Passing pointer to char array to function

Question

i had a little problem trying to pass the address for a character array to a function, here is a simple example of what i am trying to do :

char a[20] = {"hello"};
printit( &a );

can you please give me the declaration of he printit function ( and maybe why ), i was expecting something like :

void printit( char ** value );
or void printit( char * value[] );

to work, but it is not.

*Error messages :

void printit( char ** value ); => cannot convert parameter 1 from 'char (*)[20]' to 'char **'
void printit( char * value[] ); => cannot convert parameter 1 from 'char (*)[20]' to 'char *[]'

thanks in advance.

Regards, max.

Just telling us that it didn't work is not helpful. You need to tell us exactly what you tried, exactly what you expected it to do, and exactly what happened. — David Schwartz
– David Schwartz, Commented Jun 11, 2013 at 9:22
For a print function of a string, the only declaration that makes sense is this: void printit (const char* str);. It is not clear to me whether printit is your own user-defined function or part of a library, however. — Lundin
– Lundin, Commented Jun 11, 2013 at 9:28
@Max Maybe you wanted this. Check out the selected answer. &t is a pointer to the array as whole, hence int (*)[10] — Suvarna Pattayil
– Suvarna Pattayil, Commented Jun 11, 2013 at 11:50

Nicholaz · Accepted Answer · 2013-06-11 09:27:28Z

7

Your parameter &a is a pointer to an array of 20 chars, hence:

void printit(char (*value)[20]);  // value is a pointer to an array of 20 chars

.

However, more commonly (especially with strings) would be to change the call into

printit(a);   // a will be passed as pointer to first elelemt, i.e. 'a' can be used as pointer to char

and define printit as

void printit(char *value)
{
     printf("The string is: %s", value);
}

answered Jun 11, 2013 at 9:27

Nicholaz

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2 Comments

Max Over a year ago

as i understand , char a[2];// a is a pointer to the beginning of the array , so why do i have to give the size in the function declaration, and is there a way to pass it without the [20]

Nicholaz Over a year ago

if you just pass 'a' as the parameter (as in my 2nd example) you can use it like a pointer to the first element of the array. If you pass &a you need the [20].

RFon · Accepted Answer · 2013-06-11 09:24:41Z

1

*Edited .. made a mistake

You are not declaring an array of char* you are declaring an array of char. Adding asterisk to your variable declaration and removing the & should make it work:

void printit(char** arr){
  string tmp(arr[0]);
  cout<<tmp<<endl;
}


char* a[20] = {"hello"};
printit( a );

answered Jun 11, 2013 at 9:24

RFon

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