This question already has answers here:
Space for Null character in c strings
(5 answers)
Closed 3 years ago.
I think I'm going insane because I cannot find an explanation to why C is combining my chars.
I've made you guys a test programm...
#include <stdio.h>
#include <stdlib.h>
int main()
{
char alphabet_big[26] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char alphabet_small[26] = "abcdefghijklmnopqrstuvwxyz";
printf("%s\n", alphabet_small);
return 0;
}
Results: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZV
Why is C combining alphabet_small and alphabet_big? That's not making sense. And why is there a "V" at the end of the char?
I hope someone can provide me an answer to this "problem".
Best regards.
Keep in mind that a C String is defined as a null terminated char array.
Change the declaration and initialization statement here: (for both statements.)
char alphabet_big[26] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";//forces compiler to use only 26 char
//regardless of the count of initializers
//(leaving no room for NULL terminator)
To
char alphabet_big[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";//allows compiler to set aside
^^ //the proper space, no matter how many initializers
The first produces undefined behavior when using with any of the string functions, such as strcpy, strcmp, and in this case printf with the "%s" format specifier.
The first produces the following, which is not is not a C string:
|A|B|C|D|E|F|G|H|I|J|K|L|M|N|O|P|Q|R|S|T|U|V|W|X|Y|Z|?|?|?|
While the 2nd produces the following, which is a C string:
|A|B|C|D|E|F|G|H|I|J|K|L|M|N|O|P|Q|R|S|T|U|V|W|X|Y|Z|\0|?|?|
Note - The ? symbols used in above illustration depict memory locations that are not owned by the program, and for which the contents are unknown, or may not even exist. A program attempting to access these locations would be invoking undefined behavior.
Normally the library functions expect to find a NUL byte at the end of a string, and the compiler is happy to add it for you automatically except you've told it that alphabet_big has only 26 bytes, essentially avoiding that extra NUL byte, so it combines with what's next.
Remove the 26 and let the compiler count for you.
Related
This question already has answers here:
What's the rationale for null terminated strings?
(20 answers)
Closed 3 months ago.
I am new to C and recently encountered this problem.
I have two pieces of code:
#include <stdio.h>
#include <string.h>
int main()
{
char x = 'a';
// char *y=&x;
printf("%ld\n", strlen(&x)); // output: 1
return 0;
}
#include <stdio.h>
#include <string.h>
int main()
{
char x = 'a';
char *y=&x;
printf("%ld\n", strlen(&x)); //output: 7
return 0;
}
What exactly happened when I added the variable y that it changed the result?
As said by others, you may not use strlen on a single character, as there is no guarantee of a null terminator.
In the first case, you were lucky that the 'a' was followed by a null byte in the mapping of the stack.
In the second case, possibly due to the definition of the variable y, the 'a' was followed by six nonzero bytes then a null.
Note that this is just a supposition. The behaviors could be different for a debug or a relase build. Such erratic phenomena are typical of undesirable memory accesses.
strlen() expects as string as its parameters. String in C are a sequence of chars terminated with a null char.
Your variable x is just a character without terminating null char. It is not a valid argument for strlen(). As such, the behavior is dangerous and undefined.
The code triggers undefined behavior.
It's not possible to reason about it very well, since whatever happens happens. It's probably going to treat the single-character variable as a string, and read the next byte to look for the terminator.
Since that byte is not, in fact, in a valid variable, the behavior is undefined and anything could happen. Don't do this.
I have been working with strings in C. While working with ways to declare them and initialize them, I found some weird behavior I don't understand.
#include<stdio.h>
#include<string.h>
int main()
{
char str[5] = "World";
char str1[] = "hello";
char str2[] = {'N','a','m','a','s','t','e'};
char* str3 = "Hi";
printf("%s %zu\n"
"%s %zu\n"
"%s %zu\n"
"%s %zu\n",
str, strlen(str),
str1, strlen(str1),
str2, strlen(str2),
str3, strlen(str3));
return 0;
}
Sample output:
Worldhello 10
hello 5
Namaste 7
Hi 2
In some cases, the above code makes str contain Worldhello, and the rest are as they were intialized. In some other cases, the above code makes str2 contain Namastehello. It happens with different variables I never concatenated. So, how are they are getting combined?
To work with strings, you must allow space for a null character at the end of each string. Where you have char str[5]="World";, you allow only five characters, and the compiler fills them with “World”, but there is no space for a null character after them. Although the string literal "World" includes an automatic null character at its end, you did not provide space for it in the array, so it is not copied.
Where you have char str1[]="hello";, the compiler determines the array size by counting the characters, including the null character at the end of the string literal.
Where you have char str2[]={'N','a','m','a','s','t','e'};, there is no string literal, just a list of individual characters. The compiler determines the array size by counting those. Since there is no null character, it does not provide space for it.
