How char array behaves for longer strings? - c

I asked this question as one of multiple questions here. But people asked me to ask them separately. So why this question.
Consider below code lines:
char a[5] = "geeks"; //1
char a3[] = {'g','e','e','k','s'}; //d
printf("a:%s,%u\n",a,sizeof(a)); //5
printf("a3:%s,%u\n",a3,sizeof(a3)); //j
printf("a[5]:%d,%c\n",a[5],a[5]);
printf("a3[5]:%d,%c\n",a3[5],a3[5]);
Output:
a:geeksV,5
a3:geeks,5
a[5]:86,V
a3[5]:127,
However the output in original question was:
a:geeks,5
a3:geeksV,5
The question 1 in original question was:
Does line #1 adds \0? Notice that sizeof prints 5 in line #5 indicating \0 is not there. But then, how #5 does not print something like geeksU as in case of line #j? I feel \0 does indeed gets added in line #1, but is not considered in sizeof, while is considered by printf. Am I right with this?
Realizing that the output has changed (for same online compiler) when I took out only those code lines which are related to first question in original question, now I doubt whats going on here? I believe these are undefined behavior by C standard. Can someone shed more light? Possibly for another compiler?
Sorry again for asking 2nd question.

char a[5] = "geeks"; //1
Here, you specify the array's size as '5', and initialize it with 5 characters.
Therefore, you do not have a "C string", which by definition is ended by a NUL. (0).
printf("a:%s,%u\n",a,sizeof(a)); //5
The array itself still has a size of 5, which is correctly reported by the sizeof operator, but your call to printf is undefined behaviour and could print anything after the arrray's contents - it will just keep looking at the next address until it finds a 0 somewhere. That could be immediately, or it could print a 1000000 garbage characters, or it could cause some sort of segfault or other crash.
char a3[] = {'g','e','e','k','s'}; //d
Because you don't specify the array's size, the compiler will, through the initialization syntax, determine the size of the array. However, the way you chose to initialize a3, it will still only provide 5 bytes of length.
The reason for that is that your initialization just is an initialization list, and not a "string". Therefore, your subsequent call to printf also is undefined behaviour, and it is just luck that at the position a3[5] there seems to be a 0 in your case.
Effectively, both examples have the very same error.
You could have it different thus:
char a3[] = "geeks";
Using a string literal for initialization of the array with unspecified size will cause the compiler to allocate enough memory to hold the string and the additional NUL-terminator, and sizeof (a3) will now yield 6.

"geeks" here is a string literal in C.
When you define "geeks" the compiler automatically adds the NULL character to the end. This makes it 6 characters long.
But you are assigning it to char a[5]. This will cause undefined behaviour.
As mentioned by #DavidBowling, in this case the following condition applies
(Section 6.7.8.14) C99 standard.
An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array
the elements "geeks" will be copied into the array 'a' but the NULL character will not be copied.
So in this case when you try to print the array, it will continue printing until it encounters a \0 in the memory.
From the further print statements it is seen that a[5] has the value V. Presumably the next byte on your system is \0 and the array print stops.
So, in your system, at that instance, "geeksV" is printed.

Related

Does specifying array size for a user input string in C matter?

