LL Search Function

I'm writing a simple LL search function, which will search through the linked list for a Name. If found, it will print out the Name and ID associated with the name, and continue searching.

I stepped through the function in VStudio and it recognizes what it is comparing and the input, but it is not saying they are equal for some reason, and I cannot figure out why.

Code:

void LList::search(char Name[]) {
	int num_found=0;
	People * temp;//allocate memory for struct
	temp = first;//fill struct with contents of first struct in list
	if (entries == 0) {
		cout << "List is empty, nothing to search!\n";
		return;
	}//if
	cout << "Search Results for "<< Name << ":\n";
	while (temp != NULL) {
		if (temp->Name == Name) {//if name found
			cout << temp->Name << " " << temp->ID << endl;
			num_found++;
		}//if
		temp = temp->Next;//traverse the list
	}//while
	cout << "Found " << num_found << " entries.\n";
}//search

I've been testing things with simple statements:
list.add_entry("A", 1);
list.search("A");

My add_entry, print_list, delete_entry functions all seem to work fine, but this simple one has something going on I am having trouble finding. Any ideas?

Code:

void LList::search(char Name[]) {
	int num_found=0;
	People * temp;//allocate memory for struct
	temp = first;//fill struct with contents of first struct in list
	if (entries == 0) {
		cout << "List is empty, nothing to search!\n";
		return;
	}//if
	cout << "Search Results for "<< Name << ":\n";
	while (temp != NULL) {
		if (temp->Name == Name) {//if name found
			cout << temp->Name << " " << temp->ID << endl;
			num_found++;
		}//if
		temp = temp->Next;//traverse the list
	}//while
	cout << "Found " << num_found << " entries.\n";
}//search

You can't compare c-style strings (null terminated char arrays) using operator==. You need to use strcmp. You probably would benefit from using std::string containers instead which would support the use of operator==.

You're comparing pointers, not the data that is "pointed to".

Soma

Hah duh, thanks.