Checking contents of char variable - C Programming - c

Question

This might seem like a very simple question, but I am struggling with it. I have been writing iPhone apps with Objective C for a few months now, but decided to learn C Programming to give myself a better grounding.
In Objective-C if I had a UILabel called 'label1' which contained some text, and I wanted to run some instructions based on that text then it might be something like;
if (label1.text == #"Hello, World!")
{
NSLog(#"This statement is true");
}
else {
NSLog(#"Uh Oh, an error has occurred");
}
I have written a VERY simple C Program I have written which uses printf() to ask for some input then uses scanf() to accept some input from the user, so something like this;
int main()
{
char[3] decision;
Printf("Hi, welcome to the introduction program. Are you ready to answer some questions? (Answer yes or no)");
scanf("%s", &decision);
}
What I wanted to do is apply an if statement to say if the user entered yes then continue with more questions, else print out a line of text saying thanks.
After using the scanf() function I am capturing the users input and assigning it to the variable 'decision' so that should now equal yes or no. So I assumed I could do something like this;
if (decision == yes)
{
printf("Ok, let's continue with the questions");
}
else
{
printf("Ok, thank you for your time. Have a nice day.");
}
That brings up an error of "use of undeclared identifier yes". I have also tried;
if (decision == "yes")
Which brings up "result of comparison against a string literal is unspecified"
I have tried seeing if it works by counting the number of characters so have put;
if (decision > 3)
But get "Ordered comparison between pointer and integer 'Char and int'"
And I have also tried this to check the size of the variable, if it is greater than 2 characters it must be a yes;
if (sizeof (decision > 2))
I appreciate this is probably something simple or trivial I am overlooking but any help would be great, thanks.

Answer 1

Daniel Haviv's answer told you what you should do. I wanted to explain why the things you tried didn't work:
if (decision == yes)
There is no identifier 'yes', so this isn't legal.
if (decision == "yes")
Here, "yes" is a string literal which evaluates to a pointer to its first character. This compares 'decision' to a pointer for equivalence. If it were legal, it would be true if they both pointed to the same place, which is not what you want. In fact, if you do this:
if ("yes" == "yes")
The behavior is undefined. They will both point to the same place if the implementation collapses identical string literals to the same memory location, which it may or may not do. So that's definitely not what you want.
if (sizeof (decision > 2))
I assume you meant:
if( sizeof(decision) > 2 )
The 'sizeof' operator evaluates at compile time, not run time. And it's independent of what's stored. The sizeof decision is 3 because you defined it to hold three characters. So this doesn't test anything useful.
As mentioned in the other answer, C has the 'strcmp' operator to compare two strings. You could also write your own code to compare them character by character if you wanted to. C++ has much better ways to do this, including string classes.
Here's an example of how you might do that:
int StringCompare(const char *s1, const char *s2)
{ // returns 0 if the strings are equivalent, 1 if they're not
while( (*s1!=0) && (*s2!=0) )
{ // loop until either string runs out
if(*s1!=*s2) return 1; // check if they match
s1++; // skip to next character
s2++;
}
if( (*s1==0) && (*s2==0) ) // did both strings run out at the same length?
return 0;
return 1; // one is longer than the other
}

Answer 2

You should use strcmp:
if(strcmp(decision, "yes") == 0)
{
/* ... */
}

Answer 3

You should be especially careful with null-terminated string in C programming. It is not object. It is a pointer to a memory address. So you can't compare content of decision directly with a constant string "yes" which is at another address. Use strcmp() instead.
And be careful that "yes" is actually "yes\0" which will take 4 bytes and the "\0" is very important to strcmp() which will be recognized as the termination during the comparison loop.

Answer 4

Ok a few things:
decision needs to be an array of 4 chars in order to fit the string "yes" in it. That's because in C, the end of a string is indicated by the NUL char ('\0'). So your char array will look like: { 'y', 'e', 's', '\0' }.
Strings are compared using functions such as strcmp, which compare the contents of the string (char array), and not the location/pointer. A return value of 0 indicates that the two strings match.
With: scanf("%s", &decision);, you don't need to use the address-of operator, the label of an array is the address of the start of the array.
You use strlen to get the length of a string, which will just increment a counter until it reaches the NUL char, '\0'. You don't use sizeof to check the length of strings, it's a compile-time operation which will return the value 3 * sizeof(char) for a char[3].
scanf is unsafe to use with strings, you should alternatively use fgets(stdin...), or include a width specifier in the format string (such as "3%s") in order to prevent overflowing your buffer. Note that if you use fgets, take into account it'll store the newline char '\n' if it reads a whole line of text.

Answer 5

To compare you could use strcmp like this:
if(strcmp(decision, "yes") == 0) {
// decision is equal to 'yes'
}
Also you should change char decision[3] into char decision[4] so that the buffer has
room for a terminating null character.
char decision[4] = {0}; // initialize to 0

Answer 6

There's several issues here:
You haven't allocated enough storage for the answer:
char[3] decision;
C strings are bytes in the string followed by an ASCII NUL byte: 0x00, \0. You have only allocated enough space for ye\0 at this point. (Well, scanf(3) will give you yes\0 and place that NUL in unrelated memory. C can be cruel.) Amend that to include space for the terminating \0 and amend your scanf(3) call to prevent the buffer overflow:
char[4] decision;
/* ... */
scanf("%3s", decision);
(I've left off the &, because simply giving the name of the array is the same as giving the address of its first element. It doesn't matter, but I believe this is more idiomatic.)
C strings cannot be compared with ==. Use strcmp(3) or strncmp(3) or strcasecmp(3) or strncasecmp(3) to compare your strings:
if(strcasecmp(decision, "yes") == 0) {
/* yes */
}

Answer 7

C has lots of lib functions to handle this but it pays to know what you are declaring.
Declaring
char[3] decision;
is actually declaring a char array of length 3. So therefor attempting a comparison of
if(decision == "yes")
is comparing a literal against and array and therefor will not work. Since there is no defined string type in C you have to use pointers, but not directly, if you don't want to. In C strings are in fact arrays of char so you can declare them both ways eg:
char[3] decision ;
* char decision ;
Both will in point of fact work but you in the first instance the compiler will allocate the memory for you, but it will ONLY allocate 3 bytes. Now since strings in C are null terminated you need to actually allocate 4 bytes since you need room for "yes" and the null. Declaring it the second way simply declares a pointer to someplace in memory but you have no idea really where. You would then have to allocate memory to contain whatever you are going to put there since to do otherwise will more then likely cause a SEGFAULT.
To compare what you get from input you have two options, either use the strcomp() function or do it yourself by iterating through decision and comparing each individual byte against "Y" and "E" and "S" until you hit null aka \0.
There are variations on strcomp() to deal with uppercase and lowercase and they are part of the standard string.h library.