Two identical chains do not identify themselves as equals in C - c

I've two character strings, one with the first line of a file (which is "WORKING"), and a second with the word "WORKING". The problem is that when I try to put them in an IF, it says they're not the same!
I've tried to read both of them with printf command but they're the same. I've also tried to use '\n' in the second string but there's not any change.
Here's the code, have a look:
FILE *fl;
fl=fopen("test.txt", "r");
char line_working[100];
fscanf(fl, "%s\n", line_working);
fclose(fl);
printf("%s", line_working); //HERE IT PRINTS: WORKING
char* workinger="WORKING";
printf("\n%s", workinger); //HERE IT ALSO PRINTS: WORKING
getch();
if(workinger==line_working){
printf("OK");
getch();
}
And nothing happens...

if(workinger==line_working){
compares the pointers.
workinger is a pointer and the array line_working used in the expression (comparison) gets converted to the pointer to its first element, which is equal to &line_working[0]. So it does address comparison. But this is not what you want. Unfortunately, this comparison is perfectly valid in C. So compiler can't help you here.
Use strcmp() to compare C strings.

== compares the pointer addresses.
To compare null-terminated character arrays (aka C-strings), use strcmp:
if(strcmp(workinger, line_working) == 0)

Your code does not compare the strings that you correctly try to compare with the assumption that the result should be they are equal, but the memory location where the strings are stored. By using '==' you do compare the value of workinger (which is a pointer so the variables value is a memory address) and line_working (which is an array so the corresponding value equals the memory address of the 1st element)
if (workinger==line_working)
To compare the strings that are stored at the corresponding memory locations you should make use of the string compare functions (see this question How do I properly compare strings? )

Related

If statement in C always returning false [duplicate]

This question already has answers here:
Issue with main arguments handling
(3 answers)
Closed 7 years ago.
I am fairly new to C, so am not overly familiar with it's syntax, however I have debugged my code and researched for the correct syntax, and it seems to be correct, I have also changed the scope of the variables to see if this was causing the error.
The if statement should compare two variables, which both hold strings, I have even printed both the variables out to ensure they are the same, however it is still skipping straight to the else section of the if statement. Can anyone give me any pointers on why it will not run the if statement, it just skips straight to 'incorrect'.
The correctWord variable is defined at a different section in the code.
Find full code here.
-UPDATE-
I have now updated the syntax of the code, however it is still returning false.
char correctWord[20];
void userGuess(){
char userWordGuess[20];
printf("Anagram: ");
printf(anagramWord);
printf("Your Guess: ");
scanf("%s",userWordGuess); //Reads in user input
printf(correctWord);
printf(userWordGuess);
if(strcmp(userWordGuess, correctWord) == 0){
printf("Congratulations, you guessed correctly!");
}else{
printf("Incorrect, try again or skip this question");
}
}
You cannot compare strings in C using ==, because this compares the addresses of the strings, not the contents of the string. (which you certainly don't require, and obviously, the addresses of the two strings are not equal too.)
C has a pretty nice function for it : strcmp() which returns 0 if both the strings are equal.
Try using this in your if condition:
if (!strcmp(userWordGuess,correctWord))
{
//Yay! Strings are equal. Do what you want to here.
}
Be sure to #include <string.h> before using strcmp().
In C, you can't compare strings using ==. You will end up comparing the addresses of the strings, which is not the same.
You need to call the strmcp() function, which will return 0 if its arguments (two strings) are equal.
So the code should be if(strcmp(userWordGuess, correctWord) == 0).
You're comparing addresses of different arrays, which will always be unequal.
You need to use strcmp or some other strings library function to compare strings character by character.
userWordGuess == correctWord will compare the pointers (i.e. the locations in memory of the arrays), which are probably not equal.
For string comparision in C, use strcmp (or strncmp):
if (!strcmp(userWordGuess, correctWord)){
/*Strings are equal*/
Use
if(strcmp(userWordGuess, correctWord) == 0) // strings are equal
{
printf("Congratulations, you guessed correctly!");
}
else // not equal
{
printf("Incorrect, try again or skip this question");
}
if both string are equal than if condition will run. otherwise it wil run else
The strings are not first-class citizens in the C language. The strings are represented as either arrays of characters or pointers to such arrays.
In both cases, the variable you use to access the string is a synonym for the address in memory of the first character of the string.
What you compare with userWordGuess == correctWord is not the strings but their addresses in memory. Since userWordGuess and correctWord are two different arrays of characters, their addresses in memory are always different and their comparison will always produce FALSE.
In order to compare the actual string values you have to use the standard function strcmp() or one of its variants (find them at the bottom of the documentation page).
Change in the code:
/** Need to include the header that declares the strcmp() function */
#include <string.h>
char correctWord[20];
void userGuess(){
char userWordGuess[20];
/** stripped some lines here ... */
/** compare the strings, not their addresses in memory */
if (strcmp(userWordGuess, correctWord) == 0) {
/** the rest of your code */
What you are doing here is comparing two pointers. userWordGuess and correctWord point each to the beginning of an array of characters (which is what you defined at the beginning of your example code).
So if you want to compare the two arrays of chars you can use the strcmp function defined in string.h
It is important that you learn the relation between arrays and pointers. Pointer arithmetic is as well important here. Check this out: Arrays, Pointers, Pointer Arithmetic

