problem in char* &

[ibtkm]

hi every body
in this code

void toPointer(char a[],char* &b[])
{
for(int i = 0;i < sizeof(a) / 4;i++)
b[i] = &a[i];

}

i have error(this error is for char* &b[])
how can i fix it?

[aidyszh]

Use char* b[] is ok.

[ibtkm]

it's important for me to use char* &b[]
because i want to change b[] in main function

[aidyszh]

Use char* b[] can keep the change made in function toPointer().

[guestgulkan]

void toPointer(char a[],char* &b[]) //ERROR - arrays of references not legal

[ibtkm]

my problem is that i want change b array in main function
all of you know if we use (char* b[]) char* b in main function doesn't change
so we must use &.
how can i use this & for fix this problem?
tnx a lot

[filipe]

ibtkm wrote:

all of you know if we use (char* b[]) char* b in main function doesn't change

Why would you think that? As long as you're passing a non const pointer OR reference to the object, there's nothing stopping you from writing into it.

[guestgulkan]

You are confused about how C/C++ passes arrays between functions..
In C/C++ arrays are passed by reference NOT by copy
So as aidyszh already said you just do char* b[]

Another thing - this also isn't right (pt is a follow on of your array confusion)
for(int i = 0;i < sizeof(a) / 4;i++)
because you are not passing a copy of the a array - C/C++ actually passes a pointer -
so sizeof(a) is actually calculationg the size of a char*

And why the division by 4?


if you look in the articles section there is a article all about arrays.

[everid]

I understand OP's question, at least i think i understand. but might be not quite.
1. if OP wants to change the pointer itself (not the value that the pointer points to), a char*&, or char** is a must.
2. in OP's case, it doesn't work, because no way to take a reference of an array in C++.
3. an alternative is to put your array dynamically on the heap, rather than statically on stack. then you can use char** .
4. for 3. OP need a smart pointer to handle the ownership issue, if necessary.

void seearr(int* a[]){
  cout<<&a<<endl;
}

int main(int argc, char* argv[]){
  int b1;
  int b2;
  int* ap[]={&b1, &b2};
  cout<<&ap<<endl;
  seearr(ap);
}

main::ap and seearr::a, have definitely different address, a is only a copy of ap.
but, there is no way to change ap if it is an static array, that's why no reference to array is allowed.
you need to use "new".

[guestgulkan]

everid wrote:

2. in OP's case, it doesn't work, because no way to take a reference of an array in C++.

You can a reference to an array - you cannot have arrays of references

int (&ref_to_array) [6]; //OK - reference to an array_of_6_integers.

int &ref [6]; //ERROR - this is an attempt to create an array of 6  references_to_integers

[everid]

i need to add: "when passing as a function argument."