One potential consequence of failing to terminate a string with a null character is that printf will continue reading memory beyond the string and printing characters from the values it finds. When the compiler has placed other character arrays after such an array you are printing, characters from those arrays may appear in the output.
If you allow space for a null character in str and provide a zero value in str2, your program will print strings in an orderly way:
#include <stdio.h>
#include <string.h>
int main(void)
{
char str[6] = "World"; // 5 letters plus a null character.
char str1[] = "hello";
char str2[] = {'N', 'a', 'm', 'a', 's', 't', 'e', 0}; // Include a null.
char *str3 = "Hi";
printf("%s %zu\n%s %zu\n%s %zu\n%s %zu\n",
str, strlen(str),
str1, strlen(str1),
str2, strlen(str2),
str3, strlen(str3));
return 0;
}
Undefined behavior in non-null-terminated, adjacently-stored C-strings
Why do you get this part:
Worldhello 10
hello 5
...instead of this?
World 5
hello 5
The answer is that printf() prints chars until it hits a null character, which is a binary zero, frequently written as the '\0' char. And, the compiler happens to have placed the character array containing hello right after the character array containing World. Since you explicitly forced the size of str to be 5 via str[5], the compiler was unable to fit the automatic null character at the end of the string. So, with hello happening to be (not guaranteed to be) right after World, and printf() printing until it sees a binary zero, it printed World, saw no terminating null char, and continued right on into the hello string right after it. This resulted in it printing Worldhello, and then stopping only when it saw the terminating character after hello, which string is properly terminated.
This code relies on undefined behavior, which is a bug. It cannot be relied upon. But, that is the explanation for this case.
Run it with gcc on a 64-bit Linux machine online here: Online GDB: undefined behavior in NON null-terminated C strings
#Eric Postpischil has a great answer and provides more insight here.
From the C tag wiki:
This tag should be used with general questions concerning the C language, as defined in the ISO 9899 standard (the latest version, 9899:2018, unless otherwise specified — also tag version-specific requests with c89, c99, c11, etc).
You've asked a "how?" question about something that none of those documents defines, and so the answer is undefined in the context of C. You can only experience this phenomenon through undefined behaviour.
how are they are getting combined?
There is no such requirement that any of these variables are "combined" or are immediately located after each other; trying to observe that is undefined behaviour. It may appear to coincidentally work (whatever that means) for you at times on your machine, while failing at other times or using some other machine or compiler, etc. That's purely coincidental and not to be relied upon.
In some cases, the above code assigns str with Worldhello and the rest as they were intitated.
In the context of undefined behaviour, it makes no sense to make claims about how your code functions, as you've already noticed, the functionality is erratic.
I found some weird Behaviour with them.
If you want to prevent erratic behaviour, stop invoking undefined behaviour by accessing arrays out of bounds (i.e. causing strlen to run off the end of an array).
Only one of those variables is safe to pass to strlen; you need to ensure the array contains a null terminator.
This question already has answers here:
What happened when we do not include '\0' at the end of string in C?
(5 answers)
Closed 4 years ago.
so I know that in C, that '\0' is the null character, used to terminate strings. I've been looking online to see what it actually does, and I've run programs with and without it in my strings to see the difference in its use and non-use. I can't find any.
What can I not do when my strings lack the '\0' character?
Code that worked for me without a \0:
char a[10] = {'a','b','c','d','e','f','g'};
int x = strlen(a);
char b[10] = {'h','i','j','k','l'};
int y = strcmp(a,b);
printf("%d\n",x);
printf("%d\n",y);
According to the standard (C11 §7.1.1), if it doesn't end with a null byte it is not technically a 'string', it is simply a char array. (Make no mistake c-strings are char arrays as well, but they end in a terminator.)
You won't be able to use many of the string functions strcat, strcmp, strcpy, strlen, printf, etc, without a lot of your own built-in safeguards. Note, you can perform similar operations if you keep track of the length of the string manually and use functions like memcpy, strnlen, sprintf, etc, but you aren't technically working with c-string when doing so, you're simply working with arrays.
#include <stdio.h>
int main(void)
{
char username;
username = '10A';
printf("%c\n", username);
return 0;
}
I just started learning C, and here is my first problem. Why is this program giving me 2 warnings (multi-character constant, overflow in implicit constant conversion)?
And instead of giving 10A as output, it is giving just A.
You are trying to stuff multiple characters into a single set of '', and into a single char variable. You need "" for string literals, and you'll need an array of characters to hold a string. And to print a string, use %s.
Putting all of this together, you get:
#include <stdio.h>
int main(void)
{
char username[] = "10A";
printf("%s\n", username);
return 0;
}
Footnote
From Jonathan Leffler in the comments below regarding multi-character constants:
Note that multi-character constants are a part of C (hence the warning, not an error), but the value of a multi-character constant is implementation defined and hence not portable. It is an integer value; it is larger than fits in a char, so you get that warning. You could have gotten almost anything as the output — 1, A and a null byte could all be plausible.