I am writing a code to take a user's input from the terminal as a string. I've read online that the correct way to instantiate a string in C is to use an array of characters. My question is if I instantiate an array of size [10], is that 10 indexes? 10 bits? 10 bytes? See the code below:
#include <stdio.h>
int main(int argc, char **argv){
char str[10] = "Jessica";
scanf("%s", &str);
printf("%c\n", str[15]);
}
In this example "str" is initialized to size 10 and I am able to to print out str[15] assuming that when the user inputs a a string it goes up to that index.
My questions are:
Does the size of the "str" array increase after taking a value from scanf?
At what amount of string characters will my original array have overflow?
.
When you declare an array of char as you have done:
char str[10] = "Jessica";
then you are telling the compiler that the array will hold up to 10 values of the type char (generally - maybe even always - this is an 8-bit character). When you then try to access a 'member' of that array with an index that goes beyond the allocated size, you will get what is known as Undefined Behaviour, which means that absolutely anything may happen: your program may crash; you may get what looks like a 'sensible' value; you may find that your hard disk is entirely erased! The behaviour is undefined. So, make sure you stick within the limits you set in the declaration: for str[n] in your case, the behaviour is undefined if n < 0 or n > 9 (array indexes start at ZERO). Your code:
printf("%c\n", str[15]);
does just what I have described - it goes beyond the 'bounds' of your str array and, thus, will cause the described undefined behaviour (UB).
Also, your scanf("%s", &str); may also cause such UB, if the user enters a string of characters longer than 9 (one must be reserved for a terminating nul character)! You can prevent this by telling the scanf function to accept a maximum number of characters:
scanf("%9s", str);
where the integer given after the % is the maximum input length allowed (anything after this will be ignored). Also, as str is defined as an array, then you don't need the explicit "address of" operator (&) in scanf - it is already there, as an array reference decays to a pointer!
Hope this helps! Feel free to ask for further clarification and/or explanation.
One of C's funny little foibles is that in almost all cases it does not check to make sure you are not overflowing your arrays.
It's your job to make sure you don't access outside the bounds of your arrays, and if you accidentally do, almost anything can happen. (Formally, it's undefined behavior.)
About the only thing that can't happen is that you get a nice error message
Error: array out-of-bounds access at line 23
(Well, theoretically that could happen, but in practice, virtually no C implementation checks for array bounds violations or issues messages like that.)
See also this answer to a similar question.
An array declares the given number of whatever you are declaring. So in the case of:
char str[10]
You are declaring an array of ten chars.
Does the size of the "str" array increase after taking a value from scanf?
No, the size does not change.
At what amount of string characters will my original array have overflow?
An array of 10 chars will hold nine characters and the null terminator. So, technically, it limits the string to nine characters.
printf("%c\n", str[15]);
This code references the 16th character in your array. Because your array only holds ten characters, you are accessing memory outside of the array. It's anyone's guess as to if your program even owns that memory and, if it does, you are referencing memory that is part of another variable. This is a recipe for disaster.

Why I get a length of 7 when I apply strlen() to a pointer to a single char? [duplicate]

This question already has answers here:
what should strlen() really return in this code?
(4 answers)
Closed 6 years ago.
The C code:
char c = 'a';
char *p = &c;
printf("%lu\n",strlen(p));
And I get a result 7 and I have no idea how this 7 come out.
The variable p points to a single character, not to a null terminated string. So when you call strlen on it, it attempts to access whatever memory is after c. This invokes undefined behavior.
What's happening in this particular case is that after a in memory there are six non-zero bytes followed by one zero byte, so you get 7. You can't however depend on this behavior. For example, if you add more local variables before and after a, even unused ones, you'll probably get a different result.
Remember that strings in C are really called null-terminated byte strings. All strings are terminated with a single '\0' character, meaning that a single-character string actually is two characters: The single character plus the terminator.
When you have the pointer pointing to c in your code, you don't have two characters, only the single characters contained in c. You don't know if there is a terminator after that character in memory, so when strlen looks for that terminator it will pass the character and go out into memory not belonging to any string to look for it, and you will have undefined behavior.
To try an illustrate what you have, take a look at this "graphical" representation:
+---+ +-----+----------------------
| p | --> | 'a' | indeterminate data...
+---+ +-----+----------------------
That's basically how it looks like in memory. The variable p points to the location where your character is stored, but after that in memory is just indeterminate data. This will be seemingly random, and you can not tell where there will be a byte corresponding to a string terminator character.
There's no way to say why strlen get the value 7 from, except that it finds six non-terminator bytes in the indeterminate data after your character. Next time you run it, or if you run it on a different system, you might get a completely different result.
Because strlen finds first null character starting from passed address. You pass address of only character, so strlen tries to look forward in memory until first null char in memory after variable c.
Anyway, you cannot be sure about result and it depends on compiler and all code you wrote. Moreover, you program can even fail with memory exception.

