How to make a wrong user input not bug the program?

Question

Example, I declared the menu option as int, then the ear bug goes in there and type in a letter or a special character and this will make the program crash. The question is: is there any way to solve this?

Example:

int opc;
printf("Digite uma opção");
scanf("%d", &opc);

When the guy type letter, broken number or special character will bug.

Answer 1

Checks the amount returned by scanf():

if (scanf("%d", &opc) != 1) /* erro */

Even better to use fgets() followed by sscanf():

char tmp[20];
if (!fgets(tmp, sizeof tmp, stdin)) /* erro */;
if (sscanf(tmp, "%d", &opc) != 1) /* erro */;

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Edit

In the latter code I entered the variable tmp to have a temporary place to store the user input in characters until it becomes a number.

The fgets() reads those user characters and puts them in tmp. The amount returned by fgets() is the address of tmp, except in case of error when fgets() return NULL. Is this what the if check, if there was error during processing fgets().

if (!fgets(...)) /* ... */;
// estas duas pseudo instruções são iguais
if (fgets(...) == NULL) /* ... */;

The sscanf() works like the scanf(). The difference is in the input. The scanf() receives input from stdin (keyboard); the sscanf() receives input from a string (the string the user has written and is stored in tmp).
In either case the value that the scanf() (or sscanf()) return is the value of completed assignments, or EOF in case of error.
In our case we have only 1 possible assignment, so we have only 3 possible return values for the sscanf(). Or return 1 which means that the instruction went well he assigned the value to the variable opc; or return 0 which means that you could not assign value (for example, you typed "twenty"); or devovle EOF to indicate internal error of the function.
In other words, anything except 1 we don’t care.

int chk = sscanf(tmp, "%d", &opc);
if (chk == EOF)  /* erro interno */;
if (chk == 0) /* input com erro */;
if (chk == 1) /* tudo ok */;