Trouble getting a switch function to loop properly - c

Question

I'm writing a program to 'encrypt' an inputted string of text by using a switch statement to correlate the given character with a symbol, and output that symbol in the place of the character. I put it in a while loop, the idea being that it would loop the full switch function each time until the received character is EOF. On a guess, I believe it is looping through just the first character, because I don't advance the getchar() statement, but I'm not sure how to do that so any help would be greatly appreciated. I say this because if I use return instead of break, it closes the while loop and only takes that first letter, if I use a break then it spams the first 'encrypted' char.
#include <stdlib.h>
#include <stdio.h>
/* C program to encrypt a given text message, assuming all lowercase */
int main() {
int Input, Encrypted;
printf("Please type your message\n");
Input = getchar();
while (Input != EOF) {
switch (Input) {
case 'a':printf("!"); break;
case 'b':printf("#"); break;
case 'c':printf("#"); break;
case 'd':printf("$"); break;
case 'e':printf("%"); break;
case 'f':printf("^"); break;
case 'g':printf("&"); break;
case 'h':printf("*"); break;
case 'i':printf("`"); break;
case 'j':printf("~"); break;
case 'k':printf("-"); break;
case 'l':printf("_"); break;
case 'm':printf("="); break;
case 'n':printf("+"); break;
case 'o':printf("["); break;
case 'p':printf("{"); break;
case 'q':printf("]"); break;
case 'r':printf("}"); break;
case 's':printf(";"); break;
case 't':printf(":"); break;
case 'u':printf("|"); break;
case 'v':printf(","); break;
case 'w':printf("<"); break;
case 'x':printf("."); break;
case 'y':printf(">"); break;
case 'z':printf("'");break;
return 0;
}
}
return 0;
}

Answer 1

The simplest solution would be to remove the line
Input = getchar();
and to replace the line
while (Input != EOF) {
with:
while ( (Input=getchar()) != EOF && Input != '\n' ) {
Alternatively, if you find this while condition too confusing, you could also use an infinite loop, instead, like this:
#include <stdlib.h>
#include <stdio.h>
int main( void )
{
printf("Please type your message\n");
for (;;) //infinite loop, equivalent to while(true)
{
int c;
c = getchar();
if ( c == EOF || c == '\n' )
break;
switch ( c )
{
case 'a':printf("!"); break;
case 'b':printf("#"); break;
case 'c':printf("#"); break;
case 'd':printf("$"); break;
case 'e':printf("%%"); break;
case 'f':printf("^"); break;
case 'g':printf("&"); break;
case 'h':printf("*"); break;
case 'i':printf("`"); break;
case 'j':printf("~"); break;
case 'k':printf("-"); break;
case 'l':printf("_"); break;
case 'm':printf("="); break;
case 'n':printf("+"); break;
case 'o':printf("["); break;
case 'p':printf("{"); break;
case 'q':printf("]"); break;
case 'r':printf("}"); break;
case 's':printf(";"); break;
case 't':printf(":"); break;
case 'u':printf("|"); break;
case 'v':printf(","); break;
case 'w':printf("<"); break;
case 'x':printf("."); break;
case 'y':printf(">"); break;
case 'z':printf("'"); break;
}
}
return 0;
}
Note that most character sets (such as ASCII) store the characters a to z consecutively. With these character sets, you don't need the long switch statement. Instead, you can simplify it to the following:
#include <stdlib.h>
#include <stdio.h>
int main( void )
{
printf("Please type your message\n");
for (;;) //infinite loop, equivalent to while(true)
{
const char map[] = "!##$%^&*`~-_=+[{]};:|,<.>'";
int c;
c = getchar();
if ( c == EOF || c == '\n' )
break;
if ( 'a' <= c && c <= 'z' )
putchar( map[c-'a'] );
}
return 0;
}