Passing by Struct pointer is returning wrong value

Question

I am new to C but I am trying to edit a struct in a different function. I can print the value of the struct correctly from the target function readCoordinate but when execution returns to the main function the values of the struct changes to a seemingly random integers.

This is the full code below.

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

typedef struct {
   int  X;
   int  Y;
} Coordinate;

void buff_clr(void)
{
    char junk;
    do{
        junk=getchar();
    }while(junk!='\n');
}

int main()
{
    Coordinate carrierStart = {0,0};

    printf("Input Start coordinate for Carrier (example - B2) \n");
    readCoordinate(&carrierStart);
    printf("%d, %d \n", carrierStart.X, carrierStart.Y);

}

void readCoordinate(Coordinate *cood){
    char str[2];
    scanf("%s",&str);
    buff_clr();

    if (isalpha(str[0]) )
    {
        str[0] = tolower(str[0]);
        findChar(str[0], &cood->X);

        cood->Y = (str[1] - '0') - 1;
    }
    else
    {
        printf("\n Please Enter Valid Coordinate");
        exit(EXIT_SUCCESS);
    }
    
    printf("%d, %d \n", cood->X, cood->Y);
    return;
}

void findChar(char find, int *x){
    char vertical_letters[10] = {'a','b','c','d','e','f','g','h','i','j'};

    for (int i = 0; i < 10; i++)
    {
        if (find == vertical_letters[i])
        {
            *x = i;
            return;
        }
    }
    *x = -1;
}

input of b2 should print 1,1 which is correct in function (readCoordinate) but wrong in Main.

str[1]-'0' your array can only hold 1 input character and terminating 0 byte. If you enter more than 1 character, you access the array out of bounds. — Gerhardh
– Gerhardh, Commented Jun 29, 2021 at 11:19
IMO the culprit is a misunderstanding of the char str[2]; declaration. What do you think it means, exactly? — CiaPan
– CiaPan, Commented Jun 29, 2021 at 11:24

Jonathan Leffler · Accepted Answer · 2021-06-29 14:07:37Z

4

The part

    char str[2];
    scanf("%s",&str);

is wrong because

&str has type char(*)[2], but %s in scanf expects char*. Passing data having the wrong type like this invokes undefined behavior.
char str[2]; is too short to store 2-character strings like b2 because there is no room for the terminating null character. You have to allocate enough elements.
You should specify the maximum length to read (the length of buffer minus one) to prevent buffer overrun.

It should be:

    char str[3];
    scanf("%2s",str);

edited Jun 29, 2021 at 14:07

Jonathan Leffler

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answered Jun 29, 2021 at 11:18

MikeCAT

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1 Comment

CiaPan Over a year ago

...and they should test the return value to make sure anything meaningful has been read into str[], too. Also, they should test strlen(str)==2 to make sure the input was actually 2-characters long, as expected.

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