The mechanics of populating an array [closed]

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Closed 6 years ago.
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I tried to populate an array of structs defined as follows:
typedef struct{
char directive[5];
}directive_nfo_t;
By using the following:
directive_nfo_t directive_list[]=
{
{"ALIGN"},{"ASCII"},{"BSS"},{"BYTE"},{"END"},{"EQU"},{"ORG"}
};
To my surprise, the first few elements were corrupted like so:
[0]= ALIGNASCIIBSS
[1]= ASCIIBSS
[2]= BSS
...
Until I made the following change:
typedef struct{
char directive[6]; <-- made char array +1
}directive_nfo_t;
Then the first few arrays were correct like so:
[0]= ALIGN
[1]= ASCII
[2]= BSS
...
My question is what happens in the background to explain this behavior?
Regards.
In C, a string is a sequence of character values followed by a 0-valued terminator; the string "ASCII" is represented by the character sequence 'A', 'S', 'C', 'I', 'I', 0. Thus, you need a six-element array to store the string.
For a string that's N characters long, you need an array of N+1 characters to store it.
When you explicitly initialize a char array as string literal in the way you do:
char some_array[] = {"ALIGN"};
the compiler actually populates the 0th to 4th "position" (total of 5 positions) with the characters inside quotation marks, but also the fifth position with \0 without requiring you do it explicitly (if it has space enough). So the size equals 6. You exceed the boundaries if you don't count the \0 character into the size calculation and restrict the size to 5. Compiler would omit the terminating character.
In your case it looks as if the first element of the next member "overwrote" what should have been the omitted \0 character of the previous, since you haven't reserved a place for it. In fact the "mechanics of populating the array" boils down to the compiler writing as much data as could fit inside the boundaries. The address of the first position of the next member string logically corresponds to your assignment, although the \0 from the previous is missing.
Since your printf() format tag was %s, the function printed the characters until it reached the first \0 character, which is in fact undefined behavior.
That's why
char directive[6];
was correct size assignment in your code.
If the char array is big enough, C compiler automatically places a '\0' after the text.
If it is just large enough for the text, that terminator is omitted, which is what has happened here.
If there isn't even room for the text, the compiler will say something like "too many initialisers" or "array bounds overflow".
The struct array elements are adjacent in memory. The first two items lack a terminator, so the second item printed only stops at the terminator after the third item. The first item, is also printed until it reaches that same terminator. By making the array size 6, the compiler was able to place a terminator after every item.
Unlike in C++, C allows you to (unintentionally) shoot yourself in the feet, by allowing to omit NUL terminating character '\0' in the char array initializer when there is no room for it. Your case can be narrowed down to a simple array definition such as:
char str[5] = "ALFAP";
which is a syntatic shortcut to:
char str[5] = {'A', 'L', 'F', 'A', 'P'};
It may be kind of misleading, because in different context the same "ALFAP" represets string literal, that always has the ending NUL character:
char* str = "ALFAP" // here, "ALFAP" always contains NUL character at the end
My question is what happens in the background to explain this behavior? Regards.
You have an array of struct directive_nfo_t type and each struct directive_nfo_t holds array of five characters (in your first example).
The output that you were getting when you have 5 character array in directive_nfo_t type was basically due to two things-
Array elements are stored in consecutive memory locations.
In C, the abstract idea of a string is implemented with just null terminated array of characters.
When you have declared an array of directive_nfo_t type, each element of directive_nfo_t is stored in consecutive memory location and each element has 5 character array(which are also stored in consecutive locations) in it. And in your Initialization list({"ALIGN"},{"ASCII"},{"BSS"},{"BYTE"},{"END"},{"EQU"},{"ORG"}) for the array, you have used all the 5 characters in storing your data in first two elements of directive_nfo_t ("ALIGN" and "ASCII"). As, in C, functions which operate on character array to implement abstract idea of string, assume that a string will be terminated by using a null character at the end. Therefore, in the first two elements of directive_nfo_t array, the printf will keep on printing characters until it reaches null character(which it will find in element storing character array "BSS"). After printing ALIGN, printf will access the first character of second element of the array of directive_nfo_t (character A of ASCII). It occurred because there was not space for null character in the first element of array of directive_nfo_t type and compiler wouldn't add characters beyond array size as it does array bound check. From the third element of you array, you have enough space for null character and hence, printf works as expected.
You will get UNDEFINED BEHAVIOR if you allocate less memory to store your character array and use those functions which assume null terminated character array. Always set the size of the character array to MAX + 1 when you want to store maximum MAX characters in your array.

