I need to define the characters in an array and print the string...But it always prints as string7 (in this case, test7)...What am I doing wrong here?
#include <stdio.h>
int main() {
char a[]={'t','e','s','t'};
printf("%s\n",a);
return 0;
}
Why this behavior?
Because you did not \0 terminate your array, so what you get is Undefined behavior.
What possibly happens behind the scenes ?
The printf tries to print the string till it encounters a \0 and in your case the string was never \0 terminated so it prints randomly till it encounters a \0.
Note that reading beyond the bounds of allocated memory is Undefined behavior so technically this is a UB.
What you need to do to solve the problem?
You need:
char a[]={'t','e','s','t',`\0`};
or
char a[]="test";
Because your "string", or char[], is not null-terminated (i.e. terminated by \0).
then, printf("%s", a); will attempt to print every character starting from the start of a and keep printing until it sees until it sees a \0.
That \0 is outside your array, and depends on the initial state of the memory of your program, which you pretty much don't have control.
to fix this, use
char a[]={'t','e','s','t','\0'};
The string you printing must be null terminated...so your string declaration should be,
char a[]={'t','e','s','t', '\0'};
Related
As we know a string terminates with '\0'.
It's because to know the compiler that string ended, or to secure from garbage values.
But how does an array terminate?
If '\0' is used it will take it as 0 a valid integer,
So how does the compiler knows the array ended?
C does not perform bounds checking on arrays. That's part of what makes it fast. However that also means it's up to you to ensure you don't read or write past the end of an array. So the language will allow you to do something like this:
int arr[5];
arr[10] = 4;
But if you do, you invoke undefined behavior. So you need to keep track of how large an array is yourself and ensure you don't go past the end.
Note that this also applies to character arrays, which can be treated as a string if it contains a sequence of characters terminated by a null byte. So this is a string:
char str[10] = "hello";
And so is this:
char str[5] = { 'h', 'i', 0, 0, 0 };
But this is not:
char str[5] = "hello"; // no space for the null terminator.
C doesn't provide any protections or guarantees to you about 'knowing the array is ended.' That's on you as the programmer to keep in mind in order to avoid accessing memory outside your array.
C language does not have native string type. In C, strings are actually one-dimensional array of characters terminated by a null character '\0'.
From C Standard#7.1.1p1 [emphasis mine]
A string is a contiguous sequence of characters terminated by and including the first null character. The term multibyte string is sometimes used instead to emphasize special processing given to multibyte characters contained in the string or to avoid confusion with a wide string. A pointer to a string is a pointer to its initial (lowest addressed) character. The length of a string is the number of bytes preceding the null character and the value of a string is the sequence of the values of the contained characters, in order.
String is a special case of character array which is terminated by a null character '\0'. All the standard library string related functions read the input string based on this rule i.e. read until first null character.
There is no significance of null character '\0' in array of any type apart from character array in C.
So, apart from string, for all other types of array, programmer is suppose to explicitly keep the track of number of elements in the array.
Also, note that, first null character ('\0') is the indication of string termination but it is not stopping you to read beyond it.
Consider this example:
#include <stdio.h>
int main(void) {
char str[5] = {'H', 'i', '\0', 'z'};
printf ("%s\n", str);
printf ("%c\n", str[3]);
return 0;
}
When you print the string
printf ("%s\n", str);
the output you will get is - Hi
because with %s format specifier, printf() writes every byte up to and not including the first null terminator [note the use of null character in the strings], but you can also print the 4th character of array as it is within the range of char array str though beyond first '\0' character
printf ("%c\n", str[3]);
the output you will get is - z
Additional:
Trying to access array beyond its size lead to undefined behavior which includes the program may execute incorrectly (either crashing or silently generating incorrect results), or it may fortuitously do exactly what the programmer intended.
It’s just a matter of convention. If you wanted to, you could totally write code that handled array termination (for arrays of any type) via some sentinel value. Here’s an example that does just that, arbitrarily using -1 as the sentinel:
int length(int arr[]) {
int i;
for (i = 0; arr[i] != -1; i++) {}
return i;
}
However, this is obviously utterly unpractical: You couldn’t use -1 in the array any longer.
By contrast, for C strings the sentinel value '\0' is less problematic because it’s expected that normal test won’t contain this character. This assumption is kind of valid. But even so there are obviously many strings which do contain '\0' as a valid character, and null-termination is therefore by no means universal.
One very common alternative is to store strings in a struct that looks something like this:
struct string {
unsigned int length;
char *buffer;
}
That is, we explicitly store a length alongside a buffer. This buffer isn’t null-terminated (although in practice it often has an additional terminal '\0' byte for compatibility with C functions).
Anyway, the answer boils down to: For C strings, null termination is a convenient convention. But it is only a convention, enforced by the C string functions (and by the C string literal syntax). You could use a similar convention for other array types but it would be prohibitively impractical. This is why other conventions developed for arrays. Notably, most functions that deal with arrays expect both an array and a length parameter. This length parameter determines where the array terminates.
I'm learning C and am currently experimenting with storing strings in variables. I put together the following to try different stuff.
