Problems with type char arrays

[gabyam]

I have to write a simple program for school in which a teacher has to enter the following information on the students: number, name, and GPA. At the end, the program has to print on screen the number of students that have a GPA of 7 or above. Here is what I did:

#include <stdio.h>

int main ()
{
	int num, prom, cont, tot, opc;
	char nombre [20];
		
	cont = 0;
	tot = 0;

	do
	{
		printf("      MENU\n\n");
		printf("1. Insert student info\n");
		printf("2. Show results\n");
		printf("3. Exit\n\n");

		printf("Pick one: ");
		scanf("%d", & opc);

		switch (opc)
		{
			case 1: 
				printf("Number: ");
				scanf("%d", & num);

				printf("Name: ");
				scanf("%s", &nombre);

				printf("GPA: ");
				scanf("%d", & prom);

				if (prom >= 7)
				{
					cont = cont + 1;
					}

				else 
				{
					}
			break;

			case 2: 
				printf("\n\n%d de %d have a GPA of 7 or higher.\n\n", cont, tot);
			break;

			case 3:
				fflush (stdin);
				getchar();
				return 0;
			break;

			default:
				printf("\n\nInvalid option.");
			break;
			}//switch-case
		}//do-while

	while (opc != 3);
}

But the compiler shows the following error:

aprobados.cpp: In function ‘int main()’:
aprobados.cpp:28: warning: format ‘%s’ expects type ‘char*’, but argument 2 has type ‘char* (*)[20]’

I have used arrays once before and I used them just as I did now, and it didn't give me any trouble. Can someone please explain what I did wrong?

[hanst99]

Do away with the & before nombre. Arrays already are pointers, so &nombre is actually a char**.

[gabyam]

Thank you very much! The program works just fine now.