C / C++ / MFC

himangshuS wrote:

char *str="abc";

«_Superman_» wrote:

What I believe is that declaring char* ptr = "abc" is exactly the same as const char* ptr = "abc".

«_Superman_» wrote:

And also when i tried what the OP says it didn't change the value even when giving "ggg".

fun(char* str)
{
   str[0] = 'A';
}

«_Superman_» wrote:

Viral fever

himangshuS wrote:

char *str="abc";
fun(str);
cout<<str;

.
.
fun(char *str)
{
str="gggg" ;
}



but="" its="" still="" showing="" abc...why?
now="" if="" i="" change="" the="" function="" definition="" to="" as="" follow
fun(char="" ;
}<="" blockquote="">

You are altering the pointer value that is passed into fun. Like all other value parameters, that updated value is not passed back out to update the thing whose value you passed in.

If fun was change to either of these, then you could alter the pointer. But

fun(char*& str)
{
   str="ggg";
}

fun(char** str)
{
   (*str)="ggg";
}

Hopefully the string literal "ggg" will stick around outside the scope of fun.

You could alter the string that str points at using this:

fun(char* str)
{
   strcpy(str, "ggg");
}

But I really don't think that's safe either, as (in this case), str points at a string literal, which is likely to reside in read-only space.

I see you're using STL streams...why not go the next step and use STL strings as well. Then you can forget the division between pointer values and what they point at. See the code below - str is obviously a value, would you expect it to be updated by this function?

fun(std::string str)
{
   str = "ggg";
}

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

himangshuS wrote:

fun(char *str)
...
Now if I change the function definition to as follow
fun(char *str)

char str[]= "abc";
//...
void fun(char str[])
{
 strcpy(str, "ggg");
}

prerananit wrote:

code for recognising character pattern in vc++

UINT RecognisePattern(LPVOID p)
{
 //stuff goes here
}