C Outputting array from another function - c

Question

I am trying to copy the array winner from my function 'enter', so that i am able to just output it on the 'previous' function. When picking the option for the previous option I have gotten nothing outputting. Its only the last function named 'previous' that is not working, but to produce the problem the majority of the code is needed.
#include <stdio.h>
#include <string.h>
char enter(char names[][20]);
void menu();
void previous(char winner[][8]);
int main()
{
char names[16][20];
int i;
printf("Please enter the names of the players:\n");
/*Making the user enter 16 times*/
for (i = 0; i < 16; i++)
{
scanf("%9s", &names[i]);
fflush(stdin);
}
/*Clearing Screen*/
system("cls");
menu(names);
return names[16][20];
}
void menu(char names[][20], char winner[][8])
{
int choice;
printf("Please select one of the following options:\n\n"
"Press 1 to enter game results\n"
"Press 2 to display the current round\n"
"Press 3 to display the players advancing to the next round\n"
"Press 4 to display the previous round\n"
"Press 5 to exit the program\n");
scanf("%d", &choice);
if(choice == 1)
{
enter(names);
}
system("cls");
if(choice == 3)
{
previous(winner);
}
}
char enter(char names[][20])
{
int result;
int score1;
int score2;
int p, c, j, l, i;
char winner[8][8];
system("cls");
for(i = 0; i < 8; i++)
{
printf("\n\n%s vs %s",names[i],names[i+8]);
score1 = 0;
score2 = 0;
for(j = 0; j < 5; j++)
{
printf("\n\nEnter game %d results, press 1 if %s won or"
" 2 if %s won :\n",(j+1), names[i], names[i+8]);
scanf("%d", &result);
if(result == 1)
{
score1++;
}
if(result == 2)
{
score2++;
}
printf("\n\n1Current score is %d-%d", score1, score2);
if(score1 == 3)
{
printf("\n\n%s adavances to the next round!",names[i]);
strncpy(winner[i], names[i], 10);
printf("\n\nPress Enter to Continue");
getch();
system("cls");
break;
}
if(score2 == 3)
{
printf("\n\n%s adavances to the next round!",names[i+8]);
strncpy(winner[i], names[i+8], 10);
printf("\n\nPress Enter to Continue");
getch();
system("cls");
break;
}
}
}
system("cls");
printf("The players advancing to the next round are:\n\n");
for(p = 0; p < 8; p++)
{
for(c = 0; c < 8; c++)
{
printf("%c",winner[p][c]);
}
printf("\n");
}
printf("\n\nPress Enter to Continue");
getch();
system("cls");
menu(names, winner);
return winner[8][8];
}
void previous(char winner[][8])
{
int i, j;
for(i = 0; i < 8; i++)
{
for(j = 0; j < 8; j++)
{
printf("%c",winner[i][j]);
}
printf("\n");
}
}

Answer 1

There is no data for the array winner in your program! At least not when you call it for the first time.
The signature for the menu function is:
void menu(char names[][20], char winner[][8]);
but you call it from main like this:
menu(names);
The winner parameter is missing. This shouldn't happen, but you have declared a prototype for this function, namely:
void menu();
Unfortunately, C treats the empty parens as meaning "whatever parameters you pass", not as function that takes no parameters. That means that your function call slips by. The fix is to provide the correct signature for the prototype and also to pass a suitable winner array from main.
Strangely, your enter function provides a local array winner. This array will always be a new array when you call enter. That's probably not what you want. As is, your program should have one names and one winner array. (You can pass these arrays around, but you should make sure that tese arrays are consistent. Don't create new arrays when you really want to operate on existing ones.)
You also call your menu recursively. That means the you go ever deeper into the call structure without real benefit. Dont do that; use a loop instead: do display the menu while the user hasn't chosen "quit". (There are applications for recursive functions, but this isn't one.)