C中的指针问题？ 对输入进行排序的函数，作为字符串返回并在 main 中打印

Question

I want to write a function that will take two integers, three times, and then return them ordered by the first integer and (for now) print them in main (though eventually I plan/hope to switch to a file-based structure to store and organize data), but I think I might have an issue with my pointers cause even when I skip concatenations (which looks like might also be another separate issue), everything Ive tried has main print a string (or no string) which never matches the input, but the print statements suggest all the looped assignments are working properly.

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

const char * entry()
{
    int n;
    int level;
    char habit1entry[6];
    char habit2entry[6];
    char habit3entry[6];
    for (int c = 0; c< 3; c++){
        printf("Habit #\n");
        scanf("%d", &n);
        printf("Level:\n");
        scanf("%d", &level);
        switch (n)
        {
            case 1:;
                sprintf(habit1entry, "|%d|%d|\n", n,level);
                printf("n = %d\n",n); 
                printf("%s\n",habit1entry);
                continue;
            case 2:;
                sprintf(habit2entry, "|%d|%d|\n", n,level);
                printf("n = %d\n",n);
                printf("%s\n",habit2entry);
                continue;
            case 3:;
                sprintf(habit3entry, "|%d|%d|\n", n,level);
                printf("n = %d\n",n);
                printf("%s\n",habit3entry);
                continue;
        }
    }
    strcat(habit2entry,habit3entry);
    printf("%s\n",habit2entry);
    strcat(habit1entry,habit2entry);
    printf("%s\n",habit1entry);
    char *fullEntry=habit3entry;
    printf("%s\n",fullEntry);

    return strdup(&fullEntry[0]);
}
int main(){
    const char * dataEntry = entry();
    //strcpy(dataEntry,entry());
    printf("Data:\n%s",dataEntry);
}

heres an example of the output(after the correct prints inside the switch cases) for an input of 3 2 1 1 2 2: "

|2|2|

|1|1|
|2|2|
|2|2|
|
|2|2|
|
* stack smashing detected * : ./a.out terminated
Aborted (core dumped) "

ps Sorry if this all sounds silly, this is my first C project (and first real stack overflow post, plz b gentl) coming from jumping around between java, python and clojure and I would like to take an operating systems class that allows you to start without knowing C but expects you to pick it up on your own and its hard finding material that explains C concepts in a scope that matches my background knowledge and current learning constraints in terms of time available for taking deep dives through explanations that for me have ended up mostly being either hopelessly esoteric, incredibly case-specific or overly-simplistic/redundant/unhelpful explanations of programming concepts I picked up in other languages. Dont mean to complain or harp on and its probably good to get practice with different methods of asking questions and finding answers for problems like these, but the learning curve for understanding things like this (setting up the compiler/json files involved spending hours only to discover that mcafee was deleting my exes which I became convinced was a symptom of a virus, only to have the behavior stop after I restarted for a minor routine windows update and I have no idea why) outside of a traditional framework sometimes seems more like a wall and I'm worried that maybe I should revise my approach to avoid wasting too much of my time banging my head against a series of very sturdy walls. any and all advice is greatly appreciated.

Answer 1

Abstracting form the logic of the program, you have plenty of issues there:

1. You do not provide enough space for the strings
2. Your switch is not very related to your for loop
3. Names of the variables do not matter for you - but they matter for the program . Be more careful.
4. probably more but I forgot already

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

const char * entry()
{
    int n;
    int level;
    char habit1entry[21] = "";
    char habit2entry[14] = "";
    char habit3entry[7] = "";
    for (int c = 1; c < 4; c++){
        printf("Habit #\n");
        scanf("%d", &n);
        printf("Level:\n");
        scanf("%d", &level);
        switch (c)
        {
            case 1:;
                sprintf(habit1entry, "|%d|%d|\n", n,level % 10);
                printf("n = %d\n",n); 
                printf("He1: %s\n",habit1entry);
                continue;
            case 2:;
                sprintf(habit2entry, "|%d|%d|\n", n,level % 10);
                printf("n = %d\n",n);
                printf("He2 = %s\n",habit2entry);
                continue;
            case 3:;
                sprintf(habit3entry, "|%d|%d|\n", n,level % 10);
                printf("n = %d\n",n);
                printf("He3 = %s\n",habit3entry);
                continue;
        }
    }
    strcat(habit2entry,habit3entry);
    printf("H2 + H3 = %s\n",habit2entry);
    strcat(habit1entry,habit2entry);
    printf("H1 + H2 = %s\n",habit1entry);
    char *fullEntry=habit1entry;
    printf("FE: %s\n",fullEntry);

    return strdup(fullEntry);
}
int main(){
    const char * dataEntry = entry();
    //strcpy(dataEntry,entry());
    printf("Data:\n%s",dataEntry);
}

Answer 2

Welcome to the weird and wonderful world of C.

I have not actually compiled and run your program yet, just had a quick read through and though I give you my first thoughts.

The way your program is written is primed to generate stack overflows. You have three (very little) character arrays defined on the stack habitxentry, so your sprintf's will most certainly blow your stack unless both the Habit and Level inputs are less than 10. Habit is alright because your switch only allows 1, 2 or 3. Your switch does nothing if Habit is anything else.

As a side note: sprintf is not really the function to use in our security minded world. snprintf is a better choice. Not really an issue here per se as you are not passing in user supplied data but still, it's not a good habit to cultivate.

Next you strcat your character arrays together, virtually guarantying a stack violation, but lets assume this works; you are concatenating 2 and 3 into habit2entry and then 1 and 2 into habit1entry.

Next you are creating a pointer to habit3entry (not habit1entry) and returning a duplicate.

By doing so you are allocating heap in a mildy obscure manner. The callee will be responsible for freeing this memory. I always preferred to explicitly malloc the memory and then strcpy (or memcpy) the data in. Now when you grep your code, you only have to look for malloc. Also, someone using the function will notice the malloc, see you have returned to pointer and realize that freeing it will now be his problem.

In order to avoid these problems some programmers leave it to the caller to supply a buffer to the function. The reasoning is that a function is supposed to do one thing and one thing only. In this case you are doing two things, you allocate memory and you fill that memory.

In your switch statement I noticed that each of your case labels are followed by an empty statement. the semicolon at the end of that line is not necessary: write "case 1:" not "case 1:;" You also use continue at the end of each block. This is allowed but "break" is more appropriate. In this case it will have the same effect but normally you have more statements after the switch. Now the difference will become apparent. Continue will jump straight to the top of the loop, break will break out of the switch and continue executing there.

Hope this gives you some insight.

Good luck.