down the code, the innermost for loop exits if there is a different character or if one string ends. The line if (!*p || !*q) k += i; increases k only if one of the strings hit the end.

Upvotes: 1

Related Questions