Very new to c++ and I have the following code:
char input[3];
cout << "Enter input: ";
cin.getline(input,sizeof(input));
cout << input;
And entering something like abc will only output ab, cutting it one character short. The defined length is 3 characters so why is it only capturing 2 characters?
Remember that c-strings are null terminated. To store 3 characters you need to allocate space for 4 because of the null terminator.
Also as the #MikeSeymour mentioned in the comments in c++ its best to avoid the issue completely and use std::string.
You can thank your favorite deity that this fail-safe is in, most functions aren't that kind.
In C, strings are null-terminated, which means they take an extra character than the actual data to mark where the string actually ends.
Since you're using C++ anyway, you should avoid bare-bones char arrays. Some reasons:
buffer overflows. You managed to hit this issue on your first try, take a hint!
Unicode awareness. We're living in 2015. Still using 256 characters is unacceptable by any standard.
memory safety. It's way harder to leak a proper string than a plain old array. strings have strong copy semantics that cover pretty much anything you can think of.
ease of use. You have the entire STL algorithm list at your disposal! Use it rather than rolling your own.
Related
I am studying c++, In a blog they introduced the concept of copy function. When I tried the same in my system, the result is not matching to what I expected. Please let me know what I did wrong here in the below code.
#include <iostream>
main(){
std::string statement = "I like to work in Google";
char compName[6];
statement.copy(compName, 6, 18);
std::cout<<compName;
}
I expected Google but actual output is Googlex
I am using windows - (MinGW.org GCC-6.3.0-1)
You are confusing a sequence of characters, C style string, and std::string. Let's break them down:
A sequence of characters is just that, one character after another in some container (in your case a C style array). To a human being several characters may look like a string, but there is nothing in your code to make it such.
C style string is an array of characters terminated by a symbol \0. It is a carry over from C, as such a compiler will assume that if even if you don't tell it otherwise the array of characters may potentially be such a string.
C++ string (std::string) is a template class that stores strings. There is no need to worry how it does so internally. Although there are functions for interoperability with the first two categories, it is a completely different thing.
Now, let's figure out how a compiler sees your code:
char compName[6];
This creates an array of characters with enough space to store 6 symbols. You can write C style strings into it as long as they are 5 symbols or less, since you will need to also write '\0' at the end. Since in C++ C style arrays are unsafe, they will allow you to write more characters into them, but you cannot predict in advance where those extra characters will be written into memory (or even if your program will continue to execute). You can also potentially read more characters from the array... But you cannot even ask the question where that data will be coming from, unless you are simply playing around with your compiler. Never do that in your code.
statement.copy(compName, 6, 18);
This line writes 6 characters. It does not make it into a C style string, it is simply 6 characters in an array.
std::cout<<compName;
You are trying to output to the console a C style string... which you have not provided to a compiler. So a an operator<< receives a char [], and it assumes that you knew what you were doing and works as if you gave it C string. It displays one character after another until it reaches '\0'. When will it get such a character? I have no idea, since you never gave it one. But due to C style arrays being unsafe, it will have no problem trying to read characters past the end of an array, reading some memory blocks and thinking that they are a continuation of your non-existent C style sting.
Here you got "lucky" and you only got a single byte that appeared as an 'x', and then you got a byte with 0 written in it, and the output stopped. If you run your program at a different time, with a different compiler, or compiled with different optimisations you might get a completely different data displayed.
So what should you have done?
You can try this:
#include <iostream>
#include <string>
int main()
{
std::string statement = "I like to work in Google";
char compName[7]{};
statement.copy(compName, 6, 18);
std::cout<<compName;
return 0;
}
What did i change? I made an array able to hold 7 characters (leaving enough space for a C style string of 6 characters) and i have provided an empty initialisation list {}, which will fill the array with \0 characters. This means that when you will replace the first 6 of them with your data, there will be a terminating character in the very end.
Another approach would be to do this:
#include <iostream>
#include <string>
int main()
{
std::string statement = "I like to work in Google";
char compName[7];
auto length = statement.copy(compName, 6, 18);
compName[length] = '\0';
std::cout<<compName;
return 0;
}
Here i do not initialise the array, but i get the length of the data that is written there with a .copy method and then add the needed terminator in the correct position.
What approach is best depends on your particular application.