#include <stdio.h>
int main() {
char *name = "Tristan";
char today[] = "January 1st, 2016";
char newyear[] = {'H','a','p','p','y',' ','N','e','w',' ','Y','e','a','r','!','\n'};
printf("Hello world!\n");
printf("My name is %s.\n", name);
printf("Today is: %s.\n", today);
printf(newyear);
return 0;
}
After compiling this code and running it, I get the following results:
Hello world!
My name is Tristan.
Today is: January 1st, 2016.
Happy New Year!
January 1st, 2016
Now this is pretty much what I would expect, by why would "January 1st, 2016" get printed out again at the end of the program's output?
If I take the "\n" out of the "newyear" array, it will not do this.
Would someone please explain why this is?
newyear misses a trailing null byte, so printfing it is undefined behavior.
Only string literals implicitly append a null byte. You explicitly initialize every character, so no null byte is appended.
Undefined behavior means that something the standard does not define in this occasion will happen. That includes nothing happening, you bursting into tears, or, yes, printing some string twice.
Just add an additional character, i.e., a null byte to the array to resolve the problem:
char newyear[] = {'H','a','p','p','y',' ','N','e','w',' ','Y','e','a','r','!','\n', '\0'};
Note that no sane person initializes an automatic char array with a string like that. Just stick to string literals! (I think you did it just for learning purposes, though.)
Remember that strings in C are terminated by the special '\0' character.
Not having this terminator at the end of data that is treated as a string will lead to undefined behavior as the string functions pass the end of the data searching for the terminator.
This because you are defining newyear directly as a char array and not through the string literal "" syntax. This prevents the compiler from adding a trailing \0 character which is required to mark the end of a string.
Since both newyear and today reside on stack, in this case they have contiguous storage there so printf keeps after the \n of newyear and prints contents of memory until a \0 is found.
newyear should finish with a '\0' instead of the newline, to be a C string. You can then put the newline in the printf statement, like the others:
char newyear[] = {'H','a','p','p','y',' ','N','e','w',' ','Y','e','a','r','!','\0'};
//...
printf("%s.\n", newyear);
Or, you can add the string terminator to the array, and use the printf as you did:
char newyear[] = {'H','a','p','p','y',' ','N','e','w',' ','Y','e','a','r','!','\n','\0'};
//...
printf(newyear);
In your first two examples, a string defined as "my string" automatically has the '\0' appended, by the compiler.
Consider the case:
char s1[] = "abc";
s1[3] = 'x';
printf("%s", s1);
As I know, printf prints characters until it finds the null character and then stops.
When I overwrite the null character by 'x', why does printf print the s1 array correctly? How does it find the null character?
Your printf call invokes undefined behaviour because s1 doesn't have zero (aka null byte) terminator.
s1 is an array of 4 characters and over writing the null byte is not an issue.
After
s1[3] = 'x';
s1 will become:
[a][b][c][x]
But you can't print it as a string. A string in C is, by definition, a sequence of bytes terminated with a null byte. It just happens to work this time but you should never rely on that.
It means only that after this array there is by accident a null character in the memory.:)
You can try the following example
char s0[] = "xxx";
char s1[] = "abc";
char s2[] = "yyy";
s1[3] = 'x';
printf("%s",s1);
and see the result.
The printf function will print all the characters untill it encounters a nul character.
In your case, you have started accessing beyond the memory that was allocated and accessing memory beyond what is allocated is undefined behavior
In this case it accidently happen to be nul.
If it printed "abcx". It means that there was already a null in s1[4]. The value on the stack depends on previous operations. So it may always be a zero in that position but what is more likely to happen is that there is a zero while you are debugging the code and nothing goes wrong, but then in release a zero is not put in that position and you end up with a difficult bug to debug.
Undefined by the language definition does not mean undefined in an implementation. eg MS Visual Studio when compiling in Debug mode will set memory to predictable values to aid debugging.
When and why will an OS initialise memory to 0xCD, 0xDD, etc. on malloc/free/new/delete?
I did a lot of searching around for this, couldn't find any question with the same exact issue.
Here is my code:
void fun(char* name){
printf("%s",name);
}
char name[6];
sscanf(input,"RECTANGLE_SEARCH(%6[A-Za-z0-9])",name)
printf("%s",name);
fun(name);
The name is grabbed from scanf, and it printed out fine at first. Then when fun is called, there is a segmentation fault when it tries to print out name. Why is this?
After looking in my scrying-glass, I have it:
Your scanf did overflow the buffer (more than 6 byte including terminator read), with ill-effect slightly delayed due to circumstance:
Nobody else relied on or re-used the memory corrupted at first, thus the first printf seems to work.
Somewhere after the first and before the second call to printf the space you overwrote got re-used, so the string you read was no longer terminated before encountering not allocated pages.
Thus, a segmentation-fault at last.
Of course, your program was toast the moment it overflowed the buffer, not later when it finally crashed.
Morale: Never write to memory you have not dedicated for that.
Looking at your edit, the format %6[A-Za-z0-9] tries to read up to 6 characters exclusive the terminator, not inclusive!
