char* and char[]?

[IanWright]

Just wondering if someone can clear things up for me a little. I'm fairly new to C/C++ and just getting in a bit of a tangle with pointers.

My problem is I have a vector of a defined structure, which needs to store some characters. When I try and print out these characters later, it appears to be uninitialised. I believe the problem is because I'm using a char*, who's memory contents then just gets discarded later on, and hence I can't get the characters anymore.

e.g.

typedef struct _info{
    char* label;
} Info;

a little later on some parsing takes place...

char *value;
 
for (....)
{
  char *label;
  Info info;
  label = value;
  info.label = label;
 
  // info gets added to a vector
}
 

When I come to print info later I get ▌▌▌▌▌▌▌▌▌▌

Is my understanding of this correct? If so, then I guess that label within the Info structure needs to be a char[]. Is there a simple way to then copy the 'value' within the for loop into this char array as my attempts have just led to compiler errors... ?

Thanks very much

[IanWright]

I'm going to reply to my own question for the benefit of anyone else...

I think explaining the problem is right, and the useful little way to do this is to use strcpy...

char dest[50];
char* pointer = "hello world";
 
strcpy(dest, pointer);
 

pointer can now be played around with freely without changing the value in dest.

[weaknessforcats]

A char* is a pointer to a single char.

A char[] is also a pointer to a char but you have to specify the char(s) or it won't compile:

char arr[] ="ABC";  /* OK. arr is an array of 4 char */
char arr1[] = {'A', 'B', 'C'};   /* OK. arr1 is an array of 3 char */
char arr2[];            /* ERROR: No char(s) specified.
 

Read this:

First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:

int array[5];
 

That is an array of 5 elements each of which is an int.

int array[];
 

won't compile. You need to declare the number of elements.

Second, this array:

int array[5][10];
 

is still an array of 5 elements. Each element is an array of 10 int.

int array[5][10][15];
 

is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.

int array[][10];
 

won't compile. You need to declare the number of elements.

Third, the name of an array is the address of element 0

int array[5];
 

Here array is the address of array[0]. Since array[0] is an int, array is the address of an int. You can assign the name array to an int*.

int array[5][10];
 

Here array is the address of array[0]. Since array[0] is an array of 10 int, array is the address of an array of 10 int. You can assign the name array to a pointer to an array of 10 int:

int array[5][10];
 
int (*ptr)[10] = array;
 

Fourth, when the number of elements is not known at compile time, you create the array dynamically:

int* array = new int[value];
int (*ptr)[10] = new int[value][10];
int (*ptr)[10][15] = new int[value][10][15];
 

In each case value is the number of elements. Any other brackets only describe the elements.

Using an int** for an array of arrays is incorrect and produces wrong answers using pointer arithmetic. The compiler knows this so it won't compile this code:

int** ptr = new int[value][10];    //ERROR
 

new returns the address of an array of 10 int and that isn't the same as an int**.

Likewise:

int*** ptr = new int[value][10][15];    //ERROR
 

new returns the address of an array of 10 elements where each element is an array of 15 int and that isn't the same as an int***.

With the above in mind this array:

int array[10] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Wheras this array:

int array[5][2] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Kinda the same, right?

So if your disc file contains

0 1 2 3 4 5 6 7 8 9

Does it make a difference wheher you read into a one-dimensional array or a two-dimensional array? No.

Therefore, when you do your read use the address of array[0][0] and read as though you have a
one-dimensional array and the values will be in the correct locations.

[Rajesh V]

In your post, you have said that

int array[][10]; 
 

won't compile because we need to declare the number of elements.

But, if i do like this,

int array[10][];
 

The compiler gives the error - declaration of ‘a’ as multidimensiona l array must have bounds for all dimensions except the first

Also, when i do like this,

int array[][10] = { { 0 } };

It gets compiled successfully.

Can you please explain what's happening?

[Andr3w]

Well, for the last used declariation with the zero you initialize the variable to be like this:

int index[1][10] = { 0 };
 

Also you don't need a "duo" of curly brackets for that like you used... Anyway I think that weaknessforcats meant the following use:

#include <stdio.h>
 
int Size(int index[][3]);
int main(int argc, char **argv)
{
// the first dimention (currently has a value of 1 can be any value
// the function will accept it, so it would be valid to write
// index[10][3]. the function would be ok with it!
    int index[1][3];
 
    Size(index);
    return 0;
}
 
int Size(int index[][3])
{
    return;
}
 

I hope this example helped you understand what's going on! Should you have any questions post!

[weaknessforcats]

In your post, you have said that


Code: ( text )
int array[][10];


won't compile because we need to declare the number of elements.

But, if i do like this,


Code: ( text )
int array[10][];


The compiler gives the error - declaration of ‘a’ as multidimensiona l array must have bounds for all dimensions except the first

Yep.

int array[10][];
 

can't compile becuse you failed to specify the size of the elements.

[weaknessforcats]

int array[][10] = { { 0 } };

It gets compiled successfully.

Can you please explain what's happening?

You are using initialization syntax. This is the same as:

int array[1][10] = { { 0 } };
 

Rememvber:

int arr[] = {1,2,3,4,5};
 

is the same as:

int arr[5] = {1,2,3,4,5};