When inserting pointer to a character into the stream insertion operator, the pointer is required to point to null terminated string.
compName does not contain the null terminator character. Therefore inserting inserting (a pointer to an element of) it into a character stream violates the requirement above.
Please let me know what I did wrong here
You violate the requirement above. As a consequence, the behaviour of your program is undefined.
I expected Google but actual output is Googlex
This is because the behaviour of the program is undefined.
How to terminate it?
Firstly, make sure that there is room in the array for the null terminator character:
char compName[7];
Then, assign the null terminator character:
compName[6] = '\0';
I am working on a very basic REPL for Arduino. To get parameters, I need to split a String into parts, separating using spaces. I do not know how I would store the result. For example, pinmode 1 input would result in a list: "pinmode", 1, "input". The 1 would have to be an int. I have looked at other Stack Overflow answers, but they require a char input.
Don't use String. That's the reason all the other answers use char, commonly called C strings (lower case "s"). String uses "dynamic memory" (i.e., the heap), which is bad on this small microcontroller. It also add 1.6k to your program size.
The simplest thing to do is save each received character into a char array, until you get the newline character, '\n'. Be sure to add the NUL character at the end, and be sure your array is sized appropriately.
Then process the array using the C string library: strcmp, strtoul, strtok, isdigit, etc. Learning about these routines will really pay off, as it keeps your program small and fast. C strings are easy to print out, as well.
Again, stay away from String. It is tempting to beginners, because it is easy to understand. However, it has many subtle, complicated and unpredictable ways to make your embedded program fail. This is not a PC with lots of RAM and a swap file on a hard drive.
We will have to define an array for storing the string either way.
char[10];
And so suppose I want to store smcck in this array. What is the advantage of using gets(a)? My teacher said that the extra space in the array is wasted when we use cin.getline(a, 20), but that applies for gets(a) too right?
Also just an extra question, what exactly is stored in the empty "boxes"of an array?
gets() is a C function,it does not do bounds checking and is considered dangerous, it has been kept all this years for compatibility and nothing else.
You can check the following link to clear your doubt :
http://www.gidnetwork.com/b-56.html
Don't mix C features with C++, though all the feature of C works in C++ but it is not recommended . If you are working on C++ then you should probably avoid using gets(). use getline() instead.
Well, I don't think gets(a) is bettet because it does not check for the size of the string. If you try to read a long string using it, it may cause an buffer overflow. That means it will use all the 10 spaces you allocated for it and then it will try to use space allocated for another variables or another programs (what is going to make you publication crash).
The cin.getline() receives an int as a parameter with tells it to not read more than the expected number of characters. If you allocate a vector with only 10 positions and read 20 characters it will cause the same problem I told you about gets().
About the strings representation in memory, if you put "smcck" on an array
char v[10];
The word will take the first 5 positions (0 to 4), the position 5 will be taken by a null character (represented by '\0') that will mark the end of the string. Usually, what comes next in the array does not matter and are kept the way it were in the past.the null terminated character is used to mark where the string ends, so you can work it safely.
In C++ primer it is given that a null character is added at the end of every string literal. Why does a compiler do so?
Wikipedia:
"At the time C (and the languages that it was derived from) was developed, memory was extremely limited, so using only one byte of overhead to store the length of a string was attractive. The only popular alternative at that time, usually called a "Pascal string" (though also used by early versions of BASIC), used a leading byte to store the length of the string. This allowed the string to contain NULL and made finding the length need only one memory access (O(1) (constant) time).
However, C designer Dennis Ritchie chose to follow the convention of NULL-termination, already established in BCPL 'to avoid the limitation on the length of a string caused by holding the count in an 8- or 9-bit slot, and partly because maintaining the count seemed, in our experience, less convenient than using a terminator'..."
It is the best way to find end of the string from a chunk of memory!
And the whole string library functions believe strings are null terminated ;)
Because C strings are null-terminated.
i am wondering :char *cs = .....;what will happen to strlen() and printf("%s",cs) if cs point to memory block which is huge but with no '\0' in it?
i write these lines:
char s2[3] = {'a','a','a'};
printf("str is %s,length is %d",s2,strlen(s2));
i get the result :"aaa","3",but i think this result is because that a '\0'(or a 0 byte) happens to reside in the location s2+3.
how to make a not null-terminated c string? strlen and other c string function relies heavily on the '\0' byte,what if there is no '\0',i just want know this rule deeper and better.
ps: my curiosity is aroused by studying the follw post on SO.
