help with strings in C - c

I'm studying for an exam and I need some help with strings
Assume the following declarations, and further assume that string.h is uncluded
char rocky[21], bw[21], boris[21];
int result;
a.) Write a scanf statement that would enable the string Beauregard to be read into rocky
my answer= scanf("%s", &rocky);
b.) Assuming that the text Beauregard is the only thing on th eline of standard input, write a statement to read in the text and store it in rocky using an alternative to scanf
my answer= gets (Beauregard);
strcpy(rocky);
c.) assuming that the text read in is Beauregard, what is the value of result after the following statement is executed?
result=strlen(rocky);
my answer= i have no clue..
d.) what does the following statment do?
strcpy(boris, rocky);
answer= makes a copy of the string..(dont know much more than that)
e.) what does the following statement do? What are the values of rocky and bw?
strncpy(bw,rocky,3);
my answer= not a clue
help is much appreciated, and an explanation would also help :)
Thanks!

a. Arrays and pointers are closely related in C. In particular, the name of an array decays to a pointer to the first element of the array, so your answer should be
scanf("%s", rocky); /* note the lack of an & in front of rocky */
b. gets(Beauregard) doesn't really make sense. The gets function reads a string from standard input (think, "keyboard") and stores it in the character array pointed to by the argument you pass it. So you're supposed to assume the user will type "Beauregard", and you should read it into the rocky array with
gets(rocky);
c. strlen returns the length of the string, not including the trailing \0 character, so in this case, 10.
d. strcpy just copies the contents of the rocky array into the boris array, so they'd both contain "Beauregard".
e. strncpy works like strcpy, but only copies up to n characters (where n is the last argument, so in this case, bw would contain "Bea" without a terminating null character.
Note that several of these statements are really bad ideas in any real program. There is never a reason to use gets, for example, as any use of gets opens up security flaws. You should always use fgets instead. The scanf function can be used safely if you specify the width, but you haven't done so here. I mention these things just in case your teacher has covered them and you've forgotten.

strlen returns the length of the C string. strlen("Beauregard"); would return 10 because the string is 10 characters long.
strcpy just copies a string, you're right.
strncpy allows you to specify the maximum number of characters you want. So if you pass it 3, you'll get 3 characters and the null terminator on the end of your string.

strlen(rocky) is going to return the string length of what is pointed to by rocky. The number of letters that make up 'Beauregard'.
strncpy(bw, rocky, 3) copies the first 3 letters from the string pointed to by rocky into bw.
You should read man pages for strlen, strcpy and strncpy.

c.) strlen counts until it reaches the terminating null character, \0. So the answer would be 10, because there are 10 letters in Beauregard.
Strlen
d.) Yes, it copies a string. More specifically, it copies rocky into boris. I'm not sure what else they would want you to give as part of an answer there...
e) It copies the first 3 characters of rocky into bw. However, it does NOT add a terminating null character. strncpy

Related

Properties of strcpy()

I have a global definition as following:
#define globalstring "example1"
typedef struct
{
char key[100];
char trail[10][100];
bson_value_t value;
} ObjectInfo;
typedef struct
{
ObjectInfo CurrentOrderInfoSet[5];
} DataPackage;
DataPackage GlobalDataPackage[10];
And I would like to use the strcpy() function in some of my functions as following:
strcpy(GlobalDataPackage[2].CurrentOrderInfoSet[0].key, "example2");
char string[100] = "example3";
strcpy(GlobalDataPackage[2].CurrentOrderInfoSet[0].key, string);
strcpy(GlobalDataPackage[2].CurrentOrderInfoSet[0].key, globalstring);
First question: Are the global defined strings all initiated with 100 times '\0'?
Second qestion: I am a bit confused as to how exactly strcpy() works. Does it only overwrite the characters necessary to place the source string into the destination string plus a \0 at the end and leave the rest as it is or does it fully delete any content of the destination string prior to that?
Third question: All my strings are fixed length of 100. If I use the 3 examples of strcpy() above, with my strings not exceeding 99 characters, does strcpy() properly overwrite the destination string and NULL terminate it? Meaning do I run into problems when using functions like strlen(), printf() later?
Fourth question: What happens when I strcpy() empty strings?
I plan to overwrite these strings in loops various times and would like to know if it would be safer to use memset() to fully "empty" the strings prior to strcpy() on every iteration.
Thx.
Are the global defined strings all initiated with 100 times '\0'?
Yes. Global char arrays will be initilizated to all zeros.
I am a bit confused as to how exactly strcpy() works. Does it only overwrite the characters necessary to place the source string into the destination string plus a \0 at the end and leave the rest as it
Exactly. It copies the characters up until and including '\0' and does not care about the rest.
If I use ... my strings not exceeding 99 characters, does strcpy() properly overwrite the destination string and NULL terminate it?
Yes, but NULL is a pointer, it's terminated with zero byte, sometimes called NUL. You might want to see What is the difference between NUL and NULL? .
Meaning do I run into problems when using functions like strlen(), printf() later?
Not if your string lengths are less than or equal to 99.
What happens when I strcpy() empty strings?
It just copies one zero byte.
would like to know if it would be safer to use memset() to fully "empty" the strings prior to strcpy() on every iteration.
Safety is a broad concept. As far as safety as in if the program will execute properly, there is no point in caring about anything after zero byte, so just strcpy it.
But you should check if your strings are less than 99 characters and handle what to do it they are longer. You might be interested in strnlen, but the interface is confusing - I recommend to use memcpy + explicitly manually set zero byte.

