Thread: Can someone help me? Probably noob question

Hi guys, i started programming recently and i have made a program recently which is pretty simple:

Code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
char nome1[100];
char decision[100];
int age1;


printf("Welcome to the Hex Project\n\a");
printf("What is your name?\n");

scanf(" %s", &nome1);
printf("Hi %s. This is my first project i hope you like it!\n", nome1);

printf("How old are you?\n");
scanf(" %d", &age1);

printf("So, are you willing to check my program with a lot of bugs? (yes/no)?\n");
scanf(" %c", &decision);

if(age1 >= 18){

if(decision == 'yes'){
printf("Thank you for your time and welcome!");

}

}else{
printf("There is nothing to see here!\n\a");

}


return 0;

Can someone tell me why when i run the program and i awnser yes to the question "So, are you willing to check my program with a lot of bugs? (yes/no)" why it does not appear "Thank you for your time and welcome" . Thank you for your time ^^

Some additional comments:

Code:

scanf(" %s", &nome1);

You don't need the '&' here. Since you're passing an array, the array name by itself will act as a pointer to its first element.

Also note that you're not limiting the input, so this line of code is susceptible to a buffer overflow. One way to limit the number of characters "scanf()" will read would be to hard-code the limit:

Code:

scanf("%99s", ...);

I would suggest, however, you use a function more appropriate for reading a string, such as "fgets()": FAQ > Get a line of text from the user/keyboard (C) - Cprogramming.com

Code:

char decision[100];

// ...

scanf(" %c", &decision);

"%c" is not what you use when you want to read a string, as you have already shown you know.