Program stops when i input any character [closed] - c

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*the code is from a book named
Programming with c by Byron gottfried *
*when i try to input a character, the program stops *
int main()
{
char line[80];
int count;
printf(" Enter a line of text below:\n");
scanf("%[^\n], &line");
for(count=0;line[count]!='\0';++count){
if(((line[count]>='0') && (line[count] < '9'))||
((line[count] >= 'A') && (line[count]< 'Z'))||
((line[count]>= 'a' ) && (line[count] <'z' )))
putchar(line[count]+1);
else if(line[count] =='9' ) putchar('0');
else if(line[count] == 'Z')putchar('A');
else if(line[count] == 'z')putchar('a');
else putchar('.');
}
return 0;
}

Answer 1

The line:
scanf("%[^\n], &line");
Should be:
scanf("%[^\n]", line);
i.e. Put the closing " in the format string in the correct place, and no & before line.
Moreover, on some platforms the stdout buffer is not flushed until it contains a newline or becomes full, so you should add a fflush( stdout) call. You will get away with it on Windows.
You might also consider simplifying the code by using the ctype.h functions.
Since this is a simple transcription error (the book referenced does not contain this error) rather than a programming question, and the question has been closed, this purely an aside, but I'd suggest the following implementation.
#include <stdio.h>
#include <ctype.h>
int main()
{
char line[80];
int count;
/* read in the entire string */
printf("Enter a line of text below:\n");
scanf("%[^\n]", line);
/* encode each individual character and display it */
for (count = 0; line[count] != '\0'; ++count)
{
char plaintext = line[count];
char encoded = '.';
if (isupper(plaintext))
{
encoded = (((plaintext + 1) - 'A') % 26) + 'A';
}
else if (islower(plaintext))
{
encoded = (((plaintext + 1) - 'a') % 26) + 'a';
}
else if (isdigit(plaintext))
{
encoded = (((plaintext + 1) - '0') % 10) + '0';
}
putchar(encoded);
}
return 0;
}