Problems inserting values in vector pointer

Question

I’m having trouble trying to perform a programming exercise that I was given, where I am asked to "create a vector type pointer variable with 20 character occurrences".

#include <stdio.h>
#include <conio.h>

int main()
{
char *v[20];

for(int i = 0; i < 20; i++) {
char c;
v[i] = &c;
scanf("%c", &c);
*v[i] = c;
}

for(int i = 0; i < 20; i++) {
        printf("%c ", *v[i]);
}

getch();
}

There is no error when compiling, but there are some problems: It exits the first for no after typing the 20 characters, but only 10 of them; The result of printf is a lot of empty spaces

My experience with pointers is almost nil, I asked some friends who understand more, but I could not find a solution.

Answer 1

I’ve made some changes to your code, I don’t know if it exactly achieves the goal, but I believe it can help.

#include <stdio.h>

int main()
{
    char *v;
    char c[20];
    int i;  
    for(i = 0; i < 20; i++) {
        scanf("%c", &c[i]);
        getchar();

    }
    v = c;

    for(i = 0; i < 20; i++) {
            printf("letra  i = %c \n", *(v+i));
    }

return 0;
}

I used a v pointer pointing to a char vector. The getchar() within the for is picking up the enter that is given. That’s why your program was taking only half of the elements, the other half was taken as enter.

At last for, it is pointer arithmetic, v points to the first element of c, that is to say c[0], then when you do *(v+i) is saying it will access the address of v + the address of i. i is not a specific value, it is a unit of the data type that v points out, ie, char, then if i = 1, v+i == c[1] as it is adding a char unit to the address of v.

I hope I’ve helped!

Answer 2

I think your code should be this:

#include <stdio.h>
#include <conio.h>

int main() {
    char v[20];

    for(int i = 0; i < 20; i++) {
        char c;
        scanf("%c", &c);
        v[i] = c;
    }

    for(int i = 0; i < 20; i++) {
        printf("%c ", v[i]);
    }

    getch();
}

I tested it here and it worked. I just had to remove the #include <conio.h> and put a return 0; in place of getch();.

Remember that in C every vector is a pointer, so char v[20]; is a pointer to a character. It just so happens that this pointer points to an area of memory already allocated on the execution stack (stack) for 20 characters.

Already char *v[20]; is a pointer to another character pointer, which points to an area of memory in the stack where 20 other pointers fit. That’s not what you want!

The code v[i] = &c; is particularly dangerous. You take the memory address of the variable c and puts it in the character vector. This operation has no sense, besides the memory address size does not fit in the vector position, so the resulting value will be truncated. The result will be that you are polluting your vector with garbage.

Already *v[i] = c; means you take the i-th character of v, interprets it as a pointer and the override, and puts the value of c in the resulting memory position. That’s not what you want, you don’t want to interpret the i-th character as if it were a pointer. Most likely this causes a segmentation failure in the program or something else disastrous.

Now, looking at the code I fixed, note that you almost don’t need to de-reference explicit pointers, just work with arrays. The only detail is the scanf, where you get the address of c only.

Another detail of your code: It will always try to read the 20 characters, even if what you want to type is less than that. To resolve this detail, if you want, you will have to check if the user has typed \n and stop.