将指向char数组的指针传递给函数

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.

Answer 1

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);
}

Answer 2

*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 );