this function is returning this error:
I'm trying to be able to scan in a string so that i can use an integer of like 2 numbers for my queue.. as in 10/11/12.. right now i can only use single integer values...Code:q.cpp In function 'int main()': q.cpp:86: cannot convert 'int' to char* for argument 1 to insert char*
here is the source:
i also have my remove function, i'm assuming it still works..Code:void insert(char character_to_be_inserted[NAME_LENGTH]) /************* INSERT ***********************/ { QUEUE_NODE *temp_new_node_pointer; /* Used to point to a newly created node before it is linked into the queue */ temp_new_node_pointer = (QUEUE_NODE *) malloc(sizeof(QUEUE_NODE)); temp_new_node_pointer->queued_data[NAME_LENGTH] = character_to_be_inserted[NAME_LENGTH]; if (Q == NULL) /* This is the Queue_empty condition */ { Q = temp_new_node_pointer; Q->next_in_queue = Q; } else { temp_new_node_pointer->next_in_queue = Q->next_in_queue; Q->next_in_queue = temp_new_node_pointer; Q=temp_new_node_pointer; } } main() /******************** MAIN ******************************/ { ios::sync_with_stdio(); char operators_choice; do { printf("\n Enter choice (lower case is acceptable) --- (I)nsert, (R)emove, or (Q)uit: "); fflush(stdout); operators_choice=getchar(); getchar(); /* Second getchar just discards the carriage return used to enter the oeprator's choice */ switch (operators_choice) { case 'i': case 'I': printf(" Enter character to be enqueued: "); fflush(stdout); insert(cin.get()); cin.ignore(80,'\n'); break; /* return used to enter the enqueued character */ case 'r': case 'R': if (Q == NULL) { printf(" Queue already empty\n"); fflush(stdout); } else { printf(" The character removed was '%c'\n", removed_from_queue()); fflush(stdout); } break; default: if (operators_choice == 'Q' || operators_choice == 'q') break; printf("\n Please enter your choice of 'I', 'i', 'F', 'f', 'Q', or 'q', only.\n"); fflush(stdout); } } while ( !(operators_choice == 'Q' || operators_choice == 'q') ); }
thanks



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