Thread: I have the answer, I'd like to know why.

  1. #1
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    I have the answer, I'd like to know why.

    Hey Guys,

    I'm trying to understand this function 'f'. The code for function f was provided in the book. I wrote the main() function that calls f. The question is "What is the value of f("abcd", "babc")? Here's the code I wrote to try and figure this out:

    Code:
    #include <stdio.h>
    // http://knking.com/books/c2/answers/c13.html
    
    
    int f(char *s, char*t);
    
    
    int main(void){
        int dang;
        dang = f("abcd", "babc");
        printf("dang=%s\n", dang);
        return 0;
    }
    
    
    int f(char *s, char*t){
        //printf("s=%s, t=%s\n",s,t);
        char *p1, *p2;
        for(p1=s; *p1; p1++){
    
    
            for(p2=t; *p2; p2++){
                if(*p1==*p2) break;
            }
            if(*p2== '\0') break;
        }
        printf("p1=%s, s=%s\n",p1,s); //prints p1=d, s=abcd 
        return p1-s;
    }
    The Answer: The value of f is 3. The reason: The length of the longest prefix of the string s that consists entirely of characters from the string t. Or, equivalently, the position of the first character in s that is not also in t.

    How do I print out the answer '3'? In the final printf statement, p1=d, s=abcd. d-abcd = 3? I just don't get it.

    In the 2nd question, the value of f("abcd", "bcd") is 0 which I also don't understand. Can someone please help me understand what is happening.

  2. #2
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    The function f() returns an int and you assign the return value to an int (dang), so you also need to print dang as an int in main() (hint: use the %d format specifier).

    To answer your second question, what is the first position in the string "abcd" that contains a character that is not also in the string "bcd"? It's the first position, right? That's position 0 (indices are zero-based in C).

    By the way, this is the same thing the strspn() function in the standard library does.

  3. #3
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    p1 is a pointer. It's value is the address of 'd', not 'd' itself. Same with s. It is a pointer and it's value is the address of 'a' in the string "abcd". Subtracting s from p1 gives the "distance" between the pointers, which is the answer.

    About f("abcd", "bcd"), the answer is 0 since the very first letter of s is not in t, so the scan stops at that point.
    Code:
    #include <stdio.h>
     
    int f(const char *s, const char *t) {
        const char *p = s;
        for (const char *q; *p; ++p) {
            for (q = t; *q && *p != *q; ++q) ;
            if (!*q) break;
        }
        return p - s;
    }
     
    int main() {
        struct {
            const char *s, *t;
            int n;
        } tests[] = {
            {"abcd",       "babc", 3},
            {"aaaa",       "x",    0},
            {"xxxxxxtxxx", "qx",   6},
            {"xxxx",       "x",    4},
            {"abcd",       "bcd",  0}
        };
        int size = sizeof tests / sizeof tests[0];
     
        for (int i = 0; i < size; ++i) {
            printf("[%s] [%s]\n", tests[i].s, tests[i].t);
            int n = f(tests[i].s, tests[i].t);
            printf("%d", n);
            if (n != tests[i].n)
                printf("  WRONG: should be %d", tests[i].n);
            putchar('\n');
        }
     
        return 0;
    }
    Ordinary language is totally unsuited for expressing what physics really asserts.
    Only mathematics can say as little as the physicist means to say. - Bertrand Russell

  4. #4
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    Hey christop, thank you so much. I can't believe so many of my errors are with silly things like %s vs %d. OMG, so embarrassing. I will look at your reasoning in a bit. I see John.c also made a comment. I'm working on another exercise at the moment and will come back to this.

  5. #5
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    Hey John, thank you for the explanation. I got it now thanks to you and Christop. "It's value is the address of 'd', not 'd' itself. Same with s."

    I've read the chapter 3 times and it never covered the content to answer this question. Must have been a prior chapter...

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