Undefined behavior

This same function works in one program and crashes in another at the line : strcpy(buf, str);
Is this an undefined behavior?

Code:

---

int function(char * buf)
{
    char * szLocalIP = functionreturningchar*();
    char str[260];
    strcpy(str, szLocalIP);

    char * pch = strrchr(str, '.');

    str[pch-str+1] = '\0';
    strcat(str, "0");

    strcpy(buf, str); // crash
}

---

Well it is if you call it with say

function("hello");

or

char tooSmall[3];
function(tooSmall);


You need to show the context for a decent answer.

Sorry.

Code:

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    char * ip;
    function(ip);

---

In other words, you're trying to append to an uninitialised pointer.

Pass a pointer to some memory you can write to.

Thanks.

ip is an uninitialised pointer. Obtaining its value gives undefined behaviour. Dereferencing it (which also requires obtaining its value) also gives undefined behaviour.

Before calling function(), you need to ensure that ip points to the (first character of) a buffer that is long enough to hold whatever you will be copying to it. Otherwise your function will (probably) be writing data to some random area of memory. When modern operating systems catch programs doing that, they tend to forceably terminate the program (aka force it to crash).

It works with initializing with calloc().
But what's wrong with initializing like this?
It will crash though I'm writing less character to it than it was initialized with.

Code:

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char * ip = "111222333444555";
function(ip);

---

"111222333444555" is a string literal. It's probably stored by the compiler in an area of memory that's read only. I.e. you'd be safer having const char *ip = "111222333444555";

Edit:
If you want something you can modify try char ip[] = "111222333444555";

Quote:

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Originally Posted by Hodor

"111222333444555" is a string literal. It's probably stored by the compiler in an area of memory that's read only.

---

What's really wrong with modifying a string literal is that the standard says that doing so gives undefined behaviour. That's really all a C programmer needs to care about.

The string literal might not even be in memory as the programmer understands it. On some systems, for example, string literals are (literally) retrieved from the executable file on demand (and the executable file itself is locked so it can't be modified while being run). On those systems, changing the string literal has no effect that the program can observe, since the literal will revert to the same string whenever it is retrieved. Admittedly, such systems are not often encountered these days.

Quote:

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Originally Posted by Ducky

But what's wrong with initializing like this?
It will crash though I'm writing less character to it than it was initialized with.

Code:

---

char * ip = "111222333444555";
function(ip);

---

---

Try initializing like this instead:

Code:

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char ip[] = "111222333444555";

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Then the memory ip points to will not be read-only. It is a shorthand for typing

Code:

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char ip[] = {'1', '1', '1', '2', ..., '\0'};

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I see this often. Is this undefined behavior?

Code:

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char * buff;
buff = " Hello";
printf("%s", buff);

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Quote:

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Originally Posted by Ducky

I see this often. Is this undefined behavior?

Code:

---

char * buff;
buff = " Hello";
printf("%s", buff);

---

---

No till you try to modify string using your buff pointer.
if you declare pointer as const char* compiler will wanr about such attempts

Thanks.

I declared as const char and modified it and it still compiles and executes. How so?

Code:

---

    const char * buf;
    buf = "hello";
    printf("%s\n", buf);

    buf = "hello world";
    printf("%s\n", buf);

---

Quote:

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Originally Posted by Ducky

Thanks.

I declared as const char and modified it and it still compiles and executes. How so?

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You didn't modify the data the const char * points to. Just reassigned it. That's allowed

Kurt

You need to understand the difference between "const char *buf" (where the individual chars are const) and "char * const buf" (where the pointer is const, but the char's are not).