syntax for array of pointers in C++

Question

So, I see that the practice for dynamic allocating of an array of pointers looks like this:

int **array = new int *[10];

And indeed, using syntax:

int *array[] = new int *[10];

results in error:

/Users/Malachi/Projects/playground/playground.gcc/src/pgccc-5/main.cpp:8: error: definition of variable with array type needs an explicit size or an initializer
    const char* test_array[];
                ^

I'm always more comfortable using pure pointer syntax anyway. However, what bothers me is lines like this:

int main(int argc, char *argv[])

are valid. I'm accustom to empty array brackets [] more or less aliasing out to a pointer type. It seems to me char *argv[] is subject to almost exactly the same constraints as my int *array[], so why is the syntax permitted in one scenario but not the other?

EDIT: It appears the simpler case of int array[] = new int[10] exhibits the same behavior

"so why is the syntax permitted in one scenario but not the other?" Because that is what the C designers decided in the 1970's. Not the best idea. — Bo Persson
– Bo Persson, Commented Apr 21, 2017 at 22:31

Edgar Rokjān · Accepted Answer · 2017-04-21 22:45:32Z

1

This one:

int *array[] = new int *[10];

is not a valid syntax. The reason the left side has a type of an array of pointers to int, and the right side has a type of a pointer to a pointer to int. So the assignment is not legal due to the different types of left and right sides.

On the other hand, arrays decay into pointers. It means, that when you declare a function in the form of:

void foo(int* arr[])

the compiler sees it as:

void foo(int** arr)

The rule above applies only for functions, but not for assignments like in the first example.

answered Apr 21, 2017 at 22:45

Edgar Rokjān

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Malachi Over a year ago

So, if the designers of this paradigm chose to make int* arr[] everywhere represent a "sparse" array (aka pointer to a pointer to int) then we'd have a conundrum what syntax to use to declare an array of pointers to int on the stack?

luk32 · Accepted Answer · 2017-04-21 22:46:48Z

0

It's because function parameter declaration is something different than variable declaration.

An array can decay into a pointer for the first dimension.

You can explicitly express that function expects an array rather than a pointer through the declaration using [] notation in e.g int main(int argc, char *argv[]). They type doesn't matter:

void f(int* i[]) {}

is legal as well. This says "I want an array of pointers to ints". This is more expressive than:

void f(int** i) {}

edited Apr 21, 2017 at 22:46

answered Apr 21, 2017 at 22:40

luk32

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R Sahu · Accepted Answer · 2017-04-21 22:47:01Z

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I'm accustom to empty array brackets [] more or less aliasing out to a pointer type.

That's valid only in the declaration of a function argument.

void foo(int a[]);

is the same as:

void foo(int* a);

However, when declaring or defining variables, they are not the same.

int a[] = {1, 2, 3}; // Valid. Array of 3 ints

is not the same as

int* a = {1, 2, 3}; // Invalid syntax.

Exception

You can use a string literal to intialize a char array or char const*.

char s1[] = "string 1";
char const* s2 = "string 2";

However, you can't use (not in C++ anyway):

char* s2 = "string 2";

edited Apr 21, 2017 at 22:47

answered Apr 21, 2017 at 22:39

R Sahu

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