Passing 2D array of strings to a function

Question

I can initialize a 2D array of strings using the following:

const char * const aListOfStrings[] {"test","test1"};

However, if I want to do the same thing when calling a function, such as:

void function( char const * const aListOfStrings[] ) {}

void CallWithArrayOfStrings()
{
    function( {"test", "test1"} );
}

The compiler politely refuses with the following error:

error: cannot convert '<brace-enclosed initializer list>' to 'char**'

I am able to pass a single string to a function as in:

void function( char const *someString ) {}

void CallWithString()
{
    function( "test" );
}

I seem to understand how to initialize arrays when they are variables, but apparently I don't seem to understand how it works in the context of calling functions. I have always assumed that initializing a variable is the same as initializing a function parameter? Have I been living a lie my whole life? ;)

Thank you

UPATE: Removed C tag. C arrays require an '=' sign before the brace initializer, and the error is different. It appears the problem is very similar however... Also fixed the closing quote.

const char * const aListOfStrings[] {"test","test1}; you missed double quotes at the end. — Krishna Kanth Yenumula
– Krishna Kanth Yenumula, Commented Dec 25, 2020 at 2:54

Jarod42 · Accepted Answer · 2020-12-25 13:39:22Z

1

const char * const aListOfStrings[] {"test","test1"}; is a shorthand for

const char * const aListOfStrings[2] {"test","test1"};

Size is deduced.

But arguments of signature are fixed (unless template is used), moreover you cannot pass C-array by value, so:

void function(char const * const aListOfStrings[]) is equivalent to

void function(char const * const* aListOfStrings).

To have your syntax, you might use the (strange) syntax

void function(char const * const (&aListOfStrings)[2]);

Demo

or the template version

template <std::size_t N>
void function(char const * const (&aListOfStrings)[N]);

Demo

answered Dec 25, 2020 at 13:39

Jarod42

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3 Comments

Kevin Stallard Over a year ago

Thanks for taking time Jarod in answering this, but I need to keep the size deduction. Casting works (see my comment above).

Jarod42 Over a year ago

Template version allows deduction (see link demo). With your casting, you don't have size information...

Kevin Stallard Over a year ago

That makes sense. I did not think it through...the aListOfStrings[] as a parameter can't know how big it is at compile time, the only way is using a template. Thanks @Jarod42

Jack · Accepted Answer · 2020-12-25 03:07:34Z

1

That's because your compiler is not guessing the type of your inline argument, passed to function, so you must explicitly declare it before, eg:

const char* data[] = { "test", "test1" };
function(data);

But C++ has std::array, std::initalizer_list, std::vector, there's no reason to use an array of pointers to char unless you have a specific requirement.

answered Dec 25, 2020 at 3:07

Jack

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1 Comment

Kevin Stallard Over a year ago

I do have a specific requirement. Using a container isn't an option for this specific instance. Is it possible to cast it in some way? And just answered my own question: function2( ( char const * const []){"test","test2"} ); does the trick nicely.

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