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I was learning the basics of C programming , and I wanted to test some lines for strings.

This is my code:

int main(){ char a[] = "abc"; strcpy(a,"pqrst"); printf("%s; %d",a, sizeof(a)); }

I expected the code to output size=6 (p, q, r, s, t and '\0'), but instead, it still prints size=4. How does this work?

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    Your strcpy-call has a buffer-overrun (the source is two byte longer than the destination) => UB => everything's over. Commented Aug 7, 2014 at 21:10
  • Use strlen to determine length of a string. sizeof is compile-time value, it just replaced by size memory, "allocated" for "a" variable. And strcpy causes buffer overrun, of course. Commented Aug 7, 2014 at 21:13
  • it is size 4 because that is its real size, the 3 characters you put into it + the null character '\0'. you can't re-size an array, it has a fixed length. Commented Aug 7, 2014 at 21:16
  • Hey, thank you for your reply. But from where is it printing the other bytes if a is only 4 byte long. In the output, I got pqrst; 4. How s this possible? How are the other two characters accessed? Commented Aug 7, 2014 at 21:17
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    @user3920047: That is part of the wonderful world of undefined behavior. One possible outcome of undefined behavior is the behavior that you expect. But it can also change wildly due to relatively unrelated code elsewhere in your program. In this case, you're overwriting what would be other variables on the stack. Commented Aug 7, 2014 at 21:18

4 Answers 4

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sizeof is computed at compile time, based on the declaration of a, which has 4 characters (3 + 1 null terminator). It should be noted sizeof an array and length of the string in an array aren't the same thing.

Moreover, the copy has overflowed the buffer. You have to create a large enough array to hold the string you want to copy over.

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Nice answer. Hmmm --- "sizeof an array and length of the string in an array are hardly ever the same thing" - I do not think they are ever the same for a valid array and valid string.
Yeah I didn't think so either. Edited. Thanks!
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sizeof(a) is evaluated at compile-time. The type of a, partially determined from the char a[] part of the declaration and partially from the "abc" initializer, is “array of 4 chars”, therefore sizeof(a) evaluates to 4. The value of the elements of a have no influence on the result.

Incidentally, the strcpy call in your program causes a buffer overflow. Extra characters are written somewhere in memory and may cause unpredictable behavior.

If you copied the string "z" to a with strcpy(a, "z");, there would be no undefined behavior, strlen(a) would then evaluate to 1, but sizeof(a) would still be 4.

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Yea because strlen measures till the \0 character. But, why didn't C return error of attempting to access array index that was out of bounds? Thank You.
in C there's no array bound checking.
@macfij: At least implementations holding your hands that way are really few and far between (It's not forbidden, and there are some, mostly to test for such errors).
@user3920047 First, the C compiler does not store extra information about a that would allow to know its size. sizeof(a) has been transformed into 4 at compile-time, but when the program is running there is no link between a and 4. Second, the compiler does not generate extra instructions to check that reads and writes from a are in-bounds. Most C compilers generate code that is space- and time-efficient by avoiding the storage of such metadata and the checks that would detect the misuse of a.
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This line

char a[] = "abc";

creates space on the stack for a string of 4 characters. It's the same as doing:

char a[4] = "abc";

When you do:

strcpy(a, "pqrst");

It basically does:

int len = strlen("pqrst") + 1; for (int i=0; i<len; ++i) a[i] = "pqrst"[i];

Clearly, that code will overwrite the bounds of the a array.

Basically though, it sounds like you're expecting C to do extra work for you. That's the opposite of what C will do.

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So true about "... expecting C to do extra work for you. That's the opposite of what C will do."
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Your strcpy-call has a buffer-overrun (the source is two byte longer than the destination), leading to undefined behavior (UB).

Invoking UB means there's nothing left to reason about, on any execution path invoking it (that includes all paths here), even before you get to it.

If you fix the UB, sizeof is evaluated at compile-time for all but VLAs, giving size of the argument: Array of (3 elements "abc" + 1 implicit terminator "\0") char.

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