How to convert a const char * to std::string
and these word in that post :
"This is actually trickier than it looks, because you can't call strlen unless the string is actually nul terminated."
If it's not null-terminated, then it's not a C string, and you can't use functions like strlen - they will march off the end of the array, causing undefined behaviour. You'll need to keep track of the length some other way.
You can still print a non-terminated character array with printf, as long as you give the length:
printf("str is %.3s",s2);
printf("str is %.*s",s2_length,s2);
or, if you have access to the array itself, not a pointer:
printf("str is %.*s", (int)(sizeof s2), s2);
You've also tagged the question C++: in that language, you usually want to avoid all this error-prone malarkey and use std::string instead.
A "C string" is, by definition, null-terminated. The name comes from the C convention of having null-terminated strings. If you want something else, it's not a C string.
So if you have a string that is not null-terminated, you cannot use the C string manipulation routines on it. You can't use strlen, strcpy or strcat. Basically, any function that takes a char* but no separate length is not usable.
Then what can you do? If you have a string that is not null-terminated, you will have the length separately. (If you don't, you're screwed. You need some way to find the length, either by a terminator or by storing it separately.) What you can do is allocate a buffer of the appropriate size, copy the string over, and append a null. Or you can write your own set of string manipulation functions that work with pointer and length. In C++ you can use std::string's constructor that takes a char* and a length; that one doesn't need the terminator.
Your supposition is correct: your strlen is returning the correct value out of sheer luck, because there happens to be a zero on the stack right after your improperly terminated string. It probably helps that the string is 3 bytes, and the compiler is likely aligning stuff on the stack to 4-byte boundaries.
You cannot depend on this. C strings need NUL characters (zeroes) at the end to work correctly. C string handling is messy, and error-prone; there are libraries and APIs that help make it less so… but it's still easy to screw up. :)
In this particular case, your string could be initialized as one of these:
A: char s2[4] = { 'a','a','a', 0 }; // good if string MUST be 3 chars long
B: char *s2 = "aaa"; // if you don't need to modify the string after creation
C: char s2[]="aaa"; // if you DO need to modify the string afterwards
Also note that declarations B and C are 'safer' in the sense that if someone comes along later and changes the string declaration in a way that alters the length, B and C are still correct automatically, whereas A depends on the programmer remembering to change the array size and keeping the explicit null terminator at the end.
What happens is that strlen keeps going, reading memory values until it eventually gets to a null. it then assumes that is the terminator and returns the length that could be massively large. If you're using strlen in an environment that expects C-strings to be used, you could then copy this huge buffer of data into another one that is just not big enough - causing buffer overrun problems, or at best, you could copy a large amount of garbage data into your buffer.
Copying a non-null terminated C string into a std:string will do this. If you then decide that you know this string is only 3 characters long and discard the rest, you will still have a massively long std:string that contains the first 3 good characters and then a load of wastage. That's inefficient.
The moral is, if you're using the CRT functions to operator on C strings, they must be null-terminated. Its no different to any other API, you must follow the rules that API sets down for correct usage.
Of course, there is no reason you cannot use the CRT functions if you always use the specific-length versions (eg strncpy) but you will have to limit yourself to just those, always, and manually keep track of the correct lengths.
Convention states that a char array with a terminating \0 is a null terminated string. This means that all str*() functions expect to find a null-terminator at the end of the char-array. But that's it, it's convention only.
By convention also strings should contain printable characters.
If you create an array like you did char arr[3] = {'a', 'a', 'a'}; you have created a char array. Since it is not terminated by a \0 it is not called a string in C, although its contents can be printed to stdout.
The C standard does not define the term string until the section 7 - Library functions. The definition in C11 7.1.1p1 reads:
A string is a contiguous sequence of characters terminated by and including the first null character.
(emphasis mine)
If the definition of string is a sequence of characters terminated by a null character, a sequence of non-null characters not terminated by a null is not a string, period.
What you have done is undefined behavior.
You are trying to write to a memory location that is not yours.
Change it to
char s2[] = {'a','a','a','\0'};