Error Copying 1 string to another in C [closed]

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• I was trying to copy the contents of 1 string to another (a into b) . I deliberately took the second string(b) to be smaller than the 1st one(a) .

• I copied the contents of the first one into second

  . I added WATCH on both of them . In the Debug tab , I found out that while copying the original string gets destroyed and the new one also DISPLAYED LARGER than its size.

     #include<stdio.h>
     int main()
     {
   char a[10]="What?";
   char b[2];
   int i;
   for(i=0;i<6;i++)
   {
      b[i]=a[i];
   }
   printf("This is %s",a);
   printf("\n this is b now: ",b);
   return 0;
  

  }

• I have attached the screenshot for the same. I took a = a string of size 10 . a="WHat?" then I took a string b[2]

• After copying , I printed both a and b . I expected the output to be , a = "WHat?" , b="WH" But the output is coming something else.(See the screenshot)

Why did the original string get destroyed ? Has the pointer changed ? But I have made it a constant pointer .It can't be changed.

Here is the Screen shot to the problem I am facing : https://www.dropbox.com/s/8xwxwb27qis8xww/sjpt.jpg

Please Help Somebody !!

Answer 1

As pointed out in other answers, you are writing outside the bounds of the array. The original string a changes because it happens to be exactly after b in memory as you can see in the debug window.

Before the loop, memory looks like this:

 b  a
|00|WHat?00000|

After the loop, memory looks like this:

 b  a
|WH|at?0?00000|

This explains why

• a is changed
• the original questionmark in a is still there (you only write 6 characters - two into the location reserved for b, 4 (including null terminator) into the location of a)

Of course this is undefined behavior as already mentioned by Vlad Lazarenko, but it explains the behavior for your compiler/settings/version/etc.

A constant pointer only exists for the compiler. It ensures that you cannot explicitly manipulate its data, but if you have memory leaks, nothing can be guaranteed.

Answer 2

1. You are copying 6 bytes into an array of two bytes, essentially invoking undefined behavior.
2. You are passing array b to printf with %s specifier that expects a null-terminated string, while b is most likely not null-terminated at that point, which is another undefined behavior.

Also, a null-terminated string that can fit into 2 bytes array can essentially have only one printable character, so you should not expect b to be "WH". At best, if you fix the copying, it can only be "W" as the second character will be a termination byte (\0). If you want to have two characters, either increase the array size to 3 to allow for null terminator, or simply do not use C strings and print out two bytes using "%c%c" format string.

Answer 3

What you're doing currently is very unsafe! It might work on Windows for some godforsaken reason, but don't do this!

The C standard library has special functions for working with strings and memory, strcpy for example is for copying character arrays. I suggest you learn more about how strings work and how you can manipulate them.