Something strange with if() in c - c

Question

This is for a school project I'm working on, it's just a small part of the code but for some reason the program doesn't seem to go inside the if() no matter what the input is. I've tried anything and everything I know of (also used the isalpha() function) but it just won't run the commands inside the if().
do
{
flag=1;
gets(input.number);
printf("\n%s\n",input.number);
for(i=0;i<strlen(input.number);i++)
{
printf("yolo1\n"); //debug
if(input.number[i]<48 || input.number[i]>57) //PROBLEM HERE
{
printf("\nyolo2\n"); //debug
flag=-1;
break;
}
}
if(strlen(input.number)<1000 || strlen(input.number)>9999 || flag==-1) printf("\nINVALID INPUT\n\nARI8MOS: ");
}while(strlen(input.number)<1000 || strlen(input.number)>9999 || flag==-1);
Can you guys help me out here??? I've been staring and the code for the better part of 3 days now....

Answer 1

I presume you declared char input.number[].
Your condition in if says that you only want to get into its body if the character is NOT a digit. This is somehow contradictory to the name input.number of the variable (but perhaps you are just checking for incorrect characters here...)
To better see what is happening with the condition, you can choose to print the values of its components, like this:
printf("%c[%d]", input.number[i], input.number[i]);
printf("%d, %d, %d\n", input.number[i]<48 , input.number[i]>57, input.number[i]<48 || input.number[i]>57);
(you will se a 0 for false and 1 for true)
BTW: You could write the same condition in a more readable manner, using char constants like this:
input.number[i]<'0' || input.number[i]>'9')