C character array and its length

I am studying now C with "C Programming Absolute Beginner's Guide" (3rd Edition) and there was written that all character arrays should have a size equal to the string length + 1 (which is string-termination zero length). But this code:
#include <stdio.h>
main()
{
char name[4] = "Givi";
printf("%s\n",name);
return 0;
}
outputs Givi and not Giv. Array size is 4 and in that case it should output Giv, because 4 (string length) + 1 (string-termination zero character length) = 5, and the character array size is only 4.
Why does my code output Givi and not Giv?
I am using MinGW 4.9.2 SEH for compilation.
You are hitting what is considered to be undefined behavior. It's working now, but due to chance, not correctness.
In your case, it's because the memory in your program is probably all zeroed out at the beginning. So even though your string is not terminated properly, it just so happens that the memory right after it is zero, so printf knows when to stop.
+-----------------------+
|G|i|v|i|\0|\0|... |
+-----------------------+
| your | rest of |
| stuff | memory (stack)|
+-----------------------+
Other languages, such as Java, have safeguards against this sort of situations. Languages like C, however, do less hand holding, which, on the one hand, allows more flexibility, but on the other, give you much, much more ways to shoot you in the foot with subtle issues such as this one. In other words, if your code compiles, that doesn't mean it's correct and it won't blow up now, in 5 minutes or in 5 years.
In real life, this is almost never the case, and your string might end up getting stored next to other things, which would always end up getting printed out together with your string. You never want this. Situations like this might lead to crashes, exploits and leaked confidential information.
See the following diagram for an example. Imagine you're working on a web server and the string "secret"--a user's password or key is stored right next to your harmless string:
+-----------------------+
|G|i|v|i|s|e|c|r|e|t |
+-----------------------+
| your | rest of |
| stuff | memory (stack)|
+-----------------------+
Every time you would output what you would think is "Givi", you'd end up printing out the secret string, which is not what you want.
The byte after the last character always has to be 0, otherwise printf would not know when the string is terminanted and would try to access bytes (or chars) while they are not 0.
As Andrei said, apparently it just happened, that the compiler put at least one byte with the value 0 after your string data, so printf recognized the end of the string.
This can vary from compiler to compiler and thus is undefined behaviour.
There could, for instance, be a chance to have printf accessing an address, which your program is not allowed to. This would result in a crash.
In C text strings are stored as zero terminated arrays of characters. This means that the end of a text string is indicated by a special character, a numeric value of zero (0), to indicate the end of the string.
So the array of text characters to be used to store a C text string must include an array element for each of the characters as well as an additional array element for the end of string.
All of the C text string functions (strcpy(), strcmp(), strcat(), etc.) all expect that the end of a text string is indicated by a value of zero. This includes the printf() family of functions that print or output text to the screen or to a file. Since these functions depend on seeing a zero value to terminate the string, one source of errors when using C text strings is copying too many characters due to a missing zero terminator or copying a long text string into a smaller buffer. This type of error is known as a buffer overflow error.
The C compiler will perform some types of adjustments for you automatically. For instance:
char *pText = "four"; // pointer to a text string constant, compiler automatically adds zero to an additional array element for the constant "four"
char text[] = "four"; // compiler creates a array with 5 elements and puts the characters four in the first four array elements, a value of 0 in the fifth
char text[5] = "four"; // programmer creates array of 5 elements, compiler puts the characters four in the first four array elements, a value of 0 in the fifth
In the example you provided a good C compiler should issue at the minimum a warning and probably an error. However it looks like your compiler is truncating the string to the array size and is not adding the additional zero string terminator. And you are getting lucky in that there is a zero value after the end of the string. I suppose there is also the possibility that the C compiler is adding an additional array element anyway but that would seem unlikely.
What your book states is basically right, but there is missing the phrase "at least". The array can very well be larger.
You already stated the reason for the min length requirement. So what does that tell you about the example? It is crap!
What it exhibits is called undefined behaviour (UB) and might result in daemons flying out your nose for the printf() - not the initializer. It is just not covered by the C standard (well ,the standard actually says this is UB), so the compiler (and your libraries) are not expected to behave correctly.
For such cases, no terminator will be appended explicitly, so the string is not properly terminated when passed to `printf()".
Reason this does not produce an error is likely some legacy code which did exploit this to safe some bytes of memory. So, instead of reporting an error that the implicit trailing '\0' terminator does not fit, it simply does not append it. Silently truncating the string literal would also be a bad idea.
The following line:
char name[4] = "Givi";
May give warning like:
string for array of chars is too long
Because the behavior is Undefined, still compiler may pass it. But if you debug, you will see:
name[0] 'G'
name[1] 'i'
name[2] 'V'
name[3] '\0'
And so the output is
Giv
Not Give as you mentioned in the question!
I'm using GCC compiler.
But if you write something like this:
char name[4] = "Giv";
Compiles fine! And output is
Giv