Copying and comparing individual element from 2D array to another array in C

char first_array[5][4] = {"aaa","bbb","ccc","ddd","eee"};
char second_array[1][4];
How would I copy, for example, the third element in first_array ("ccc") and save it to second_array?
The syntax below is clearly wrong, but this is what I'm asking for:
second_array[0] = first_array[2];
Also, after copying, I also want to know how to compare elements in the two arrays. Again, the syntax below might be wrong, I'm just explaining what I'm trying to do:
if(second_array[0] == first_array[2]){ printf("yes"); } //should print yes
You can't assign to arrays in c, you can fill arrays with some library functions like strcpy(), so
second_array[0] = first_array[2];
would be
strcpy(second_array[0], first_array[2]);
you must however ensure that the destination array fits the number of characters you are copying to it.
If you try to compare two strings in c, you can't do it through the == operator, because strings in c are arrays of char which contain a sequence of non-nul characters followed by a nul character, so if you write this
if (second_array[0] == first_array[2])
even when you succeeded at copying the data, the result will be most likely false, because you are not comparing the contents of the arrays, but their addresses, so to compare them correctly there is also a function strcmp() then the correct way of comparing the strings is
if (strcmp(second_array[0], first_array[2]) == 0)
The functions above require you to include the string.h header, and also that the passed strings are strings in the c sense, i.e what I described above.
I was recently trying to do this, as well: it is not possible to do this sort of direct assignment in C.
When you write first_array[0], the compiler will read that as an address which points to the first element (character) of first_array[2], not the entire string. When you run the assignment, if it were to work, it would only set the first character.
The easiest way is to use strncpy or memcpy (or a loop to cycle through the string.

C - test string equivalence without strcmp

in one of my university assignments I am restricted in the libraries I use. I am new to C and pointers and want to see if two strings (or should I say char's) are equal.
Part of me wants to loop through every char of the 'char string' and test equivalence, but then it comes back how to test equivalence (lol).
Any help is appreciated.
edit: I am seeing this:
warning: result of comparison against a string literal is
unspecified (use strncmp instead) [-Wstring-compare]
which leads to a segmentation fault. I know it has to do with this piece of code because all I added was:
if (example.name == "testName"){
printf("here!\n");
}
Part of me wants to loop through every char of the 'char string' and test equivalence
That's exactly what you need to do. Make a function mystrcmp with the signature identical to regular strcmp,
int mystrcmp ( const char * str1, const char * str2 );
and write your own implementation.
but then it comes back how to test equivalence.
When you loop character-by-character, you test equivalence of individual characters, not strings. Characters in C can be treated like numbers: you can compare them for equality using ==, check what character code is less than or greater than using < and >, and so on.
The only thing left to do now is deciding when to stop. You do that by comparing the current character of each string to zero, which is the null terminator.
Don't forget to forward-declare your mystrcmp function before using it.
A string in C is terminated with null character(0x00 or \0).You should compare both strings in a loop character by character till null char for either of the string is reached.
Loop should be broken if characters are not equal.
EDIT:
To answer your edit in question:
You should take two character pointers pointing to both strings and then copmare them like
//loop start,loop till null for any one of the string is found
if(*ptr1 != *ptr2)
{
//break loop
}
ptr1++;ptr2++;
//end loop
if((*ptr1 == *ptr2) &&(*ptr1== 0x00))
{
//strings are equal
}
Given that this is a university assignment, you should pay heed to chars just being small integers. You should also pay heed that C strings are contiguous memory buffers terminated by a binary zero (0x00).
You should also learn about pointer math. You will learn ways to shorten the code you have to write while learning something really interesting concerning the C language and how computers work. It will certainly help you if you choose a career on lower-level programming.

C string - identical, and not matching? [duplicate]

This question already has answers here:
Using the equality operator == to compare two strings for equality in C [duplicate]
(9 answers)
Closed 9 years ago.
Got a small problem with C. Restricting myself to simple C (i.e. OS instructions), and two strings seem to not be the same. Here is my code:
char inputData[256];
int rid;
rid = read(0,inputData,256);
// Strip input
char command[rid];
int i;
for (i = 0; i<=rid-2; i++) {
command[i] = inputData[i];
}
command[rid-1] = '\0';
if (command == "exit") {
write(1,"exit",sizeof("exit"));
}
Now, if a user enters "exit" into the terminal when queried and hits enter, the if for detecting "exit" never gets run. Any ideas?
Thanks,
EDIT: I am commiting to git as I go, so the current version can be found at github.com/samheather/octo-os. It's very obviously not complete code, but it demonstrates the problem.
You can't compare strings with ==. You need to use strcmp.
if (strcmp(command, "exit") == 0) {
C strings are actually character arrays. You can think of "command" as a pointer to the first character. You want to compare every character in the string, not just the location of the first characters.
You should use strcmp to compare strings in C.
if(strcmp(command, "exit") == 0) //strcmp returns 0 if strings are equal
To quote:
A zero value indicates that both strings are equal. A value greater than zero indicates
that the first character that does not match has a greater value in str1 than in str2.
a value less than zero indicates the opposite.
As it stands right now, you're comparing the address of command with the address of the string literal "exit", which pretty much can't be the same.
You want to compare the contents, with either strcmp, or (if "only OS instructions" means no standard library functions) an equivalent you write yourself that walks through the strings and compares characters they contain.
As others said, == doesn't work with strings. The reason is that it would compare the pointers given.
In the expression
command == "exit"
command is a pointer to your array variable, while "exit" is a pointer to that string which resides in read-only data space. They can never be identical, so the comparison always is false.
That's why strcmp() is the way to go.
Use strcmp from the standard library.

Checking contents of char variable - C Programming

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

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