C structure struct pointer and functions

Question

I have recently read a code but I have a doubt in code

toTs *now = try();
now->index = 30;

Where toTs is a struct and try() is a function with return type toTs*

*now being a pointer can keep address of label try() but as try() not being a structure variable now can't access it like struct and can never access it like now->index=30.

After compiling it shows segmentation fault.
I just want to ask is above code legitimate or not.

#include <stdio.h>
#include <unistd.h>

typedef struct toT {
    int index;
} toTs;

toTs lst[3];

toTs *try() {
    int i;
    for (i = 0; i < 3; i++) {
        toTs *current = &lst[i];
        printf("%d\n", current->index);
        if (current->index == 3) {
            printf("test work");
            return current;
        }
    }
}

int main() {
    int i;

    for (i = 0; i < 3; i++) {
        lst[i].index = i;
    }

    for (i = 0; i < 3; i++) {
        printf("test %d\n", lst[i].index);
    }

    toTs *now = try();
    now->index = 30;

    printf("current %d\n", now->index);
    printf("current %d\n", lst[2].index);
}

now is a struct pointer that can point to struct variable but try() is not a struct variable nor array of data structure its a function

try is not a keyword/reserved-word in C, so no conflict there. You need to allocate memory for now inside try() or before calling try(). — जलजनक
– जलजनक, Commented Feb 25, 2022 at 6:08
Sorry i think i was not able to make you understand what i wanted to tell — OSdev
– OSdev, Commented Feb 25, 2022 at 6:10

जलजनक · Accepted Answer · 2022-02-25 06:54:57Z

You need to always return a valid toTs* from try() for the code to work.

#include <stdio.h>
#include <unistd.h>

typedef struct toT {
    int index;
} toTs;

toTs lst[4];

toTs *try() {
    int i;
    for (i = 0; i < 3; i++) {
        toTs *current = &lst[i];
        printf("%d\n", current->index);
        if (current->index == 3) {
            printf("test work");
            return current;
        }
    }
    return &lst[3]; // return a spare structure.
}

int main() {
    int i;

    for (i = 0; i < 3; i++) {
        lst[i].index = i;
    }

    for (i = 0; i < 3; i++) {
        printf("test %d\n", lst[i].index);
    }

    toTs *now = try();
    now->index = 30;

    printf("current %d\n", now->index);
    for (i = 0; i < 4; i++) {
        printf("Final %d\n", lst[i].index);
    }
}

You're confusing function pointers with plain pointers in C. Function pointers go deeper : https://en.wikipedia.org/wiki/Function_pointer

Bill Lynch · Accepted Answer · 2022-02-25 06:32:53Z

0

This code is not particularly healthy.

Given the way lst is initialized, when you call try() it will reach the end of that function without a return statement.

The C Standard says that if you use the return value of a function, without actually returning a value, that's undefined behavior.

answered Feb 25, 2022 at 6:32

Bill Lynch

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3 Comments

OSdev Over a year ago

Sorry if i have wasted your time but please forget everything about code and just tell me how this is possible "toTs *now = try(); now->index = 30;" a struct pointer pointing to a function and than writing "now->index=30" there isn't it accessing area of code

जलजनक Over a year ago

Struct pointer is not pointing to a function. A function is returning a struct pointer to save in a struct pointer variable.

OSdev Over a year ago

Thanks a lot this was my issue @SparKot and you solved it.

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