Question

Below code produces two smiley faces when I call case 2 after case 1 (in other words after one while loop). However printSentence(); works as it should be in case 1.

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

    char *enterSentence();
    void printSentence(char *);
    char *sentence;
    int willContinue = 1;

main() {
    while (willContinue) {
    int a;
    scanf("%d", &a);
    switch (a) {
           case 1:
                getchar();
                sentence = enterSentence();
                printSentence(sentence);
                break;
           case 2:
                getchar();
                printSentence(sentence);
                break;
           case 3:
                willContinue = 0; //exit
                break;
                }
    }
}

 char *enterSentence() {
         char temp[999];
         gets(temp);
         return temp;
      }

 void printSentence(char *asd) {
         puts(asd);
      }
 .
 . //more code
 .

I wonder what is the problem here, thanks for any help..

Answer 1

temp is local to the function enterSentence. It is created when the function is entered and it is destroyed when the function terminates.

When you return the address of the object (return temp;) it still exists and has that address, but it will be immediately destroyed afterwards and the calling function receives a pointer to an invalid location.

Quick and dirty solution: make temp a static object that can live since the program started till it ends

static char temp[999];

Note: static is a quick and dirty solution, as I said. It is mostly better avoided.

---

Edit

Slow and clean solution: move the temp object to the calling function and pass its pointer to the function

int main(void) {
    char temp[999];
    /* ... */
    enterSentence(temp, sizeof temp);
    /* ... */
}

size_t enterSentence(char *dst, size_t len) {
    size_t retlen;
    fgets(dst, len, stdin);
    retlen = strlen(dst);
    if (dst[retlen - 1] == '\n') dst[--retlen] = 0;
    return retlen;
}