Why calling strcpy() leads to changes of my source string?

Question

main()
{
      char a[]="abss";
      char c[]="";
      strcpy(c,a);
      printf("%s",a);
}

Why does the source string a change on using strcpy() it is checked only when string c is greater than or equal to string a??

Because `c` is not long enough to hold 5 characters. The way you defined it, it only has length 1. — JS1, Dec 19 '14 at 20:25

score 8 · Accepted Answer · answered Dec 19 '14 at 20:26

c has size 1 but you try to copy 5 characters into it. This causes undefined behaviour.

To explain what you are seeing, probably what happens is that c and a are stored next to each other in memory, so the things you write into c overflow and land in a.

score 2 · Answer 2 · answered Dec 19 '14 at 20:25

2

you don't have enough storage for c (only a 1 byte terminator) you are overwriting memory.

try char c[8]="";

answered Dec 19 '14 at 20:25

Don Weingarden

31
3

barak manos · Answer 3 · 2014-12-19T20:35:47.053

Official answer:

The memory allocated for the destination string is 1 character, and the length of the source string is 5 characters. So you are invoking undefined behavior by the C-language standard.

Practical answer:

The memory allocated for the destination string is 1 character, and the length of the source string is 5 characters. Your specific compiler has probably allocated the source string immediately after the destination string in memory. So the first character is successfully copied into the destination string, and the remaining 4 characters are copied into the source string itself.

Please note that you have yet another problem, as the destination string is no longer null-terminated.

score 0 · Answer 4 · answered Dec 19 '14 at 20:29

0

You need to allocate memory for the destination which is unallocated. In your case it should be

char[8] c;

and it should work fine

answered Dec 19 '14 at 20:29

Justinas Kanguras

129
9

Why calling strcpy() leads to changes of my source string?

4 Answers4