在C中使用指针覆盖内存的可能性？

Question

I've written my own getline function following K&R c book

void getline(char * const str) 
{
  int c;
  char* temp = str;
  while ((c=getchar()) != '\n') {
    *temp = c;
    temp++;
  }
  *temp = '\0'
}

and it's used to initialize strings

char *str1, *str2;
printf("Type string 1: ");  
getline(str1);
printf("Type string 2: ");  
getline(str2);

Just wonder that, what if the memory locations str1 and str1 point to are very close, then getline(str2) overwrites contents in string 1?

It that's possible how could I avoid it? THANKS!

Update:

Yes the program stops executing the above code snippet but the code below works:

#include <stdio.h>
main()
{
  char* str;
  char* temp = str; 
  int c;
  while ((c=getchar()) != '\n') {
    *temp = c;
    ++temp;
  }
  *temp = '\0';
  printf("%s\n", str);
}

Here the str is also uninitialized character pointer but why deoesn't it give me an error?

Answer 1

What you have is Undefined Behavior .

Explanation:

You declared two pointers to char :

char *str1, *str2;

but you haven't initialized them. They point to some "random" memory location as they are uninitialized.

Then, you pass str1 and str2 to getline and here:

char* temp = str;

temp points to where str points to. Then, in the loop,

*temp = c;

you write to this memory location. This writes into an invalid memory location. And invokes UB .

Fix:

1. You can use an automatic array with a fixed size:

    char str1[101], str2[101]; 

   Note that you should add a check in the loop in the getline function which breaks the loop when the user has entered 100 characters so that there won't be a buffer overflow .

2. A better solution would be to use dynamic memory allocation. You need to use malloc and realloc for that. These functions require the stdlib.h header.

   Fixed code (untested) :

    char* getline() { char* str; int c, size = 10, counter = 0; str = malloc(size); /* Allocate `size` memory */ if(str == NULL) { printf("malloc failed to allocate memory"); exit(-1); /* Exit the program */ /* Or return NULL; */ } while ((c = getchar()) != '\\n' && c != EOF) { /* Added check for EOF as well */ str[counter] = c; counter++; if(counter == size) { char* temp = str; /* Backup in case realloc fails */ size += 10; /* Which is the same as `size = size + 10` */ str = realloc(str, size); /* realloc size(20) memory */ if(str == NULL) /* If realloc failed */ { printf("reallocing memory failed"); str = temp; /* str is NULL, retrieve the original contents */ break; /* Break out of the loop */ } } } str = realloc(str, counter + 1); /* realloc `counter + 1` memory */ str[counter] = '\\0'; return str; } 

   and in the calling function,

    char* str1 = getline(); if(str1) puts(str1); free(str1); char* str2 = getline(); if(str2) puts(str2); free(str2); 

Answer 2

str1 and str2 are not initialized thus it'll be undefined behaviour. You can access to not allowed memory which will crash your program.

You have to allocate enough memory for each pointer and pass its size to get line function to ensure you write only in allocated memory.

Answer 3

str1 and str2 are uninitialized.Right now you write ( *temp = c; ) at invalid (or unauthorized) memory location invoking UB .

First you need to allocate memory to str1 and str2 .

str1=malloc(100);     // check return 
str2=malloc(100);

To be able to write into that memory location.

Just wonder that, what if the memory locations str1 and str1 point to are very close, then getline(str2) overwrites contents in string 1?It that's possible how could I avoid it?

And as far as you concern , memory allocated by these malloc won't overlap (will be two different contagious memory blocks) , so if you also tend to write beyond these memory location ,you will invoke undefined behaviour ( if you are lucky segmentation fault ) . So, IMHO, there won't be any case of str2 over-writing str1 .