a question from C++ primer . about Dynamically Allocated Arrays

[hzesen1221]

c++ primer says:
// C-style character string implementation
const char *pc = "a very long literal string";
const size_t len = strlen(pc +1); // space to allocate
// performance test on string allocation and copy
for (size_t ix = 0; ix != 1000000; ++ix) {
char *pc2 = new char[len + 1]; // allocate the space
strcpy(pc2, pc); // do the copy
if (strcmp(pc2, pc)) // use the new string
; // do nothing
delete [] pc2; // free the memory

we can count that string "a very long literal string" has 26 words, so len will be 25. but lets see this: char *pc2 = new char[len + 1]; . it means the size of pc2 will be 26. and then do the copy: strcpy(pc2, pc); . if we do so, we will only copy the first 26 words from pc to pc2, the remaining '\0' wont be a part of pc2. why do we do like this?? i think we should write "const size_t len = strlen(pc);" instead of "const size_t len = strlen(pc +1);" . am i right?

Thank you.

[hamsterman]

yes, you're right!
len(pc+1) would get the lingth of string " very long literal string".

[hzesen1221]

sorry im a little confused.. but is len() a function? thanks!

[brokenbot]

When posting code, please use the code brackets "ur code"


and yes it is a function... When i first started learning, the use of the "()" kinda confused me.

If anything has that in programming, that usually means it's a function

[hzesen1221]

yes and im sorry about that.. but in my code, len is just a variable,, not a function.. am i right?? i was just confused by that..

[tummychow]

len is not a function. It is a variable. But it is defined by a function.
Now then why is len going to be 25? It's going to be 26 because there are 26 characters. Then you must add one for the null.

[moorecm]

I think you have it right with len equaling 25.

const char *pc = "a very long literal string"; // 26 characters + 1 null
const size_t len = strlen(pc +1);              // 25 = strlen(" very long literal string") 

strcpy includes the null when it copies. That means that I would expect this to write beyond the end of the array:

char *pc2 = new char[len + 1];                 // allocate 25 characters + 1 null
strcpy(pc2, pc);                               //     copy 26 characters + 1 null 

References:
http://www.cplusplus.com/reference/clibrary/cstring/strlen/
http://www.cplusplus.com/reference/clibrary/cstring/strcpy/

[hzesen1221]

thats just why i asked this question. why did the author put the null beyond the end of the array? the array would not be a c-style string because he did so.. according to c++ primer , never forget the null after your c-style strings. maybe the author just wanna illustrate something, but he should explain clearly... beginner like me are always confused by unclear explanations like this.

[blackcoder41]

edit:
i have deleted my post for it is useless..

[guestgulkan]

The problem here is this particular line (as moorecm has aleady mentioned):
const size_t len = strlen(pc +1); // space to allocate
That would give an incorrect value for the length of the string. It will be one short.

So this line:
char *pc2 = new char[len + 1]; // allocate the space does not allocate enough characters.
So the string copy will cause a buffer overflow (heap corruption).


I checked the C++ primer book and hzesen1221 has copied the code the code as given in the book.

[moorecm]

I checked the C++ primer book and hzesen1221 has copied the code the code as given in the book.

Excellent. I was wondering when I replied before (no offense OP).