C数组指针分割错误

Question

I'm not familiar to pointers and I stumbled upon segmentation fault on my code, wheres if I don't use pointers this code runs perfectly.

#include <stdio.h>
#include <string.h>

int main()
{ 

    char *string[100],i,j;
    char *(*odd)[100];
    i = j = 0;

    fgets(*string, 100, stdin);

    while (*string[i] != '\0') {
        if (i % 2 == 0) {
                *odd[j++] = string[i];
        }
        i++;
    }

    *odd[j] = '\0';

    printf("Characters at odd position: %s\n",*odd[j]);

    return 0;
}

I'm guessing that I'm printing the odd array the wrong way, but I can't print it just using *odd as well.

Answer 1

char *string[100],i,j;
fgets(*string, 100, stdin);

You are trying to write to address at string[0] but it is not initialized. I guess you mean smth like

char string[100];
fgets(string, 100, stdin);

PS char *string[] is an array of pointers to char while char string[] is an array of chars, and for this case in expressions string decays to pointer to first element ie to pointer to char.

Answer 2

With pointer:

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

#define SIZEOF_BUFFER 100

int main(){ 
    char *string; // string is a pointer
    char i,j;

    char *odd; // odd is a pointer

    // allocate memory to be able to store data
    string = malloc(sizeof(char) * SIZEOF_BUFFER);

    if (string == NULL) {
        exit(-1);
    }

    odd = malloc(sizeof(char) * SIZEOF_BUFFER);

    if (od == NULL) {
        free(string);
        exit(-1);
    }

    i = j = 0;

    fgets(string, SIZEOF_BUFFER, stdin);

    while (string[i] != '\0') {
        if (i % 2 == 0) {
                odd[j++] = string[i];
        }
        i++;
    }

    odd[j] = '\0';

    printf("Characters at odd position: %s\n",odd);

    // Don't forget to free allocated memory
    free(odd);
    free(string);

    return 0;
}

Without pointer:

#include <stdio.h>
#include <string.h>

int main(){ 
    char string[100],i,j;

    char odd[100];

    i = j = 0;

    fgets(string, 100, stdin);

    while (string[i] != '\0') {
        if (i % 2 == 0) {
                odd[j++] = string[i];
        }
        i++;
    }

    odd[j] = '\0';

    printf("Characters at odd position: %s\n",odd);

    return 0;
}

also, no need of *string[strlen(*string) - 1] = '\\0'; => fgets put always a null character at the end of string

Answer 3

This is a problem with precedence of [] vs *. See http://en.cppreference.com/w/c/language/operator_precedence . To solve it do the dereferences like this:

(*odd)[j] = '\0';

This also applies to declarations; to declare string as a pointer to an array of 100 chars, change it to:

char (*string)[100];

You also must allocate memory. Here is a changed version of your code with all these changes:

#include <stdio.h>
#include <string.h>

int main(){
    char actualstring[100];
    char (*string)[100] = &actualstring, i = 0, j = 0;
    char actualodd[100];
    char (*odd)[100] = &actualodd;

    fgets(*string, 100, stdin);

    (*string)[strlen(*string) - 1] = '\0';

    while ((*string)[i] != '\0') {
        if (i % 2 == 0) {
                (*odd)[j++] = (*string)[i];
        }
        i++;
    }

    (*odd)[j] = '\0';

    printf("Characters at odd position: %s\n",*odd);

    return 0;
} 

You may find it useful to translate declarations to english and vice-versa at http://cdecl.org/

Answer 4

string is an array of pointers. You did not initialize any of those pointers. But then you write through those pointers as if they pointed somewhere.

Your code is a more complicated version of:

char *ptr;
fgets(ptr, 100, stdin);

in which (hopefully) it should be obvious what the problem is. odd has a similar problem.