C language, char=char with unexpected results

Hi everybody and thanks in advance for any help, this is the situation:
#define N 12
[..]
char vect[N][2];
char strng[2];
[..]
vect[i][0]=strng[2]; //this two lines are in a simple for cycle
vect[i][2]=strng[0];
Now, if in string[2] I have "c 2", what I expect in vect[i][0] is '2' and in vect[i][1] 'c'.
I use code::blocks and watching vect I have instead "2#", but it could be "2À" as well.
Can you help me? Where am I wrong?
Array indexes goes from zero up to the size minus one. So using e.g. strng[2] you access the third entry in the two-entry array. Accessing an array out of bounds leads to undefined behavior and the data will be indeterminate.
You should also remember that all strings in C are one more character than reported by e.g. strlen, and that extra character is a special terminator character. So if you want a two-character string, you really need three characters: Two for the string, and one for the terminator.
Rewrite these statements
vect[i][0]=strng[2]; //this two lines are in a simple for cycle
vect[i][2]=strng[0];
the following way
vect[i][0]=strng[1]; //this two lines are in a simple for cycle
vect[i][1]=strng[0];
provided that string contains two characters { 'c', '2' }.
Take into account that array string can not have string literal "c 2", because you defined it as
char strng[2];
that is it can contain only two characters.
If you want that the array would contain indeed "c 2" then you have to define it either as
char strng[3];
or as
char strng[4];
if you want to include the terminating zero.
In this case you may write
vect[i][0]=strng[2]; //this two lines are in a simple for cycle
vect[i][1]=strng[0];
Assuming strng literally contains "c 2", then your memory is the issue. strng[2] contains 3 cells iirc. 2 for holding chars and then a null terminator (ie \0). so when you try to access strng[2], (which you cant because you can only go to N-1 cells, where N is the number allocated for it) it contains undefined results, since it isnt null terminated and you are reaching beyond memory you allocated

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