'10A' is an allowed but obscure way to define a value.
In the case of an int variable,
int username = '10A';
printf("%x\n", username);
will output
313041
These are pairs of hexadecimal values - each pair is
0x31 is the '1' of your input.
0x30 is the '0' of your input.
0x41 is the 'A' of your input.
But a char type can't hold this.
In C there are no String objects. Instead Strings are arrays of characters (followed by a null character). Other answers have pointed out statically allocating this memory. However I recommend dynamically allocating Strings. Just remember C lacks a garbage memory collector (like there is in java). So remember to free your pointers. Have fun!!
You could use char *username to point to the beginning of the address and loop through the memory after. For instance use sizeof(username) to get the size and then loop printf until you have printed the amount of characters in username. However you may end up with major problems if you aren't careful...
I want to understand a number of things about the strings on C:
I could not understand why you can not change the string in a normal assignment. (But only through the functions of string.h), for example: I can't do d="aa" (d is a pointer of char or a array of char).
Can someone explain to me what's going on behind the scenes - the compiler gives to run such thing and you receive segmentation fault error.
Something else, I run a program in C that contains the following lines:
char c='a',*pc=&c;
printf("Enter a string:");
scanf("%s",pc);
printf("your first char is: %c",c);
printf("your string is: %s",pc);
If I put more than 2 letters (on scanf) I get segmentation fault error, why is this happening?
If I put two letters, the first letter printed right! And the string is printed with a lot of profits (incorrect)
If I put a letter, the letter is printed right! And the string is printed with a lot of profits and at the end something weird (a square with four numbers containing zeros and ones)
Can anyone explain what is happening behind?
Please note: I do not want the program to work, I did not ask the question to get suggestions for another program, I just want to understand what happens behind the scenes in these situations.
Strings almost do not exist in C (except as C string literals like "abc" in some C source file).
In fact, strings are mostly a convention: a C string is an array of char whose last element is the zero char '\0'.
So declaring
const char s[] = "abc";
is exactly the same as
const char s[] = {'a','b','c','\0'};
in particular, sizeof(s) is 4 (3+1) in both cases (and so is sizeof("abc")).
The standard C library contains a lot of functions (such as strlen(3) or strncpy(3)...) which obey and/or presuppose the convention that strings are zero-terminated arrays of char-s.
Better code would be:
char buf[16]="a",*pc= buf;
printf("Enter a string:"); fflush(NULL);
scanf("%15s",pc);
printf("your first char is: %c",buf[0]);
printf("your string is: %s",pc);
Some comments: be afraid of buffer overflow. When reading a string, always give a bound to the read string, or else use a function like getline(3) which dynamically allocates the string in the heap. Beware of memory leaks (use a tool like valgrind ...)
When computing a string, be also aware of the maximum size. See snprintf(3) (avoid sprintf).
Often, you adopt the convention that a string is returned and dynamically allocated in the heap. You may want to use strdup(3) or asprintf(3) if your system provides it. But you should adopt the convention that the calling function (or something else, but well defined in your head) is free(3)-ing the string.
Your program can be semantically wrong and by bad luck happening to sometimes work. Read carefully about undefined behavior. Avoid it absolutely (your points 1,2,3 are probable UB). Sadly, an UB may happen to sometimes "work".
To explain some actual undefined behavior, you have to take into account your particular implementation: the compiler, the flags -notably optimization flags- passed to the compiler, the operating system, the kernel, the processor, the phase of the moon, etc etc... Undefined behavior is often non reproducible (e.g. because of ASLR etc...), read about heisenbugs. To explain the behavior of points 1,2,3 you need to dive into implementation details; look into the assembler code (gcc -S -fverbose-asm) produced by the compiler.
I suggest you to compile your code with all warnings and debugging info (e.g. using gcc -Wall -g with GCC ...), to improve the code till you got no warning, and to learn how to use the debugger (e.g. gdb) to run your code step by step.
If I put more than 2 letters (on scanf) I get segmentation fault error, why is this happening?
Because memory is allocated for only one byte.
See char c and assigned with "a". Which is equal to 'a' and '\0' is written in one byte memory location.
If scanf() uses this memory for reading more than one byte, then this is simply undefined behavior.
char c="a"; is a wrong declaration in c language since even a single character is enclosed within a pair of double quotes("") will treated as string in C because it is treated as "a\0" since all strings ends with a '\0' null character.
char c="a"; is wrong where as char c='c'; is correct.
Also note that the memory allocated for char is only 1byte, so it can hold only one character, memory allocation details for datatypes are described bellow