scope

Hello,

Quick question about scope in C: if I have the following:

Code:

---

int* p;

{
  int x = 1;
  p = &x;
}

//do something with p

---

then there is no guarantee that *p == 1, since x is destroyed after the end of the code block.

However if I had:

Code:

---

char* p;

{
  char* x = "hello world";
  p = x;
}

//do something with p

---

would that have the same instability? If I had used malloc() instead, would that keep the value 'alive' in memory?

Thanks

> then there is no guarantee that *p == 1, since x is destroyed after the end of the code block.
Correct.

> would that have the same instability?
No, the string constant is in persistent memory. x and p are just copies of pointers to that memory.

Also note that you should always use const char* when you want a pointer to a string literal. (It is unmodifiable, but the type system for some legacy reasons lets you get away with removing the const.)

Quote:

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Originally Posted by anon

Also note that you should always use const char* when you want a pointer to a string literal. (It is unmodifiable, but the type system for some legacy reasons lets you get away with removing the const.)

---

But then why is it that:

Code:

---

char* x = "hello";
x = "world";
printf(x);

---

prints "world" without issue?

Would arrays of say, integers, face the same constraints as strings?

Thanks


edit. also,

char x[] = "hello";
x[2] = 'k';
printf(x);

produces 'heklo' as its output? how does this work?

In the first case you modify the POINTER, not the actual string.

In the second case you do modify the string, and the behaviour is undefined. For you it works. For me, it'd probably make the application crash.

Quote:

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Originally Posted by EVOEx

In the second case you do modify the string, and the behaviour is undefined. For you it works. For me, it'd probably make the application crash.

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The second case the op uses is an array not a pointer. The behavior is not undefined. If it had been a pointer then yes.

Quote:

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Originally Posted by hk_mp5kpdw

The second case the op uses is an array not a pointer. The behavior is not undefined. If it had been a pointer then yes.

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Ahh sorry, I misread it (I'll blame it on the lack of code tags :P). I stand corrected ;).