Wanna discuss in the field of programming

Discuss? You have problems or have solutions?

ok start!

ल ल सुरु गरुम् न सुरु गरुम्।

char* p;
char* q;

p = "pqrst\0";
q = "abcde\0";

while(*p==*q)
{
p++;
q++;
}

should be the same as:
while(*p++=*q++);

But this Damn visual C++ is giving a link error..Why??

bro run on turbo,where are you giving command to print value.

I'm trying to copy q into p. Logically it should work as the length of both strings are the same and at the end '\0' is passed into p which is ascii 0 and the loop should stop. But it isn't working on VCPP. I don't know if that's a microsoft problem or a bad programming practice. So just wondering if anybody else gets the desired results using a different compiler.

Pprobably you mean a run-time error. One typo you have! Correction
while(*p++ == *q++);

>>testdirector
provided that he wants to copy q into p ,I don't think that's a typo. But with the same supposition, his first code snippet should be
while(*p=*q){
p++;
q++;
}

>>sojodude
I haven't tried it in Microsoft's compiler but it should not give you a *link* error. It should rather end up in a segmentation fault (runtime error) [ I don't think Microsoft's compiler is so smart as to detect a runtime error at the linking stage ].
Reason:
p points to "pqrst\0" but the latter being a string constant is placed in that part of the memory which can only be read but can't be written. An attempt to write on it will surely cause a runtime error.

Btw, when defining a string constant you do not explicitly need to type the trailing "\0". The compiler will place a \0 at the end of the characters in double quotes.

And here is a code snippet that copies q into p, you should point p to a region in the memory which can be both read and written. An array or dynamic allocation with malloc will do.

char *p=(char *)malloc(6);
char *q="abcde";
char *r = p;

while(*p=*q){
p++;
q++;
}

printf("%s",r); //will print "abcde"

well, I failed to understand what he wants; assuming he wants a copy which he expressed in his second posting, there is a discrepancy between the original two examples. It will be manifested by different behaviours (run-time error) in the two cases during the run. Now you want to call it a typo in the first example or the second, you are free to judge (I did not read his second posting).
But here is what I have to say : when you say, p points to a constant area to which a write fails(takes an exception) is not a valid "C" statement (I do that all the time). Although the content of the string is constant, the "abcde\0" is actually a free pointer equivalent to "abcde", a constant string, i.e. a constant pointer to a string of constant letters. Can the letters be changed, yeah sure, any time. Can the address of the string be changed during the run, no, never, but can the pointer p be pointed to another address, yes, sure, again any time (that's what you are doing).

Your second example is redundant; the first is good enough.

>>test director

>when you say, p points to a constant area to which a write fails(takes an exception) is not a valid "C" statement (I do that all the time).

Where did I say that ?

>Can the letters be changed, yeah sure, any time.

Can you ? Will appreciate if you write a sample program, you OS (with version) and your compiler (With version).

>Your second example is redundant; the first is good enough.
Ya I will accept that it was indeed redundant if you can write a program that changes the letters in a string constant.

I actually found this question at

http://www.acetheinterview.com/cgi-bin/answers.cgi?action=answers&number=5&topic=000017.ubb&q_id=17

And everybody there seems so confident that the piece of code is correct. So I was a little surprized when it didn't work.

Arch, I still get the runtime error. Sorry by link I meant runtime. This is what I had.

char* p;
char* q;

p = (char*)malloc(6);
q = (char*)malloc(6);

q = "abcde";

while(*p=*q)
{
p++;
q++;
}
char* r;
r = p;
cout << "\nNew p is: " << r << "\n";

>>sojodude

char* p;
char* q;

p = (char*)malloc(6);
q = (char*)malloc(6); // u don't need this 'coz u have pointed q to a different address in the next line. The memory u just allocated becomes completely useless.

q = "abcde";

while(*p=*q)
{
p++;
q++;
}
char* r; //
r = p; //move these lines before the while loop because at this point, pointer r points to a memory region that points to the end of "abcde".

Well, you've already done so by correcting sojodude's first example. If you insist, I have to ask you to compile and run (on any OS , any good compiler) it, it is guaranteed to work.
Here it is:

#include
int main()
{

char *p, *q, *st;

st = p = "abcde";
q = "pqrst";

while(*p++=*q++);
printf("%s", st);
return 0;
}

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#include "stdio.h"
int main()
{

char *p, *q, *st;

st = p = "abcde";
q = "pqrst";

while(*p++=*q++);
printf("%s", st);
return 0;
}

The malloc version from arch119 would work.
While Testdirector version with:
--- st = p = "abcde";
--- q = "pqrst";
--- while(*p++=*q++);
should not work. It should compile fine though. You are trying to modify a string constant. Thats dangerous. A good OS should detect that danger. Thats how some hackers crack system.

GN

Well Gidilat, I ask you to prove your point. Hackers? Define hackers, otherwise hackers have nothing to do with constant strings, believe me.

Let us change the matter a little and use arrays.

#include "stdio.h"

int main()
{
char p1[] = "abcde";
char q1[] = "pqrst";
char *p, *q,*st;

st = p = p1;
q = q1;

while(*p++ = *q++);
printf("%s", st);
return 0;
}

Will it work?

what about using const operator.

#include

const int array[] = {1,2,3,4,5}; /* tell compiler to generate a protected area for it*/

int main()
{
int *intp;
int i;

intp = (int *) array;
for (i = 0; i < 5; i++)
intp[i] ++; /* Access violation, the compiler does not detect it,
* the system (processor +OS) may detect it */
return 0;
}

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