[Solved] What is the right way to explain this code? [closed]

by


You probably want this:

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

int main(void)
{
    char temp[256]; 
    char *st[3];     // array of three pointers to char

    for (int i = 0; i < 3; i++)
    {
       scanf("%255s", temp);   // prevents potential buffer overflow
       st[i] = strdup(temp);
    }

    for(int i = 0; i < 3; i++)
    {
        printf("%s\n", st[i]);
        free(st[i]);           // free strduped memory
    }
}

This program displays:

./a.out
11
22
33
11
22
33

Whereas your program displays

./a.out
11
22
33
33
33
33

this is because:

char *st;        // in your prog. you only declare one pointer

for (int i = 0; i < 3; i++)
{
   scanf("%s", temp);
   st= strdup(temp);   // here you overwrite the st pointer loosing
                       // the string strduped in the previous run of the loop
}

1

solved What is the right way to explain this code? [closed]