A little query, String in C

Recently I was programming in my Code Blocks and I did a little program only for hobby in C.
char littleString[1];
fflush( stdin );
scanf( "%s", littleString );
printf( "\n%s", littleString);
If I created a string of one character, why does the CodeBlocks allow me to save 13 characters?
C have no bounds-checking, writing out of bounds of arrays or dynamically allocated memory can't be checked by the compiler. Instead it will lead to undefined behavior.
To prevent buffer overflow with scanf you can tell it to only read a specific number of characters, and nothing more. So to tell it to read only one character you use the format "%1s".
As a small side-note: Remember that strings in C have an extra character in them, the terminator (character '\0'). So if you have a string that should contain one character, the size actually needs to be two characters.
LittleString is not a string. It is a char array of length one. In order for a char array to be a string, it must be null terminated with an \0. You are writing past the memory you have allotted for littleString. This is undefined behavior.Scanf just reads user input from the console and assigns it to the variable specified, in this case littleString. If you would like to control the length of user input which is assigned to the variable, I would suggest using scanf_s. Please note that scanf_s is not a C99 standard
Many functions in C is implemented without any checks for correctness of use. In other words, it is the callers responsibility that the arguments fulfill some rules set by the function.
Example: For strcpy the Linux man page says
The strcpy() function copies the string pointed to by src,
including the terminating null byte ('\0'), to the buffer
pointed to by dest. The strings may not overlap, and the
destination string dest must be large enough to receive the copy.
If you as a caller break that contract by passing a too small buffer, you'll have undefined behavior and anything can happen.
The program may crash or even do exactly what you expected in 99 out of 100 times and do something strange in 1 out of 100 times.

Why output length is coming 6?

I have written a simple program to calculate length of string in this way.
I know that there are other ways too. But I just want to know why this program is giving this output.
#include <stdio.h>
int main()
{
char str[1];
printf( "%d", printf("%s", gets(str)));
return 0;
}
OUTPUT :
(null)6
Unless you always pass empty strings from the standard input, you are invoking undefined behavior, so the output could be pretty much anything, and it could crash as well. str cannot be a well-formed C string of more than zero characters.
char str[1] allocates storage room for one single character, but that character needs to be the NUL character to satisfy C string constraints. You need to create a character array large enough to hold the string that you're writing with gets.
"(null)6" as the output could mean that gets returned NULL because it failed for some reason or that the stack was corrupted in such a way that the return value was overwritten with zeroes (per the undefined behavior explanation). 6 following "(null)" is expected, as the return value of printf is the number of characters that were printed, and "(null)" is six characters long.
There's several issues with your program.
First off, you're defining a char buffer way too short, a 1 char buffer for a string can only hold one string, the empty one. This is because you need a null at the end of the string to terminate it.
Next, you're using the gets function which is very unsafe, (as your compiler almost certainly warned you about), as it just blindly takes input and copies it into a buffer. As your buffer is 0+terminator characters long, you're going to be automatically overwriting the end of your string into other areas of memory which could and probably does contain important information, such as your rsp (your return pointer). This is the classic method of smashing the stack.
Third, you're passing the output of a printf function to another printf. printf isn't designed for formating strings and returning strings, there are other functions for that. Generally the one you will want to use is sprintf and pass it in a string.
Please read the documentation on this sort of thing, and if you're unsure about any specific thing read up on it before just trying to program it in. You seem confused on the basic usage of many important C functions.
It invokes undefined behavior. In this case you may get any thing. At least str should be of 2 bytes if you are not passing a empty string.
When you declare a variable some space is reserved to store the value.
The reserved space can be a space that was previously used by some other
code and has values. When the variable goes out of scope or is freed
the value is not erased (or it may be, anything goes.) only the programs access
to that variable is revoked.
When you read from an unitialised location you can get anything.
This is undefined behaviour and you are doing that,
Output on gcc (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3 is 0
For above program your input is "(null)", So you are getting "(null)6". Here "6" is the output from printf (number of characters successfully printed).

