How would I compare two char arrays in an array of structs

A struct is the same as a class, except that a struct's data is public by default.  So a struct can contain methods.

I suggest a comparison method which is a member of the struct.  The comparison method would take a char array and compare to the char array which is also a member of the struct.

here was my initial attempt

if( contact[i+1].name.compare(contact.name[i]) == -1)

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strcmp() will do the job, but the sorting algorithm is another story. I will elaborate in a moment.

qsort() will be the shortest method. first define a comparison function, there you will use the strcmp() function:

#include <stdlib.h>
int comparefn(const void *a, consr void *b){
   return strcmp((struct yourstruct *)a)->name, ((struct yourstruct *)b)->name);
}

then you can use qsort, using the new function as a callback:
qsort(arrayOfStruct,sizeof(arrayOfStruct)/sizeof(structType),sizeof(structType),comparefn);

using your code of previous questions, should be:

#include <stdlib.h>
int comparefn(const void *a, consr void *b){
   return strcmp((struct yourstruct *)a)->name, ((struct yourstruct *)b)->name);
}

then you can use qsort, using the new function as a callback:
qsort(contact,sizeof(contact)/sizeof(struct contactParameter),sizeof(struct contactParameter),comparefn);

notice my solution uses directly the char arrays (name), and doesn't require STL functions.

>> and doesn't require STL functions.

You say that like it's a bad thing. It's not ... STL is the standard template library, and is part of C++ ... not using it is passing up on a great tool.

I would suggest to use STL functions, if the struct has STL objects, but it is not the case.
I always try to avoid mixing libraries as far as possible.

Would this not work?

 strcmp(contact.name[i],contact.name[i+1])

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Yes, thanks so much, it compiled!

>> Would this not work?

If name is an array of strings, yes, but I suspect that name is a string (array of characters), so you just do :

        strcmp(contact[i].name, contact[i+1].name)

to compare the name of contact i with the name of contact i + 1.