Why does my nested if condition is not working properly in c? - arrays

Question

Here is my assignment:
Ask the user to input their name.
If their name ends with a vowel, ask them to input their age as well.
If their age is even, print Wow, you're special!. Otherwise, if their age is odd, ask them to input their birth year.
If their birth year is even, print Oh, you're still special!. Otherwise, print You will be special next year..
If their name ends with a consonant though, print You're awesome!
Here is my code:
#include <stdio.h>
#include <string.h>
int main() {
// variables
char name[100];
int age, birth;
printf("Enter name: ");
scanf("%[^\n]s", &name);
// formula for getting last letter of the name
int size = strlen(name);
int last = size - 1;
if (name[last] == 'a' || name[last] == 'e' || name[last] == 'i' ||
name[last] == 'o' || name[last] == 'u') {
printf("Enter age: ");
scanf("%d", &age);
if (age %2 == 0) {
printf("Wow, you're special!\n");
} else {
printf("Enter birth year: ");
scanf("%d", &birth);
}
if (birth % 2 != 0) {
printf("You will be special next year.\n");
} else {
printf("Oh, you're still special!\n");
}
} else {
printf("You're awesome!\n");
} // end of nested if
return 0;
}
The problem is with the birth variable: the age and birth variables seems to be connected, because when the condition of age executes the condition of birth executes as well leading to a multiple executions of conditions.

Answer 1

"%[^\n]s" is a typo. %[ and %s are two separate scanf specifiers - you do not combine them as such. What you have is a %[ specifier that reads characters until a newline, and then the format string attempts to match a literal s character.
An unbound %[ is as dangerous as gets. You most provide a maximum field-width specifier that is at most the size of your array minus one (e.g., %99[^\n]) to limit the amount of data scanf reads, and prevent a buffer overflow.
&name would be of type char (*)[100], that is a pointer to an array. %[ expects a char *, simply a pointer to char. Passing an array to a function causes it to decay to a pointer to its first element - you do not need the & operator here.
You must test the return value of scanf is the expected number of conversions, otherwise you are operating blindly, and may utilize uninitialized or otherwise indeterminate values.
You should initialize birth, otherwise its value is indeterminate. In the event that age is even, you would attempt to use this indeterminate value in birth % 2 != 0.
Failing to correct most of the above issues will lead to problems, including Undefined Behaviour.
A safer example program to study:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char name[100];
int age;
int birth = 0;
printf("Enter name: ");
if (1 != scanf("%99[^\n]", name)) {
fprintf(stderr, "Could not read <name>.\n");
return EXIT_FAILURE;
}
size_t length = strlen(name);
if (!length) {
fprintf(stderr, "<name> is empty!\n");
return EXIT_FAILURE;
}
if (!strchr("aeiou", name[length - 1])) {
puts("Name does not end in a vowel.");
return EXIT_SUCCESS;
}
printf("Enter age: ");
if (1 != scanf("%d", &age)) {
fprintf(stderr, "Could not read <age>.\n");
return EXIT_FAILURE;
}
if (age % 2 != 0) {
printf("Enter birth: ");
if (1 != scanf("%d", &birth)) {
fprintf(stderr, "Could not read <birth>.\n");
return EXIT_FAILURE;
}
}
printf("NAME: <%s> AGE: <%d> BIRTH: <%d>\n", name, age, birth);
}

Answer 2

The conditions expressed in the assignment imply that you only test the birth year if the age is even... otherwise you do not even ask for the birth age. Your tests are not properly nested.
Note that scanf("%[^\n]s", &name); is incorrect:
you must tell scanf() the maximum number of characters to store to the destination array to avoid a potential buffer overflow;
&name is not the expected type, you should just pass name, which is an array that is converted automatically to a pointer to its first element when passed as an argument or used in an expression (except as an argument to sizeof or alignof);
the s in the format is meaningless. Use scanf(" %99[^\n]", name) instead;
test the return value to detect invalid or missing input and avoid undefined behavior later in the program.
Also note that letters can be uppercase and the last character may be neither a vowel nor a consonant (eg: X Æ A-12).
Here is a modified version, more consistent with the assignment:
#include <stdio.h>
#include <string.h>
int main() {
// variables
char name[100];
int age, birth;
printf("Enter name: ");
// read up to 99 characters, skipping initial white space
if (scanf(" %99[^\n]", name) != 1) {
fprintf(stderr, "invalid or missing input for name\n");
return 1;
}
// formula for getting last letter of the name
size_t len = strlen(name);
if (len == 0) {
fprintf(stderr, "invalid empty name\n");
return 1;
}
char last = name[len - 1];
// test if last character is a vowel (trailing 'y' is a vowel)
if (strchr("aeiouyAEIOUY", last)) {
printf("Enter age: ");
if (scanf("%d", &age) != 1) {
fprintf(stderr, "invalid or missing input for age\n");
return 1;
}
if (age % 2 == 0) {
printf("Wow, you're special!\n");
} else {
printf("Enter birth year: ");
if (scanf("%d", &birth) != 1) {
fprintf(stderr, "invalid or missing input for birth\n");
return 1;
}
if (birth % 2 == 0) {
printf("Oh, you're still special!\n");
} else {
printf("You will be special next year.\n");
}
}
} else {
// the last character is not a vowel, test if it is a consonant
if (strchr("bcdfghjklmnprstvwxzBCDFGHJKLMNPQRSTVWXZ", last) {
printf("You're awesome!\n");
}
}
return 0;
}

Answer 3

of course you get two outputs, your if statements regarding age and birth are parallel. Shouldn't your birth block be nested inside of the else block for age?