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Char**

Started by gimp Apr 24, 2004 at 4:23 AM 2 replies 300+ views

Original Post

Apr 24, 2004 at 4:23 AM

I''m sure this is grossly simple: I''ve never come across char** except when recieved from main(). I''m writing a test case for some code that accepts this data. I can''t figure out how to write some test data for this function. So, how would I create a char** block of data in my code? What does it look like? Many thanks,

Jingo

Apr 24, 2004 at 4:29 AM

Well you are pretty much making an array of pointers to arrays.


void func(char** parray, unsigned int size);

char array[10], array2[10];
char *array3[2];

array3[0]=array;
array3[1]=array2;

func(array3, sizeof(array3)/sizeof(char*));

[edited by - Jingo on April 24, 2004 5:36:31 AM]

gimp

Author

gimp

Apr 24, 2004 at 4:36 AM

Thanks Jingo, the compiler liked it:


char* String1 = "Key4=Value4";
char* String2 = "Key5=Value5";
char *Elements[2];
Elements[0]=String1;
Elements[1]=String2;

[edited by - gimp on April 24, 2004 5:37:30 AM]

Jingo

Apr 24, 2004 at 4:46 AM

You should note that char* String1 = "Key4=Value4"; is a deprecated conversion in C++. It should be either of


const char[] string1 = "Key4=Value4"; 
const char* string2 = "Key4=Value4";
//Elements should be changed to something like

const char* Elements[2];

But using those will no doubt give you problems passing it to the char** function. The function should be changed to accept a char const** parameter.


void func(char const** parray, unsigned int size);

const char array[] = "eg", array2[] = "eg";
char const* array3[2] = {array, array2};
func(array3, sizeof(array3)/sizeof(char*));

[edited by - Jingo on April 24, 2004 5:48:53 AM]

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