Since you're reading 6 characters, you have to declare name to be 7 characters, so there's room for the terminating null character:
char name[7];
Otherwise, you'll get a buffer overflow, and the consequences are undefined. Once you have undefined consequences, anything can happen, including 2 successful calls to printf() followed by a segfault when you call another function.
You're probably walking off the end of the array with your printf statement. Printf uses the terminating null character '\0' to know where the end of the string is. Try allocating your array like this:
char name[6] = {'\0'};
This will allocate your array with every element initially set to the '\0' character, which means that as long as you don't overwrite the entire array with your scanf, printf will terminate before walking off the end.
Are you sure that name is zero byte terminated? scanf can overflow your buffer depending on how you are calling it.
If that happens then printf will read beyond the end of the array resulting in undefined behavior and probably a segmentation fault.
Consider following case:
#include<stdio.h>
int main()
{
char A[5];
scanf("%s",A);
printf("%s",A);
}
My question is if char A[5] contains only two characters. Say "ab", then A[0]='a', A[1]='b' and A[2]='\0'.
But if the input is say, "abcde" then where is '\0' in that case. Will A[5] contain '\0'?
If yes, why?
sizeof(A) will always return 5 as answer. Then when the array is full, is there an extra byte reserved for '\0' which sizeof() doesn't count?
If you type more than four characters then the extra characters and the null terminator will be written outside the end of the array, overwriting memory not belonging to the array. This is a buffer overflow.
C does not prevent you from clobbering memory you don't own. This results in undefined behavior. Your program could do anything—it could crash, it could silently trash other variables and cause confusing behavior, it could be harmless, or anything else. Notice that there's no guarantee that your program will either work reliably or crash reliably. You can't even depend on it crashing immediately.
This is a great example of why scanf("%s") is dangerous and should never be used. It doesn't know about the size of your array which means there is no way to use it safely. Instead, avoid scanf and use something safer, like fgets():
fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer. A terminating null byte ('\0') is stored after the last character in the buffer.
Example:
if (fgets(A, sizeof A, stdin) == NULL) {
/* error reading input */
}
Annoyingly, fgets() will leave a trailing newline character ('\n') at the end of the array. So you may also want code to remove it.
size_t length = strlen(A);
if (A[length - 1] == '\n') {
A[length - 1] = '\0';
}
Ugh. A simple (but broken) scanf("%s") has turned into a 7 line monstrosity. And that's the second lesson of the day: C is not good at I/O and string handling. It can be done, and it can be done safely, but C will kick and scream the whole time.
As already pointed out - you have to define/allocate an array of length N + 1 in order to store N chars correctly. It is possible to limit the amount of characters read by scanf. In your example it would be:
scanf("%4s", A);
in order to read max. 4 chars from stdin.
character arrays in c are merely pointers to blocks of memory. If you tell the compiler to reserve 5 bytes for characters, it does. If you try to put more then 5 bytes in there, it will just overwrite the memory past the 5 bytes you reserved.
That is why c can have serious security implementations. You have to know that you are only going to write 4 characters + a \0. C will let you overwrite memory until the program crashes.
Please don't think of char foo[5] as a string. Think of it as a spot to put 5 bytes. You can store 5 characters in there without a null, but you have to remember you need to do a memcpy(otherCharArray, foo, 5) and not use strcpy. You also have to know that the otherCharArray has enough space for those 5 bytes.
You'll end up with undefined behaviour.
As you say, the size of A will always be 5, so if you read 5 or more chars, scanf will try to write to a memory, that it's not supposed to modify.
And no, there's no reserved space/char for the \0 symbol.
Any string greater than 4 characters in length will cause scanf to write beyond the bounds of the array. The resulting behavior is undefined and, if you're lucky, will cause your program to crash.
If you're wondering why scanf doesn't stop writing strings that are too long to be stored in the array A, it's because there's no way for scanf to know sizeof(A) is 5. When you pass an array as the parameter to a C function, the array decays to a pointer pointing to the first element in the array. So, there's no way to query the size of the array within the function.
In order to limit the number of characters read into the array use
scanf("%4s", A);
There isn't a character that is reserved, so you must be careful not to fill the entire array to the point it can't be null terminated. Char functions rely on the null terminator, and you will get disastrous results from them if you find yourself in the situation you describe.
Much C code that you'll see will use the 'n' derivatives of functions such as strncpy. From that man page you can read:
The strcpy() and strncpy() functions return s1. The stpcpy() and
stpncpy() functions return a
pointer to the terminating `\0' character of s1. If stpncpy() does not terminate s1 with a NUL
character, it instead returns a pointer to s1[n] (which does not necessarily refer to a valid mem-
ory location.)
strlen also relies on the null character to determine the length of a character buffer. If and when you're missing that character, you will get incorrect results.
the null character is used for the termination of array. it is at the end of the array and shows that the array is end at that point. the array automatically make last character as null character so that the compiler can easily understand that the array is ended.
\0 is an terminator operator which terminates itself when array is full
if array is not full then \0 will be at the end of the array
when you enter a string it will read from the end of the array