need for the last '\0' in fgets

I've seen several usage of fgets (for example, here) that go like this:
char buff[7]="";
(...)
fgets(buff, sizeof(buff), stdin);
The interest being that, if I supply a long input like "aaaaaaaaaaa", fgets will truncate it to "aaaaaa" here, because the 7th character will be used to store '\0'.
However, when doing this:
int i=0;
for (i=0;i<7;i++)
{
buff[i]='a';
}
printf("%s\n",buff);
I will always get 7 'a's, and the program will not crash. But if I try to write 8 'a's, it will.
As I saw it later, the reason for this is that, at least on my system, when I allocate char buff[7] (with or without =""), the 8th byte (counting from 1, not from 0) gets set to 0. From what I guess, things are done like this precisely so that a for loop with 7 writes, followed by a string formatted read, could succeed, whether the last character to be written was '\0' or not, and thus avoiding the need for the programmer to set the last '\0' himself, when writing chars individually.
From this, it follows that in the case of
fgets(buff, sizeof(buff), stdin);
and then providing a too long input, the resulting buffstring will automatically have two '\0' characters, one inside the array, and one right after it that was written by the system.
I have also observed that doing
fgets(buff,(sizeof(buff)+17),stdin);
will still work, and output a very long string, without crashing. From what I guessed, this is because fgets will keep writing until sizeof(buff)+17, and the last char to be written will precisely be a '\0', ensuring that any forthcoming string reading process would terminate properly (although the memory is messed up anyway).
But then, what about fgets(buff, (sizeof(buff)+1),stdin);? this would use up all the space that was rightfully allocated in buff, and then write a '\0' right after it, thus overwriting...the '\0' previously written by the system. In other words, yes, fgets would go out of bounds, but it can be proven that when adding only one to the length of the write, the program will never crash.
So in the end, here comes the question: why does fgets always terminates its write with a '\0', when another '\0', placed by the system right after the array, already exists? why not do like in the one by one for-loop based write, that can access the whole of the array and write anything the programmer wants, without endangering anything?
Thank you very much for your answer!
EDIT: indeed, there is no proof possible, as long as I do not know whether this 8th '\0' that mysteriously appears upon allocation of buff[7], is part of the C standard or not, specifically for string arrays. If not, then...it's just luck that it works :-)
but it can be proven that when adding only one to the length of the write, the program will never crash.
No! You can't prove that! Not in the sense of a mathematical proof. You have only shown that on your system, with your compiler, with those particular compiler settings you used, with particular environment configuration, it might not crash. This is far from a mathematical proof!
In fact the C standard itself, although it guarantees that you can get the address of "one place after the last element of an array", it also states that dereferencing that address (i.e. trying to read or write from that address) is undefined behaviour.
That means that an implementation can do everything in this case. It can even do what you expect with naive reasoning (i.e. work - but it's sheer luck), but it may also crash or it may also format your HD (if your are very, very unlucky). This is especially true when writing system software (e.g. a device driver or a program running on the bare metal), i.e. when there is no OS to shield you from the nastiest consequences of writing bad code!
Edit This should answer the question made in a comment (C99 draft standard):
7.19.7.2 The fgets function
Synopsis
#include <stdio.h>
char *fgets(char * restrict s, int n,
FILE * restrict stream);
Description
The fgets function reads at most one less than the number of characters specified by n
from the stream pointed to by stream into the array pointed to by s. No additional
characters are read after a new-line character (which is retained) or after end-of-file. A
null character is written immediately after the last character read into the array.
Returns
The fgets function returns s if successful. If end-of-file is encountered and no
characters have been read into the array, the contents of the array remain unchanged and a
null pointer is returned. If a read error occurs during the operation, the array contents are
indeterminate and a null pointer is returned.
Edit: Since it seems that the problem lies in a misunderstanding of what a string is, this is the relevant excerpt from the standard (emphasis mine):
7.1.1 Definitions of terms
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.
From C11 standard draft:
The fgets function reads at most one less than the number of characters specified by n
from the stream pointed to by stream into the array pointed to by s. No additional
characters are read after a new-line character (which is retained) or after end-of-file. A
null character is written immediately after the last character read into the array.
The fgets function returns s if successful. If end-of-file is encountered and no
characters have been read into the array, the contents of the array remain unchanged and a
null pointer is returned. If a read error occurs during the operation, the array contents are indeterminate and a null pointer is returned.
The behaviour you describe is undefined.

Null termination of char array

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

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