Character array Assignment

Question

int main()
{ 
     char s[]="stack";
     s="overflow";
}

This is not allowed.Its gives an error.But below code works fine.

int main()
{
      char s[]="stack";
      strcpy(s,"overflow");
}

Why this happens?

can anybody explain why this happens based on memory allocation??? — user2712068, Aug 23 '13 at 18:43
When you say that the lower segment "works" you mean the compiler does not throw an error, or s appears to contain "overflow"? — Nick Beeuwsaert, Aug 23 '13 at 18:44
For the second program, you should initialize `s` to `"buffer"` to more accurately describe what is happening. — jxh, Aug 23 '13 at 18:45
Also note that in the second case, you're going off the end of the array, so you're invoking UB. — Dennis Meng, Aug 23 '13 at 18:45

score -1 · Answer 1 · answered Aug 23 '13 at 18:47

-1

The variable s represents a pointer to the string. More specifically, it refers to the memory address of the first letter in your string "stack". Due to this reason, the operation s="overflow" makes no sense. How can you set the value of s (a pointer) to a string?

Keep in mind that C is a very low-level language, so you have to be wary of things that might seem intuitive to you not behaving the way you expect them to.

answered Aug 23 '13 at 18:47

Lanaru

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It doesn't decay into a pointer. It's a character array and you can't change the base address of an array once allocated. – Uchia Itachi Aug 23 '13 at 18:50
You can't change it, but you can treat it like a pointer in some cases (such as for pointer arithmetic: `s + 1` refers to the second character in the array). – Lanaru Aug 23 '13 at 18:56
Yes, true indeed. But your `How can you set the value of s (a pointer) to a string?` statement is misleading. – Uchia Itachi Aug 23 '13 at 19:00

Character array Assignment

1 Answers1