How can I use scanf and if with strings? (Very basic) - c

Question

I just had my first class in C programing at college and I did this:
#include <stdio.h>
int main()
{
char answer='x';
printf("Should the crocodile eat the man? y/n ");
scanf("%s",&answer);
printf("%s",& answer);
if(answer == 'y')
{printf("the man is dead");}
if(answer == 'n')
{printf("the man still alive ");}
}
How can I do the same with strings instead of single characters?
I have tried many things but nothing works.

Answer 1

In C, a string is a sequence of character values followed by a zero-valued terminator. The string "hello" is represented by the sequence {'h', 'e', 'l', 'l', 'o', 0}.
Strings are stored as arrays of char - the array must be large enough to hold the entire string plus the 0 terminator, so to store the string "hello", you need an array of char that's at least 6 elements long.
It's important to remember that an array of char doesn't have to contain a string - you can use it to store character data that's not meant to be interpreted as a string. You can also store more than one string to an array, like so:
/**
* Array size is determined by number of elements in the initializer
*/
char strs[] = {'T','h','i','s',0,'i','s',0,'a',0','t','e','s','t',0};
After this, strs will contain 4 strings: "This", "is", "a", "test".
So, to store the strings "y" or "n", you need an array of char that's at least 2 elements wide.
char answer[2];
To read strings with scanf, you use the %s conversion specifier:
scanf( "%s", answer );
%s tells scanf to skip over any leading whitespace, then read a sequence of non-whitespace characters and store them to the answer array. A 0 terminator will automatically be appended to the array, making it a string.
In this case, you don't need to use the & operator on answer; under most circumstances, an expression of type "array of T" will be converted ("decay") to an expression of type "pointer to T", so the expression answer "decays" from type char [2] to type char *.
To be safe, you should specify a maximum field width so that you don't read more characters than the target array is sized to hold:
scanf( "%1s", answer );
Since we have to reserve an element for the 0 terminator, the field width specifier should be at least 1 less than the size of the target array. Since our target array is sized to hold 2 elements, the field width specifier pretty much has to be 1.
To compare strings, use the strcmp or strncmp library functions:
if ( strcmp( answer, "y" ) == 0 )
{
// process "y" answer
}
Yes, strcmp returns 0 on a match.
You could also do a comparison against individual elements in the array:
if ( answer[0] == 'y' )
{
// process "y" answer
}
Remember
If you want to read a single character from the input stream and store it to a single character object with scanf, you would use the %c conversion specifier:
char answer; // single character, not an array
scanf( " %c", &answer ); // use & operator to obtain pointer, blank skips
// leading whitespace
If you want to read a sequence of non-whitespace characters from the input stream and store them to an array of char with scanf, you would use the %s conversion specifier:
char answer[2]; // array of character
scanf( "%1s", answer ); // do not use & operator to obtain pointer, array
// expressions are converted to pointer expressions
// automatically in this case.

Answer 2

You need to create a buffer to store the string:
char answer[50]; // change 50 to the maximum length of an answer + 1
scanf("%s", answer);

Answer 3

Yes but you need to pass it a buffer for the string and use strcmp instead of == for comparison.
#include <stdio.h>
#include <string.h>
int main()
{
char answer[16];
printf("Should the crocodile eat the man? yes/no ");
scanf("%15s", answer);
printf("%s", answer);
if (strncmp(answer, "yes", 16) == 0) {
printf("the man is dead\n");
} else if (strncmp(answer, "no", 16) == 0) {
printf("the man still alive\n");
}
}