Output not showing in C - c

Question

I'm writing a code that must identify the letter 't' or 'T' in a word, before or after the middle of it.
If the first half of the word does contain a 't' or a 'T', the program should output a 1. If the first half does not contain the letter 't' or 'T', but the second half does, then the program should output a 2. Otherwise, if there is no 't' or 'T' in the word at all, the program's output should be -1. The word entered will not have more than 50 letters.
#include <stdio.h>
#include <string.h>
int main() {
char word[50];
int i = 0, length, t = 0, T = 0;
scanf("%s", word);
length = strlen(word);
t = word[i] == 't';
T = word[i] == 'T';
while(!t || !T) {
if((t || T) && i <= length / 2) {
printf("%d", '1');
} else if((t || T) && i > length / 2) {
printf("%d", '2');
//}else{
// printf("%d", '-1');
}
i++;
}
return 0;
}
If I enter any word and press enter, nothing is printed. Another thing is that when I remove the comment slashes from the two lines at the bottom, the program goes through an infinite loop.
Could someone please help?

Answer 1

This sounds like a school assignment, so I'll focus on advising/critiquing your code rather than giving a solution.
The first recommendation I have is to use a for loop instead of a while loop. A Rule of thumb in C is to only use a while loop when you actually don't have any idea how many things you need your program to look at.
You already have the length of the string, so set up your for loop to loop exactly once for each character.
Next you need to change how you are using printf. The %d format specifier is for printing integers, but you are passing it '1'. This is not an integer, it is the ascii representation of the symbol 1 (which is actually has the value 49, see the ascii table for more info)
You can either pass printf the value 1, or use the %c specifier, which expects ascii characters.
Better yet, just say printf("1");
That doesn't get you all the way there, but I think it lays the ground work so you can find the solution!

Answer 2

Condition !t || !T has no sense to be used as loop condition ...ask yourself how the loop will end ? you need just to check i is less than length
Second, the assignments t = word[i] == 't'; T = word[i] == 'T'; outside the loop have no sense ...you will be just pointing to the zero index of the string ...you should check all characters
third , the printf lines need to use %d
fourth , you appear not getting the purpose of the program printing inside loop will lead to printing many numbers and you just want to know if there is t or T you need to print single line.you may use variable int result=0; to hold the value you want and print it in the end ...of course you will need using break statement in the if((t || T) && i <= length / 2) and if((t || T) && i > length / 2) because no need for more searching
fifth, you should re-read , re-think , re-code the assignment before going bored and asking about it
sixth, there is a working version by modifying your code but you should try writing a good solution before looking at a solution as it better to solve your problems by yourself
#include <stdio.h>
#include <string.h>
int main() {
char word[50];
int i = 0, length, t = 0, T = 0;
scanf("%s", word);
length = strlen(word);
int result=0;
while( i<length) {
t = word[i] == 't';
T = word[i] == 'T';
if((t || T) && i <= length / 2) {
result=1;
break;
} else if((t || T) && i > length / 2) {
result=2;
break;
}else{
result=-1;
}
i++;
}
printf("%d",result);
return 0;
}

Answer 3

# include <stdio.h>
int main()
{
char name[20];
int age;
int siblings;
int childrens;
printf ("Hello my name is A.I, what is your name? \n");
scanf("%s", &name);
printf("how old are you : \n");
scanf("%d",&age);
printf("how many siblings you have: \n");
scanf("%d", &siblings);
printf("how many children you have: \n");
scanf("%d", &childrens);
printf("so your name is : %s \n", name);
printf("and your age is : %d \n", age);
printf("you have siblings : %d\n", siblings);
printf("so you have childrens : %d\n", childrens);
return 0;
}