Extreme troubles with full line input. C Programming Language - c

Question

I am having the absolute craziest time getting full line input to work. I will explain my problem. I need to get a full line of input, including a space, from the user entered at the keyboard. Simple right? Wrong!
MY GOAL
Store multiple strings, with spaces, into variables. If it makes a difference, I want to make the variables equal to a char pointer. So once I get the input from tempString, I want to set it to a char pointer. Like so:
char *variable1, *variable2;
//get user input
variable1 = tempString;
//get more user input
variable 2 = tempString;
//etc etc etc
Here's what I've tried.
First try
char tempString[100];
scanf("%s", &tempString);
printf("%s", tempString);
Invalid: scanf will stop reading at a white space, so "Example String" would just end up being "Example".
Second try
So I do more research. I thought I found the magic fix.
char tempSTring[100];
fgets(tempString, 100, stdin);
printf("%s", tempString);
Originally this works. However there is a massive problem. I need to get the user to enter about 8 inputs. Meaning I have to use a command like this 8 times. The problem is the program often skips over the fgets command. If I use a scanf previously, somehow the \n character is stuck in the input stream, and automatically feeds into fgets, satisfying its stdin input, and then does not prompt the user for input.
Third try
After thinking fgets was maybe my solution with a work around, I tried some tricks.
char tempSTring[100];
getc(stdin);
fgets(tempString, 100, stdin);
printf("%s", tempString);
I tried adding this getc(stdin) line. It worked for much of my program. It absorbs the \n character left behind in the stream. When it does so, great, it works. But sometimes, for some reason, the \n is NOT left in the stream, and when debugging, it looks like getc(stdin) is requesting input from the user, so it pauses my program to ask for input.
Question
These don't work for me.
How should I be doing this easy task?

Answer 1

To read (up to) 8 lines from a file, you can use either of these solutions. I decline to use variables char *variable1, *variable2, …; — that is an array seeking to escape.
POSIX getline()
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
enum { MAX_LINES = 8 };
char *lines[MAX_LINES];
int index = 0;
char *buffer = 0;
size_t buflen = 0;
while (index < MAX_LINES && getline(&buffer, &buflen, stdin) != -1)
{
lines[index++] = buffer;
buffer = 0;
buflen = 0;
}
free(buffer); // Space may be allocated before EOF is detected
for (int i = 0; i < index; i++)
printf("%d: %s", i, lines[i]);
return 0;
}
If getline() fails to allocate memory, it will report an error, so there is no need to do an explicit error check.
Standard C fgets()
Code using strdup(), another POSIX function. It isn't a part of standard C (though it is widely available). It is trivial to implement.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
enum { MAX_LINES = 8 };
char *lines[MAX_LINES];
int index = 0;
char buffer[4096];
while (index < MAX_LINES && fgets(buffer, sizeof(buffer), stdin) != 0)
{
if ((lines[index] = strdup(buffer)) == 0)
break;
index++;
}
for (int i = 0; i < index; i++)
printf("%d: %s", i, lines[i]);
return 0;
}
The test in the loop allows for the possibility of strdup() failing to allocate memory.
Notes
Both the solutions above keep the newline at the end of the input string. If you don't want that, you can zap it with:
lines[i][strcspn(lines[i], "\r\n")] = '\0';
This overwrites a carriage return or newline with a null byte, transforming DOS or Unix line endings. You then need to adjust the printing which assumes the string includes a newline. Note that the expression shown works correctly even if there is no carriage return or newline in the string.
The fgets() solution will break lines at 4095 characters, leaving the rest to be read as 'the next line'. If that's not acceptable, you have a variety of strategies open to you.
You can detect whether there is a newline and arrange to allocate more memory and read the next section of the line into the extra memory, repeating until you come across a newline or EOF.
You can read the remaining characters up to the newline or EOF:
int c;
while ((c = getchar()) != EOF && c != '\n')
;
Implementing strdup()
If for some reason your system doesn't have an implementation of strdup(), you can create a surrogate with:
#include <assert.h>
#include <stdlib.h>
#include <string.h>
char *strdup(const char *old_str)
{
assert(old_str != 0);
size_t old_len = strlen(old_str) + 1;
char *new_str = malloc(old_len);
if (new_str != 0)
memmove(new_str, old_str, old_len);
return new_str;
}

Answer 2

Here's how we old fart C programmers would do it:
#include <stdio.h>
#define MAX_LEN 100
int main( )
{
int c;
char input[MAX_LEN+1];
int i = 0;
while ( (c=getchar()) != '\n' && c != EOF && i < MAX_LEN)
input[i++] = c;
if (c == EOF || c =='\n') {
/* received input that terminated within buffer limit */
input[i] = '\0';
printf("read in your input string of: %s\n", input);
}
else {
printf("don't buffer overflow me dude!\n");
return -1;
}
return 0;
}
But nowadays people will tell you to use one of the library functions. I'm still an old fart though.
EDIT: Fixed my embarrassing mistakes pointed out by the helpful comments below.

Answer 3

You can take care of '\n' left by previous scanf by writing it like this -
scanf("%d%*c", &x); //<-- example to take int input
%*c will read from stdin and then discard it, thus '\n' would be removed from stdin.
You can achieve with scanf like this (a way for your previous attempt)-
char tempString[100];
/* As suggested by chqrile it is essential to check return of scanf */
if(scanf("%99[^\n]", tempString)!=1){
// ^^ & not required
tempString[0]='\0';
}
%99[^\n] this will read 99 characters and will stop only after encountering '\n' , thus would read input with spaces.